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When the unknown is squared
A STORY
Forty-eight squares
Meera fits a grey paving square into place. Bhavisha lifts the next one from the stack.
“We have $48$ squares,” Bhavisha says. “Enough for a rectangle, if the long side has $2$ more
squares than the short side.”
“How many along the short side, then?” Meera asks.
“Call it $x$,” Bhavisha says. “The long side is $x + 2$, so $x(x + 2) = 48$.”
Meera tries numbers under her breath. “$6$ works. $6$ times $8$ is $48$.”
“So does $-8$,” Bhavisha says. “$-8$ times $-6$ is $48$ too.”
Meera laughs. “Nobody lays $-8$ squares.”
“No,” Bhavisha says. “But the equation has two answers all the same. When the unknown is
squared, there can be two. There can never be more than two.”
“And we keep the one that fits the ground,” Meera says.
You will learn to find both answers every time, and to tell before you start how many there will
be.
Solve $x + 5 = 12$ and you get one answer, in one step. Now square the unknown instead. Try solving $x^2 = 9$ yourself, before you read on. (Answer: $x = 3$ and $x = -3$.) No single algebra move finds both at once.
Let us write that down properly. An equation of the form $a x^2 + b x + c = 0$, with $a \neq 0$, is called a QUADRATIC EQUATION. It never has more than two real roots.
This chapter shows you two ways to solve one. One way factorises it into two linear pieces. The other applies a formula, and that one works even when factorising is hard to spot. *One number, computed straight from $a$, $b$ and $c$, tells you which case you are in before you solve anything: it is called the discriminant, $b^2 - 4 a c$.* We will compute it first, every time, and know what to expect before we start.
A SCHOLAR INDIA REMEMBERS
Brahmagupta wrote his book, the Brahmasphutasiddhanta, around 628 CE. In it he gave rules for negative numbers, as numbers in their own right. That matters here, because a root can be negative. The equation $x^2 = 9$ has two roots, $3$ and $-3$. Check: $(-3)^2 = 9$. The negative root is as good an answer as the positive one.
Brahmagupta · the brahminy kite
The line meets the axis once and the parabola twice, so the power on the unknown, not the arithmetic, sets the number of answers.
Check yourself
Which condition must always hold for $a x^2 + b x + c = 0$ to be a quadratic equation in $x$?
$a \neq 0$
$b \neq 0$
$c \neq 0$
$a$, $b$, $c$ all positive
Check your answer
✓ $a \neq 0$ — (A) The coefficient of $x^2$ must be non-zero — that is exactly what keeps the squared term in the equation.
$b \neq 0$ — The linear term can be $0$, as in $x^2 - 4 = 0$, and the equation is still quadratic.
$c \neq 0$ — The constant term can be $0$, as in $x^2 - 4x = 0$, and the equation is still quadratic.
$a$, $b$, $c$ all positive — Nothing requires $b$ or $c$ to be positive, or even present — only $a \neq 0$ is required.
The discriminant $b^2 - 4 a c$ can be found before the equation is solved. What does this let a student do?
find the exact value of both roots directly
skip checking whether $a \neq 0$
decide the root count first
avoid writing the equation in standard form
Check your answer
find the exact value of both roots directly — The discriminant signals how many real roots exist and their nature, not their numeric value on its own.
skip checking whether $a \neq 0$ — Checking $a \neq 0$ happens first, to confirm the equation is quadratic at all; the discriminant is a separate, later step.
✓ decide the root count first — (C) The discriminant is computed from $a$, $b$, $c$ alone, so the number of real roots is known before either solving method is even started.
avoid writing the equation in standard form — Standard form is needed first, to read off $a$, $b$, $c$ — the discriminant cannot be found without it.
A quadratic equation is found to have exactly one real root, repeated. How many DISTINCT real roots does it have?
one
two
zero
cannot be decided without factorising
Check your answer
✓ one — (A) A repeated root is one distinct value, counted twice in the equation’s structure.
two — Repeated means the same value is counted twice, not that two different values exist.
zero — A repeated root is still a genuine real root — it is not the same as having none.
cannot be decided without factorising — The discriminant alone already fixes the root count and type before any factorising is attempted.
You have solved equations with $x$ before. Now $x$ gets squared. Try each check below. It takes a minute.
Square root of a number (Class 8, Ganita Prakash Part 1, page 8). Here: a root is a number that fits an equation, just as $7$ fits $x^2 = 49$. Check: $7^2 = 49$, so $7$ is a square root of $49$.
A situation as an equation (Class 9, Ganita Manjari, page 17). Here: a situation becomes an equation once the unknown is named $x$. Check: if $l = 3$, the wire fencing costs $200 \cdot 3 = 600$.
Splitting the middle term (Class 9, Ganita Manjari, page 81). Here: factorising starts from exactly this pair, $a + b = 7$, $a \cdot b = 12$. Check: $(x + 3)(x + 4) = x^2 + 7x + 12$, since $3 + 4 = 7$ and $3 \cdot 4 = 12$.
Expanding a bracket (Class 8, Ganita Prakash Part 1, page 140). Here: expanding turns $x(x + 2)$ into $x^2 + 2x$ before its degree is checked. Check: At $x = 3$, $x(x + 2) = 3 \cdot 5 = 15$ and $x^2 + 2x = 9 + 6 = 15$.
Naming an equation by degree (Class 9, Ganita Manjari, page 21). Here: an equation with highest power $2$ is named quadratic, like this. Check: $10x - x^2$ has degree $2$, so it is quadratic.
Area of a rectangle (Class 9, Ganita Manjari, page 139). Here: a garden’s area is length times breadth, just like this rectangle’s. Check: a rectangle $8$ by $12$ has area $96$.
If any of these felt new, read the page named before going on.
Let us take any quadratic equation. Write its terms in descending powers of $x$. Put them all on one side, equal to $0$. Do that, and you get $a x^2 + b x + c = 0$. This is the equation’s STANDARD FORM. $a$, $b$, $c$ are the three real numbers that name the whole equation.
*The equation stays quadratic only when $a \neq 0$.* That single condition keeps the $x^2$ term alive. Drop it, and $a x^2 + b x + c = 0$ collapses to $b x + c = 0$: a linear equation, with a different number of roots. Check this yourself. Set $a = 0$ in $2 x^2 + 3 x - 5 = 0$. (Answer: it leaves $3 x - 5 = 0$, a linear equation.)
Reading an equation against this pattern is the first move every method makes. Before you factorise, or reach for a formula, write the equation in standard form. Then read $a$, $b$ and $c$ straight off it.
Check yourself
Which of these is the standard form of a quadratic equation in $x$?
$a x^2 + b x + c = 0$, with $b \neq 0$
$a x + b x^2 + c = 0$, with $a \neq 0$
$a x^2 + b x + c = 0$, with $a \neq 0$
$a x^2 + b x + c$
Check your answer
$a x^2 + b x + c = 0$, with $b \neq 0$ — It is $a$, the coefficient of $x^2$, that must be non-zero — $b$ may equal $0$.
$a x + b x^2 + c = 0$, with $a \neq 0$ — Standard form lists the terms in descending powers of $x$; here $x^2$ is written after $x$.
✓ $a x^2 + b x + c = 0$, with $a \neq 0$ — (C) This is exactly the standard form: descending powers of $x$, one side equal to $0$, and $a \neq 0$.
$a x^2 + b x + c$ — Without $= 0$ this is only an expression — standard form is an equation, set to $0$.
Why does the standard form of a quadratic equation require $a \neq 0$?
if $a = 0$ the equation has no solution at all
if $a = 0$ the discriminant cannot be computed
the $x^2$ term disappears
if $a = 0$ the coefficient $c$ must also be $0$
Check your answer
if $a = 0$ the equation has no solution at all — A linear equation is not unsolvable — it simply has one root instead of up to two.
if $a = 0$ the discriminant cannot be computed — $b^2 - 4 a c$ still computes to a number when $a=0$ — what changes is that the equation is no longer quadratic.
✓ the $x^2$ term disappears — (C) Setting $a = 0$ removes the squared term entirely, so the degree of the equation drops to $1$.
if $a = 0$ the coefficient $c$ must also be $0$ — $a$ and $c$ are independent coefficients; $a=0$ says nothing about the value of $c$.
A student says $3x^2 = 5x - 2$ is already in standard form. Is the student correct?
Yes — any equation with an $x^2$ term is already in standard form
No — because the coefficient of $x^2$ is not $1$
No — it must first be rearranged to $3x^2 - 5x + 2 = 0$
Yes — because $x^2$, $x$, and a constant all appear
Check your answer
Yes — any equation with an $x^2$ term is already in standard form — Standard form specifically needs the terms collected with one side equal to $0$ — an $x^2$ term alone is not enough.
No — because the coefficient of $x^2$ is not $1$ — Standard form only requires $a \neq 0$; the coefficient of $x^2$ need not be $1$.
✓ No — it must first be rearranged to $3x^2 - 5x + 2 = 0$ — (C) Standard form needs one side equal to $0$; the equation must be rearranged before it matches that shape.
Yes — because $x^2$, $x$, and a constant all appear — The terms being present is not sufficient — they must be collected on one side, set equal to $0$.
A real number $k$ is a ROOT, or a SOLUTION, of $a x^2 + b x + c = 0$, if putting $x = k$ into it makes the equation true. That means $a k^2 + b k + c = 0$. Checking a claimed root costs you nothing more than one substitution.
Let us try one. Take $x^2 - 5 x + 6 = 0$. Try $x = 2$: $4 - 10 + 6 = 0$. It works, so $2$ is a root. Now try $x = 1$: $1 - 5 + 6 = 2 \neq 0$, so $1$ is not a root. Try $x = -1$ yourself. (Answer: $1 + 5 + 6 = 12 \neq 0$, so $-1$ is not a root either.)
*A root of the equation is the same number as a zero of the quadratic polynomial $a x^2 + b x + c$.* That idea is from an earlier chapter, carried forward under a new name. A quadratic equation has at most two real roots, never more.
A SCHOLAR INDIA REMEMBERS
Is $-3$ a root of $x^2 + x - 6 = 0$? Do not guess. Put it in and look. $(-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0$. The left side is $0$, so yes, $-3$ is a root. A negative root counts as fully as a positive one. The other root is $2$. Check it the same way.
Brahmagupta · the brahminy kite
Check yourself
A real number $k$ is called a root of $a x^2 + b x + c = 0$ when
$a k^2 + b k + c = 0$
$a k^2 + b k + c > 0$
$k$ is a factor of $c$
$k^2 = a x^2 + b x + c$
Check your answer
✓ $a k^2 + b k + c = 0$ — (A) This is exactly the definition: substituting $k$ for $x$ makes the equation true.
$a k^2 + b k + c > 0$ — A root must satisfy the equation exactly, giving $0$ — not merely a positive value.
$k$ is a factor of $c$ — Being a factor of $c$ is sometimes useful for guessing, but it is not what defines a root.
$k^2 = a x^2 + b x + c$ — A root is found by substituting $k$ in place of $x$ throughout, not by equating $k^2$ to the whole expression.
How is a root of the quadratic equation $a x^2 + b x + c = 0$ related to the quadratic polynomial $a x^2 + b x + c$?
a root of the equation is the same number as a zero of the polynomial
a root of the equation is always larger than every zero of the polynomial
a root exists only if the polynomial has no zero
a root is the value of the polynomial at $x = 0$
Check your answer
✓ a root of the equation is the same number as a zero of the polynomial — (A) Setting the polynomial to $0$ is exactly the equation, so its roots and the polynomial’s zeros are the same numbers.
a root of the equation is always larger than every zero of the polynomial — A root of the equation and a zero of the polynomial are the same number, not two different numbers to compare.
a root exists only if the polynomial has no zero — A root exists exactly when the polynomial DOES have a zero — the two go together, not apart.
a root is the value of the polynomial at $x = 0$ — A root is a value of $x$ that makes the polynomial $0$, not the value of the polynomial when $x$ itself is $0$.
Which value of $k$ is a root of $x^2 - 5x + 6 = 0$?
$k = 1$
$k = 6$
$k = 2$
$k = -2$
Check your answer
$k = 1$ — Substituting $k=1$ gives $1-5+6=2$, not $0$, so $k=1$ is not a root.
$k = 6$ — Substituting $k=6$ gives $36-30+6=12$, not $0$ — being a factor of $c$ does not make a number a root.
An equation may not arrive looking like $a x^2 + b x + c = 0$. You may need to expand brackets first. Collect every term onto one side. Only then does the equation show what it really is.
Expand every bracket, every time. Read the coefficient of $x^2$ only after simplifying, never before.
Let us watch for the trap. An $x^2$ term can look present at the start, then cancel once you collect both sides. What is left is a linear equation, not a quadratic one. Check the two equations in the worked example that follows. Watch this happen to one of them yourself. The $x^2$ you see before simplifying promises nothing about the one left after.
A number line marks the coefficient of x squared for two equations: one stays quadratic, the other collapses to linear.
Check yourself
To check whether an equation is quadratic once every bracket is expanded, what must be examined?
whether the $x^2$ coefficient survives simplifying
whether an $x^2$ term appears anywhere before simplification
whether the equation has exactly three terms
whether the highest power written is $2$
Check your answer
✓ whether the $x^2$ coefficient survives simplifying — (A) The degree of the equation is decided only after every bracket is expanded and like terms are collected.
whether an $x^2$ term appears anywhere before simplification — An $x^2$ term present before simplifying can cancel completely once both sides are expanded.
whether the equation has exactly three terms — A quadratic equation can have two, three, or more terms — the count of terms does not decide the degree.
whether the highest power written is $2$ — The power as first written can drop after simplifying, exactly as it does when the $x^2$ terms cancel.
A student expands $(x-1)^2 = x^2 - 1$ and says: ‘There is an $x^2$ on both sides, so this must be quadratic.’ Is this correct?
Yes — any equation with $x^2$ written down is quadratic
No — because the equation has no constant term
Yes — because both sides have degree $2$ before simplifying
No — the $x^2$ terms cancel, leaving a linear equation
Check your answer
Yes — any equation with $x^2$ written down is quadratic — Writing $x^2$ down is not enough; here it cancels out once both sides are simplified.
No — because the equation has no constant term — The simplified equation is $-2x + 2 = 0$, which does have a constant term — that is not why it is linear.
Yes — because both sides have degree $2$ before simplifying — Degree before simplifying is not decisive — the $x^2$ terms cancel once both sides are expanded.
✓ No — the $x^2$ terms cancel, leaving a linear equation — (D) Expanding gives $x^2 - 2x + 1 = x^2 - 1$; the $x^2$ terms cancel and only $-2x + 2 = 0$ remains.
Which of these, once fully simplified, is NOT a quadratic equation?
$(x+3)(x-1) = 2x + 100$
$2x^2 - 3 = x^2 + 5$
$(x+2)(x-2) = x^2 - 3x$
$3x^2 = 2x^2 + x + 6$
Check your answer
$(x+3)(x-1) = 2x + 100$ — Expanding gives $x^2 - 103 = 0$; the $x^2$ term survives, so this equation is quadratic.
$2x^2 - 3 = x^2 + 5$ — Simplifying gives $x^2 - 8 = 0$; the $x^2$ term survives, so this equation is quadratic.
✓ $(x+2)(x-2) = x^2 - 3x$ — (C) Expanding gives $x^2 - 4 = x^2 - 3x$; the $x^2$ terms cancel and only $3x = 4$ remains, a linear equation.
$3x^2 = 2x^2 + x + 6$ — Simplifying gives $x^2 - x - 6 = 0$; the $x^2$ term survives, so this equation is quadratic.
In the worked check, $(x-1)^2 = x^2 - 1$ turns out to be linear once expanded. What made the $x^2$ term disappear?
$x^2$ can never appear in an equation containing a squared bracket
the $x^2$ terms match and cancel on simplifying
the constant term on the right was negative
brackets always cancel with brackets when squared
Check your answer
$x^2$ can never appear in an equation containing a squared bracket — $(x-1)^2 = 0$ is a squared bracket that stays quadratic — the cancellation here was specific to this pair of sides.
✓ the $x^2$ terms match and cancel on simplifying — (B) Both sides expand to include $x^2$ with coefficient $1$, so subtracting one side from the other removes it entirely.
the constant term on the right was negative — The $x^2$ terms cancel because they match on both sides — the sign of the constant is unrelated.
brackets always cancel with brackets when squared — There is no such general rule — the cancellation happened because the $x^2$ coefficients on both sides were equal.
A day-to-day situation can be an area, a product of two ages, or a train’s speed against its journey time. Every one of them turns into a quadratic equation the same way. Let us name the unknown first: call it $x$, in words, before writing any equation. Then write every other quantity in terms of that one $x$.
The situation gives you one condition. An area must work out to a fixed number, or a product must equal one. Turn that condition into a single equation in $x$. Simplify it into the standard form $a x^2 + b x + c = 0$. That is the quadratic equation the situation satisfies.
A root drawn from a real situation may still need rejecting, once you solve it. A negative length, or a negative count of objects, is not a value the situation allows, even when it satisfies the equation perfectly. In Science, a projectile’s height at time $t$ often has this same shape, $h(t) = u t - c t^2$. Asking when it lands is the same problem, in a different dress. Try writing that landing condition as an equation yourself.
Two number lines: one shows a kept root beside a rejected negative root, the other shows no real root at all.
Check yourself
To turn a day-to-day situation into a quadratic equation, what is the first step?
guess a value of $x$ that seems to fit the situation
assume the situation is already quadratic before checking it
name the unknown quantity as $x$
write the discriminant of the situation directly
Check your answer
guess a value of $x$ that seems to fit the situation — Guessing a value skips forming the equation entirely — the method needs the equation written first.
assume the situation is already quadratic before checking it — Whether the situation is quadratic is only known after the condition is translated into an equation, not assumed beforehand.
✓ name the unknown quantity as $x$ — (C) Every other quantity in the situation is expressed using $x$ before the one stated condition is written as an equation.
write the discriminant of the situation directly — The discriminant comes from $a$, $b$, $c$ of an equation that does not exist yet at this step.
A rectangular plot has area $75$ square metres. Its length is $2$ metres more than its breadth. Taking the breadth as $x$ metres, which equation must $x$ satisfy?
$x^2 - 2x - 75 = 0$
$x(x+2) = 75x$
$2x^2 = 75$
$x^2 + 2x - 75 = 0$
Check your answer
$x^2 - 2x - 75 = 0$ — The length is $(x+2)$, not $(x-2)$; expanding $x(x+2)$ keeps the $+2x$ term positive.
$x(x+2) = 75x$ — The area condition is simply $x(x+2) = 75$ — there is no reason to multiply the right side by $x$ again.
$2x^2 = 75$ — The area is breadth times length, $x(x+2)$, not $2x^2$ — the actual product must be expanded, not replaced.
✓ $x^2 + 2x - 75 = 0$ — (D) The length is $(x+2)$ metres, so $x(x+2) = 75$ gives $x^2 + 2x - 75 = 0$.
The product of a man’s age (in years) $5$ years ago and his age $8$ years from now is $30$. Taking his present age as $x$, this gives $(x-5)(x+8) = 30$, that is $x^2 + 3x - 70 = 0$, with roots $x = 7$ and $x = -10$. Which root gives his present age?
$x = -10$, since it is the larger magnitude root
both $7$ and $-10$, since the equation has two roots
$x = 7$, since age cannot be negative
neither, since the equation must be re-solved
Check your answer
$x = -10$, since it is the larger magnitude root — Size is not the test — a root is accepted only when it fits the situation, and a negative age does not.
both $7$ and $-10$, since the equation has two roots — Not every algebraic root fits the situation; $-10$ years is not a valid present age, even though it solves the equation.
✓ $x = 7$, since age cannot be negative — (C) Only $x=7$ is a physically valid age; $x=-10$ is an algebraic root with no meaning here.
neither, since the equation must be re-solved — The equation and its roots are correct; what is needed is choosing the physically valid one, not re-solving.
Two numbers differ by $3$, and their product is $88$. Taking the smaller number as $x$, which equation represents this situation?
$x^2 - 3x - 88 = 0$
$x^2 + 3x + 88 = 0$
$x(x+3) = 3$
$x^2 + 3x - 88 = 0$
Check your answer
$x^2 - 3x - 88 = 0$ — Since $x$ is the smaller number, the larger one is $x+3$ — writing $x-3$ would make $x$ the larger number instead.
$x^2 + 3x + 88 = 0$ — The product $88$ is subtracted, not added, when the equation is written as $x^2+3x-88=0$.
$x(x+3) = 3$ — The given product is $88$, not the difference $3$ — the two facts describe different things.
✓ $x^2 + 3x - 88 = 0$ — (D) The larger number is $(x+3)$, so $x(x+3) = 88$ gives $x^2 + 3x - 88 = 0$.
When $a x^2 + b x + c$ factorises into two linear pieces, you can solve $a x^2 + b x + c = 0$ with nothing more than arithmetic. A product of two things is $0$ exactly when one of them is $0$. So once you have the two factors, you already have the two roots.
The standard route there splits the middle term $b x$ into two pieces. Their coefficients multiply to $a c$ and add to $b$. Let us group the four terms in pairs, then pull the common factor out of each pair.
Finding the right split is the one step worth doing carefully. Once you have it, the grouping and the factoring that follow are mechanical. Try the worked example next, and watch each step fall into place.
Two crossing lines multiply into a parabola that crosses the axis twice. A line squared multiplies into a parabola that only touches it.
Check yourself
To factorise $a x^2 + b x + c$ by splitting the middle term, the two split coefficients must
multiply to $b$ and add to $a c$
multiply to $c$ and add to $a$
add to $a c$ and multiply to $b$
multiply to $a c$ and add to $b$
Check your answer
multiply to $b$ and add to $a c$ — It is the product that must equal $a c$, and the sum that must equal $b$ — not the other way round.
multiply to $c$ and add to $a$ — The product target is $a c$, the product of both coefficients — not $c$ alone paired with $a$.
add to $a c$ and multiply to $b$ — Multiplying gives $a c$, and adding gives $b$ — the operations here are reversed.
✓ multiply to $a c$ and add to $b$ — (D) This is exactly the splitting rule: the two parts of $b$ must multiply to $a c$ and add back to $b$.
To factorise $x^2 + 7x + 10$, a student splits $7x$ as $3x + 4x$, since $3 + 4 = 7$. Is this split usable?
Yes — matching the sum to $7$ is the only requirement
No — the two numbers must both be even
No — the product must be $10$, not $12$
Yes — because $10$ is close enough to $12$
Check your answer
Yes — matching the sum to $7$ is the only requirement — Matching the sum is not enough; the two numbers must also multiply to $a c = 10$.
No — the two numbers must both be even — There is no requirement that the two numbers be even — only that they multiply to $10$ and add to $7$.
✓ No — the product must be $10$, not $12$ — (C) The sum matches but the product does not; $12 \neq 10$, so this split cannot be used.
Yes — because $10$ is close enough to $12$ — The product must equal $10$ exactly — there is no ‘close enough’ in this check.
Which split of the middle term correctly factorises $x^2 - x - 12$?
$-3x + 4x$
$-6x + 2x$
$-12x + 1x$
$-4x + 3x$
Check your answer
$-3x + 4x$ — $-3$ and $4$ multiply to $-12$ correctly, but they add to $1$, not the required $-1$.
$-6x + 2x$ — $-6$ and $2$ multiply to $-12$, but their sum is $-4$, not $-1$.
$-12x + 1x$ — $-12$ and $1$ do multiply to $-12$, but their sum is $-11$, not $-1$ — the sum was never checked.
✓ $-4x + 3x$ — (D) $-4$ and $3$ multiply to $-12$ and add to $-1$, matching $a c=-12$ and $b=-1$.
Which method is the more efficient choice for solving $x^2 + 5x - 2 = 0$?
the quadratic formula
factorisation, since the equation has three terms
either method works equally quickly here
factorisation, by trying $x=1$ and $x=2$ as roots
Check your answer
✓ the quadratic formula — (A) Checking factor pairs of $-2$ ($1,-2$ and $-1,2$) shows none sum to $5$, so factorisation over integers is not available here.
factorisation, since the equation has three terms — Every quadratic equation in standard form has up to three terms — that alone never guarantees an integer factorisation exists.
either method works equally quickly here — The two methods are not equally quick when no integer split exists — checking that first is what decides the faster route.
factorisation, by trying $x=1$ and $x=2$ as roots — Substituting shows neither $x=1$ nor $x=2$ satisfies the equation, so this guess does not lead to a factorisation.
Factorisation does not always hand back two different factors. Sometimes the same linear factor appears twice. Then $a x^2 + b x + c = 0$ becomes $a (x - r)(x - r) = 0$, for one number $r$.
Setting either copy of the factor to $0$ gives the same equation, $x = r$. This root is called a REPEATED, or EQUAL, root.
Let us picture the parabola this equation describes. Sketch it yourself on paper if you like. At a repeated root, the curve only touches the $x$-axis at that one point. It does not cross from one side to the other, the way it does at two distinct roots.
Check yourself
When factorisation gives the same linear factor twice, such as $(x-3)(x-3) = 0$, the equation is said to have
one repeated root, counted twice
two different roots, both equal to $3$
no real root, since the factor did not vary
an undefined root, since factorisation failed
Check your answer
✓ one repeated root, counted twice — (A) The single value $x=3$ is the root, and it is counted twice because the factor $(x-3)$ appears twice.
two different roots, both equal to $3$ — A repeated root is one value counted twice, not two different values that happen to match.
no real root, since the factor did not vary — Setting a real linear factor to $0$ always produces a real root, whether or not it repeats.
an undefined root, since factorisation failed — Factorisation did not fail — it succeeded, and happened to give the same factor twice.
$(x-5)^2 = 0$ gives the repeated root $x=5$. Why is this root counted twice even though only one value of $x$ satisfies it?
because $x=5$ must be substituted into the equation twice to confirm it
because the same factor $(x-5)$ appears twice in the factorisation
because the equation has two terms
because $5$ is an odd number of factors away from $0$
Check your answer
because $x=5$ must be substituted into the equation twice to confirm it — Substituting once is enough to confirm $x=5$ is a root; the count of two comes from the factorisation, not from checking twice.
✓ because the same factor $(x-5)$ appears twice in the factorisation — (B) The factorisation itself contains $(x-5)$ twice, which is exactly why the single value $x=5$ is counted twice.
because the equation has two terms — $(x-5)^2=0$ expands to three terms, and term count has no bearing on how many times a root is counted.
because $5$ is an odd number of factors away from $0$ — There is no such rule — the repeated count comes only from the factor $(x-5)$ appearing twice.
A student factorises $x^2 - 8x + 16$ as $(x-4)(x-4)$ and concludes the equation has two distinct roots, $4$ and $4$. What is wrong with this conclusion?
nothing is wrong — two roots always means two different numbers
the factorisation itself is wrong; it should be $(x-4)(x+4)$
the equation should have no real root, since both factors match
only one distinct root exists, $x=4$
Check your answer
nothing is wrong — two roots always means two different numbers — Two roots from a repeated factor are the same number counted twice, not two different numbers.
the factorisation itself is wrong; it should be $(x-4)(x+4)$ — $(x-4)(x-4)$ correctly expands to $x^2-8x+16$; $(x-4)(x+4)$ would give a different equation, $x^2-16$.
the equation should have no real root, since both factors match — Matching factors give a repeated REAL root, $x=4$ — they do not eliminate it.
✓ only one distinct root exists, $x=4$ — (D) Identical factors mean the same value is repeated, not two different roots that happen to coincide.
Factorisation is not always easy to spot. For some equations, no simple integer split exists at all. The QUADRATIC FORMULA reaches every real root of $a x^2 + b x + c = 0$ directly from $a$, $b$ and $c$, with no factorising step needed. This is the one formula worth learning by heart:
$x = (-b \pm \sqrt{b^2 - 4 a c}) / (2 a) \cdot$
One equation, two roots: one from the $plus$, one from the $minus$, in front of the square root.
The formula needs no separate proof. Substitute either root back into $a x^2 + b x + c$, and you will see it makes the expression $0$. That holds when the quantity under the square root is not negative. Let us try that check ourselves, on the very first worked example that uses this formula. Then try it yourself on every equation you solve with the formula next.
Check yourself
The quadratic formula for the roots of $a x^2 + b x + c = 0$ is
$x = (b \pm \sqrt{b^2-4 a c})/(2a)$
$x = (-b \pm \sqrt{b^2-4 a c})/(2a)$
$x = (-b \pm \sqrt{b^2-4 a c})/a$
$x = (-b \pm \sqrt{b^2+4 a c})/(2a)$
Check your answer
$x = (b \pm \sqrt{b^2-4 a c})/(2a)$ — The numerator begins with $-b$, not $b$; dropping the sign changes both roots.
✓ $x = (-b \pm \sqrt{b^2-4 a c})/(2a)$ — (B) This is the formula exactly: $-b$, plus-minus the square root of the discriminant, all over $2a$.
$x = (-b \pm \sqrt{b^2-4 a c})/a$ — The denominator is $2a$, not $a$ — dividing by $a$ alone gives the wrong roots.
$x = (-b \pm \sqrt{b^2+4 a c})/(2a)$ — The quantity inside the square root is $b^2 - 4 a c$, with a minus sign, not $b^2+4 a c$.
Why does the quadratic formula need no factorising step to find a root?
it gives any real root directly from $a$, $b$, $c$
it only works after the equation has already been factorised
it works only when the roots happen to be whole numbers
it replaces the need to write the equation in standard form
Check your answer
✓ it gives any real root directly from $a$, $b$, $c$ — (A) The formula computes a root straight from the coefficients; no factorisation is involved at any stage.
it only works after the equation has already been factorised — The formula is used exactly when factorising is not needed or not possible — it never depends on factorising first.
it works only when the roots happen to be whole numbers — The formula finds irrational roots just as well as whole-number ones — that is its main advantage over factorising.
it replaces the need to write the equation in standard form — Standard form must still be written first — the formula cannot be applied without knowing $a$, $b$, $c$.
Solving $x^2 - 4x - 5 = 0$ with $a=1$, $b=-4$, $c=-5$, a student substitutes and writes $x = (4 \pm \sqrt{16-20})/2$. What is the error?
there is no error; $16-20$ is correct
the numerator should be $-4$, not $4$
the denominator should be $2a=1$, not $2$
$16-20$ should be $16+20=36$
Check your answer
there is no error; $16-20$ is correct — $-4 a c$ works out to $+20$ here since $c=-5$ is negative, so it must be added, not subtracted.
the numerator should be $-4$, not $4$ — $-b=4$ is correct here; the actual mistake is inside the square root, not in the $-b$ term.
the denominator should be $2a=1$, not $2$ — $2a=2$ is correct since $a=1$; the actual mistake is inside the square root, not in the denominator.
✓ $16-20$ should be $16+20=36$ — (D) The error is in the discriminant: $-4 a c = -4(1)(-5) = +20$, not $-20$, so it should be added, giving $36$.
Once the quadratic formula gives a value for $x$, how can a student confirm it is genuinely a root?
substitute it back and check it gives $0$
check that the discriminant is a perfect square
check that the value is positive
check that it matches one factor of $c$
Check your answer
✓ substitute it back and check it gives $0$ — (A) This is the direct definition of a root, and it works whether the root is rational or irrational.
check that the discriminant is a perfect square — A non-perfect-square discriminant simply gives an irrational root — that root is still confirmed the same way, by substitution.
check that the value is positive — A root can be negative, positive, or zero — positivity is not part of the definition.
check that it matches one factor of $c$ — This check fails for irrational roots entirely, and even for integer roots it is not the actual definition of a root.
Look again at the quantity under the square root in the quadratic formula: $b^2 - 4 a c$. It matters enough to carry its own name, the DISCRIMINANT. You can find its value from $a$, $b$, $c$ alone, before attempting to solve the equation.
For $2 x^2 + 3 x - 5 = 0$, that means $3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$: one number, found in one line. Let us try another. Compute it yourself for $x^2 - 2 x - 2 = 0$, before you read further.
The discriminant is not a step inside solving the equation. It is a number you compute in advance. As the next section shows, that number already tells you what kind of answer to expect.
A number line marks three discriminants: one negative with no real root, one zero with a repeated root, one positive with two real roots.
A number line marks two discriminants. One lands on a perfect square and gives a rational root, the other does not, giving an irrational root.
A SCHOLAR INDIA REMEMBERS
What if the discriminant comes out negative? Do not rub it out. Read it. For $x^2 + 2x + 5 = 0$, $b^2 - 4a c = 4 - 20 = -16$. A negative discriminant is a result, and it tells you something. This equation has no real roots.
Brahmagupta · the brahminy kite
Check yourself
For $a x^2 + b x + c = 0$, the discriminant is
$b^2 - 4 a c$
$b^2 + 4 a c$
$\sqrt{b^2-4 a c}$
$4 a c - b^2$
Check your answer
✓ $b^2 - 4 a c$ — (A) This is the discriminant exactly as defined: the square minus four times the product of the outer coefficients.
$b^2 + 4 a c$ — The discriminant subtracts $4 a c$ from $b^2$; adding it gives a different, incorrect quantity.
$\sqrt{b^2-4 a c}$ — The discriminant is $b^2-4 a c$ itself; its square root is a separate quantity used inside the formula.
$4 a c - b^2$ — Reversing the subtraction flips the sign — $b^2-4 a c$ and $4 a c-b^2$ are negatives of each other.
The discriminant of $a x^2+b x+c=0$ can be computed before the equation is solved because
it needs the roots to already be known
it needs only $a$, $b$, $c$
it needs the equation to be factorised first
it needs the equation checked against a graph first
Check your answer
it needs the roots to already be known — The whole value of the discriminant is that it is known BEFORE the roots — needing the roots first would defeat its purpose.
✓ it needs only $a$, $b$, $c$ — (B) $b^2-4 a c$ is built entirely from the three coefficients, all of which are known before any solving begins.
it needs the equation to be factorised first — The discriminant needs only $a$, $b$, $c$ — factorising is not required to compute it.
it needs the equation checked against a graph first — No graph is needed — the discriminant is computed purely from the numbers $a$, $b$, $c$.
For $2x^2 - 3x + 1 = 0$, what is the discriminant?
$9$
$-8$
$17$
$1$
Check your answer
$9$ — $b^2=9$ alone is not the discriminant; $4 a c=8$ must still be subtracted, giving $1$.
$-8$ — $-4 a c=-8$ alone is not the discriminant; $b^2=9$ must still be added, giving $1$.
$17$ — $9+8=17$ comes from adding rather than subtracting $4 a c$ from $b^2$.
✓ $1$ — (D) $b^2-4 a c = (-3)^2 - 4(2)(1) = 9 - 8 = 1$.
Two equations share the same values of $b$ and $c$ but have different values of $a$. Must their discriminants be equal?
No — $a$ appears in $4 a c$ too
Yes — the discriminant depends only on $b$ and $c$
Yes — as long as $a$ is not $0$ in either equation
No — but only because $b$ and $c$ are being held equal, not because of $a$
Check your answer
✓ No — $a$ appears in $4 a c$ too — (A) The term $4 a c$ contains $a$, so changing $a$ while $b$ and $c$ stay fixed generally changes the whole discriminant.
Yes — the discriminant depends only on $b$ and $c$ — The term $4 a c$ contains $a$ directly, so the discriminant is not independent of it.
Yes — as long as $a$ is not $0$ in either equation — $a \neq 0$ only confirms both equations are quadratic — it does not make their discriminants equal.
No — but only because $b$ and $c$ are being held equal, not because of $a$ — Holding $b$ and $c$ equal is what makes the comparison possible at all — the change itself comes from $a$ appearing inside $4 a c$.
The sign of the discriminant $b^2 - 4 a c$ decides the NATURE of the roots of $a x^2 + b x + c = 0$. It does this before you find either root. There are three cases, and only three.
$b^2 - 4 a c > 0$ gives two distinct real roots. $b^2 - 4 a c = 0$ gives one repeated real root. $b^2 - 4 a c < 0$ gives no real root at all. We work with real roots only here, so a negative discriminant is where your search simply stops.
Knowing the case in advance turns solving from a search into a checked calculation. Look back at the equation you tried last section. Say out loud which of the three cases it falls into, before you check.
Three parabolas, one for each discriminant sign. The first crosses the x-axis twice, the second touches once, the third never meets it.
A SCHOLAR INDIA REMEMBERS
Now let the discriminant be $0$. Does $0$ mean nothing happened? No. It tells you something exact. For $x^2 - 6x + 9 = 0$, $b^2 - 4a c = 36 - 36 = 0$. The two roots are equal. Both are $3$, because $x^2 - 6x + 9 = (x - 3)^2$.
Brahmagupta · the brahminy kite
Check yourself
If $b^2-4 a c > 0$ for $a x^2+b x+c=0$, the equation has
two equal real roots
two distinct real roots
no real root
exactly one real root and one complex root
Check your answer
two equal real roots — Two equal roots come from $b^2-4 a c=0$; a positive discriminant gives two DIFFERENT roots.
✓ two distinct real roots — (B) A positive discriminant is exactly the case that gives two different real roots.
no real root — No real root comes from a negative discriminant — this case is positive, the opposite sign.
exactly one real root and one complex root — Real quadratic equations never have one real and one complex root; the roots are either both real or a complex-conjugate pair.
For $a x^2+b x+c=0$, $b^2-4 a c=0$. What does this say about the roots, without solving the equation?
the equation has two distinct real roots
the equation has no real root
the equation cannot be solved
the equation has one repeated real root
Check your answer
the equation has two distinct real roots — Two distinct roots come from a POSITIVE discriminant — here it is exactly $0$.
the equation has no real root — No real root comes from a NEGATIVE discriminant — here it is exactly $0$.
the equation cannot be solved — A repeated root is a genuine, solvable case — it is not a sign that solving has failed.
✓ the equation has one repeated real root — (D) A discriminant of exactly $0$ is the boundary case that gives a single repeated real root.
A quadratic equation is known to have no real root. What must be true of its discriminant?
$b^2-4 a c > 0$
$b^2-4 a c < 0$
$b^2-4 a c = 0$
$b^2-4 a c$ cannot be found for such an equation
Check your answer
$b^2-4 a c > 0$ — A positive discriminant gives two distinct real roots, not none — the direction here is reversed.
✓ $b^2-4 a c < 0$ — (B) No real root corresponds exactly to a negative discriminant.
$b^2-4 a c = 0$ — A discriminant of $0$ gives a repeated REAL root — that contradicts having no real root at all.
$b^2-4 a c$ cannot be found for such an equation — The discriminant can always be computed from $a$, $b$, $c$ — it is its VALUE, not its existence, that signals no real root.
For $x^2 - k x + 9 = 0$ to have equal roots, what must $k$ satisfy?
$k^2+36=0$, which has no real solution for $k$
$k-36=0$, so $k=36$
$k^2-9=0$, so $k=3$ or $k=-3$
$k^2 - 36 = 0$, so $k=6$ or $k=-6$
Check your answer
$k^2+36=0$, which has no real solution for $k$ — The discriminant condition is $k^2 - 4 a c = 0$, so $4 a c=36$ must be SUBTRACTED, not added.
$k-36=0$, so $k=36$ — $b^2=k^2$, not $k$ — the square on $k$ cannot be dropped when writing the discriminant condition.
$k^2-9=0$, so $k=3$ or $k=-3$ — The discriminant condition needs $4 a c = 4(1)(9) = 36$, not $c=9$ alone.
✓ $k^2 - 36 = 0$, so $k=6$ or $k=-6$ — (D) Equal roots need $b^2-4 a c=0$: $k^2 - 4(1)(9) = k^2 - 36 = 0$.
A rectangular park’s length must be $6$ metres more than its breadth, and its area must be exactly $500$ square metres. Taking the breadth as $x$, the equation is $x^2+6x-500=0$, with discriminant $b^2-4 a c = 36+2000 = 2036$. Since $2036 > 0$ and is not a perfect square, what does this say about the plan, before solving for $x$?
a real, irrational breadth exists
no such park can exist, since $2036$ is not a perfect square
the breadth must be found by factorisation, since the discriminant is positive
the plan needs no further checking, since the discriminant is already known
Check your answer
✓ a real, irrational breadth exists — (A) A positive, non-perfect-square discriminant means two real roots exist, and factorisation with integers cannot reach them.
no such park can exist, since $2036$ is not a perfect square — A positive discriminant means real roots exist; ‘not a perfect square’ only means those roots are irrational, not absent.
the breadth must be found by factorisation, since the discriminant is positive — Only a PERFECT-SQUARE discriminant guarantees an integer factorisation; a positive but non-square one does not.
the plan needs no further checking, since the discriminant is already known — The discriminant only confirms real roots exist — which root actually fits the physical situation still needs to be checked separately.
A negative constant term makes the discriminant larger
✕MISCONCEPTION
The minus sign inside $b^2 - 4 a c$ is easy to lose. Once you write $c$ down with its own sign, you must carry that sign through the multiplication, not drop it.
Take $2 x^2 + 3 x - 5 = 0$: here $a = 2$, $b = 3$, $c = -5$. Compute $-4 a c$ as $-4 \cdot 2 \cdot (-5)$, and the two minus signs give a plus: $-4 \cdot 2 \cdot (-5) = +40$.
Drop that sign, and $D = 9 - 40 = -31$ looks like no real root. Keep it, and $D = 9 + 40 = 49$, giving two real roots, $x = 1$ and $x = -5/2$.
Let us check which is right. $2 x^2 + 3 x - 5$ factorises as $(2 x + 5)(x - 1)$, so real roots plainly exist. Try the same computation yourself, slowly, and watch where the sign decides the answer.
First try
Here $a = 2$, $b = 3$ and $c = -5$. I wrote $D = 9 - 4 \cdot 2 \cdot 5 = 9 - 40 = -31$, so no real roots.
Second look
The minus sign is already in the formula, so $c$ keeps its own sign too. $D = 9 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$, giving two roots, $x = 1$ and $x = -5/2$.
Let $c$ carry its own sign into the discriminant, always.
If c is negative, does its sign already show up correctly in the discriminant?
Weaker. A student solves $3 x^2 + 2 x - 8 = 0$. They write $D = 2^2 - 4 \cdot 3 \cdot 8 = 4 - 96 = -92$. They read
off $c = -8$, but drop the sign when they multiply. They conclude there is no real root.
Check with a quick factorisation. $3 x^2 + 2 x - 8$ factorises as $(3 x - 4)(x + 2)$, and that gives
two real roots straight away, so the discriminant above must be wrong.
Stronger. Keep the sign of $c$ through the whole computation. $D = 2^2 - 4 \cdot 3 \cdot (-8) = 4 + 96 = 100$.
$D = 100$ is a perfect square, so $\sqrt{100} = 10$, and $x = (-2 \pm 10) / 6$ gives $x = 4/3$
and $x = -2$, matching the factorisation exactly.
You might be tempted to guess a factorisation for any quadratic equation. You might treat the formula as a fallback, only for when guessing takes too long.
That shortcut skips a check that would have saved the wasted search. $b^2 - 4 a c < 0$ means no real root exists, so no factorisation, guessed or otherwise, will ever be found. A discriminant that is positive but not a perfect square means the roots are irrational, so no integer or simple rational split turns up either. The formula is then your only route.
Let us check $D$ first, always. It costs one line of arithmetic. It tells you, before a single guess, whether guessing was ever going to work. Try it yourself on $x^2 + x + 3 = 0$: is a real root even worth searching for? (Answer: $D = 1 - 12 = -11 < 0$, so no real root exists.)
Should you always try guessing a factorisation first, and use the formula only when guessing fails?
Weaker. A student meets $x^2 + 2 x + 5 = 0$. They start guessing pairs of numbers that multiply to $5$ and add
to $2$. They try $1$ and $5$. They try $-1$ and $-5$. Nothing works, so they try a few more pairs, then
give up and reach for the formula.
Compute $D$ first, and the guessing was never going to succeed.
$D = 2^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16$, and $D < 0$ means no real root exists at all. No pair of
numbers, guessed or not, was ever going to factorise this equation.
Stronger. Compute $D$ before guessing anything. For $x^2 - 4 x + 1 = 0$:
$D = (-4)^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12$.
$D = 12$ is positive, so two real roots exist, but $12$ is not a perfect square. No integer or simple
rational split will ever be found. Go straight to the formula: $x = (4 plus.minus sqrt(12)) / 2 = 2
plus.minus sqrt(3)$.
Check yourself
A student says: ‘Any quadratic equation can be solved by guessing a factorisation — the quadratic formula is only a fallback for when guessing takes too long.’ Is this student correct?
Yes — guessing eventually works for every quadratic equation, given enough time
No — some equations cannot be factorised at all
No — but only because factorisation is always the slower method
Yes — the quadratic formula is only needed as a shortcut
Check your answer
Yes — guessing eventually works for every quadratic equation, given enough time — No amount of guessing reaches an integer factorisation when none exists, as the discriminant can show in advance.
✓ No — some equations cannot be factorised at all — (B) $b^2-4 a c<0$ rules out any real factorisation, and a non-perfect-square $b^2-4 a c$ rules out any integer or simple rational one.
No — but only because factorisation is always the slower method — The issue is not speed — some equations have no integer factorisation at all, which the discriminant reveals before any guessing starts.
Yes — the quadratic formula is only needed as a shortcut — The formula is not a shortcut for a method that would eventually work — for some equations it is the only method that works at all.
For $x^2+2x+5=0$, the discriminant is $4-20=-16$. Why will guessing a factorisation never succeed here?
guessing would work eventually, with more patience
the equation must first be checked for a repeated root
no real root exists, so no factorisation is possible
factorisation fails only because the coefficients are too large
Check your answer
guessing would work eventually, with more patience — No amount of patience helps here — a negative discriminant means no real root exists to find.
the equation must first be checked for a repeated root — A repeated root needs $b^2-4 a c=0$; here the discriminant is $-16$, so this check does not apply.
✓ no real root exists, so no factorisation is possible — (C) With no real root at all, there is no real linear factorisation to guess towards, however long the search continues.
factorisation fails only because the coefficients are too large — The coefficients here are small; it is the NEGATIVE sign of the discriminant that rules out any real factorisation.
For $x^2-2x-2=0$, the discriminant is $4+8=12$, positive but not a perfect square. What does this mean for factorisation?
no integer factorisation exists here
factorisation will work if the student tries larger integer pairs
the equation has no real root, since $12$ is not a perfect square
the discriminant must be recomputed, since it should be a perfect square
Check your answer
✓ no integer factorisation exists here — (A) A discriminant that is positive but not a perfect square gives two real, irrational roots, which no integer splitting can reach.
factorisation will work if the student tries larger integer pairs — The roots here are irrational, so no integer pair, however large, can ever produce this factorisation.
the equation has no real root, since $12$ is not a perfect square — $12>0$ still gives two real roots; ‘not a perfect square’ only means those roots are irrational, not absent.
the discriminant must be recomputed, since it should be a perfect square — There is no such requirement — a positive, non-perfect-square discriminant is a normal, valid outcome.
$x (x + 2) = 3 (x - 4)$ Let us expand both sides first, then collect everything onto one side.
$x^2 + 2 x = 3 x - 12$, so $x^2 - x + 12 = 0$ Expand both sides, then move every term to one side.
coefficient of $x^2$ is $1 \neq 0$ So this equation is quadratic.
$(x - 1)^2 = x^2 - 1$ Now a second equation, checked the same way.
$x^2 - 2 x + 1 = x^2 - 1$, so $-2 x + 2 = 0$ Expand and collect. The $x^2$ terms cancel on both sides.
this is a linear equation, not a quadratic one The coefficient of $x^2$ is $0$, once everything is collected, even though $x^2$ appeared before simplifying.
at $x = 2$: $x (x + 2) = 8$ and $3 (x - 4) = -6$, a difference of $14$, matching $x^2 - x + 12 = 4 - 2 + 12 = 14$ Check the first simplification: test one value in both the original equation and the simplified form. They agree. Try the second equation the same way yourself. At $x = 2$: $(x - 1)^2 = 1$ and $x^2 - 1 = 3$, matching $-2 x + 2 = -2$.
How do you know if a messy equation is really quadratic?
Expand every bracket Multiply out both sides completely. Do not skip a single bracket.
Collect onto one side Move every term to one side, equal to zero. Now the equation is in standard form.
Read the coefficient Look only at the number in front of x squared, after collecting. If it is not zero, the equation is quadratic.
Watch for cancelling Sometimes the x squared terms cancel once collected. Then the equation is linear, not quadratic, even though x squared appeared before.
Check yourself
Following the same check used for $x(x+2)=3(x-4)$, is $2x(x+1) = x^2+2x-5$ quadratic once simplified?
No — the $x$ terms cancel, so nothing remains
Yes — $x^2+5=0$ remains after simplifying
Yes — because both sides already show an $x^2$ term before simplifying
No — because the constant term is negative
Check your answer
No — the $x$ terms cancel, so nothing remains — The $x$ terms do cancel, but $x^2+5=0$ is left over — an $x^2$ term still survives.
✓ Yes — $x^2+5=0$ remains after simplifying — (B) Expanding gives $2x^2+2x = x^2+2x-5$; the $x$ terms cancel but $x^2+5=0$ remains, with a non-zero $x^2$ coefficient.
Yes — because both sides already show an $x^2$ term before simplifying — Presence before simplifying is not decisive, as the earlier cancelling example showed — only the result AFTER simplifying decides this.
No — because the constant term is negative — The sign of the constant term is unrelated to whether the equation is quadratic — only the surviving $x^2$ coefficient matters.
Using the same method, is $(x+3)^2 = x^2+6x+9$ quadratic once simplified?
Yes — an $x^2$ term is written on both sides
No — every term cancels to $0=0$
No — because the coefficient of $x$ is too large
Yes — because expanding a square always gives a quadratic
Check your answer
Yes — an $x^2$ term is written on both sides — Writing $x^2$ on both sides is not decisive; here every term cancels completely, leaving nothing.
✓ No — every term cancels to $0=0$ — (B) Expanding the left side gives exactly the right side, so every term cancels and no equation in $x$ remains at all.
No — because the coefficient of $x$ is too large — The size of the coefficient of $x$ plays no role — what matters is that both sides are identical and cancel entirely.
Yes — because expanding a square always gives a quadratic — Expanding a square usually gives a quadratic, but here it exactly matches the other side, so everything cancels.
In the worked check, expanding $x(x+2)=3(x-4)$ gives $x^2+2x = 3x-12$. What is the next correct step?
leave the equation as $x^2+2x=3x-12$ and read off $a$, $b$, $c$ directly
collect all terms on one side to get $x^2-x+12=0$
divide both sides by $x$ to simplify
add $12$ to both sides only
Check your answer
leave the equation as $x^2+2x=3x-12$ and read off $a$, $b$, $c$ directly — Coefficients can only be read off once the equation is arranged with one side equal to $0$, which has not happened yet.
✓ collect all terms on one side to get $x^2-x+12=0$ — (B) Moving every term to one side is exactly what standard form requires, before $a$, $b$, $c$ can be read off.
divide both sides by $x$ to simplify — Dividing by $x$ risks losing the root $x=0$ if it exists, and is not part of reaching standard form.
add $12$ to both sides only — Adding $12$ to both sides alone still leaves $x$ terms on both sides — the rearrangement is not finished.
breadth $= x$ metres, length $= (x + 4)$ metres Let us name the unknown first. The length is stated in terms of the breadth, so call the breadth $x$.
$x (x + 4) = 96$ Area is length times breadth, and here the area is $96$ square metres.
$x^2 + 4 x - 96 = 0$ Expand, then move every term to one side. This is the quadratic equation the breadth must satisfy.
check $x = 8$: $8 (8 + 4) = 8 \cdot 12 = 96$ *Check it: does a breadth of $8$ metres satisfy the equation?* Then the length is $12$ metres, and $8 \cdot 12 = 96$ matches the garden’s area. The equation checks out.
breadth
length
Aarav marks out a rectangular garden bed with four pegs, its length longer than its breadth.
A rectangle, eight metres by twelve, shows the garden’s breadth and length. Together they give ninety-six square metres.
How do you turn an area problem into a quadratic equation?
Name the unknown Call the smaller length $x$. Write every other length in terms of that same $x$.
Write the area condition Multiply length by breadth. Set the product equal to the area you are given.
Simplify to standard form Expand the brackets and move every term to one side. This is the quadratic equation $x$ must satisfy.
Check your answer Substitute your value of $x$ back in. Confirm the area comes out right, and reject a negative length.
Check yourself
Following the garden method, a rectangular hall has area $150$ square metres and its length is $5$ metres more than its breadth. Taking the breadth as $x$ metres, which equation results?
$x^2-5x-150=0$
$x^2+5x+150=0$
$x^2+5x-150=0$
$x(x+5)=150x$
Check your answer
$x^2-5x-150=0$ — The length is $5$ metres MORE than the breadth, so it is $(x+5)$, not $(x-5)$.
$x^2+5x+150=0$ — Moving the area $150$ across the equation keeps it negative, giving $-150$, not $+150$.
✓ $x^2+5x-150=0$ — (C) The length is $(x+5)$ metres, so $x(x+5)=150$ gives $x^2+5x-150=0$.
$x(x+5)=150x$ — The area condition is simply $x(x+5)=150$ — there is no reason to multiply the right side by $x$ again.
Solving $x^2+5x-150=0$ from the hall problem gives $x=10$ and $x=-15$. Which value is the breadth?
$x=-15$ metres, taken as a magnitude of $15$ metres
both values, since the equation gives two roots
neither value, since a breadth this large is unrealistic
$x=10$ metres, since a breadth cannot be negative
Check your answer
$x=-15$ metres, taken as a magnitude of $15$ metres — The equation’s actual root is $-15$, not $15$ — a negative root is rejected, not converted into a positive one.
both values, since the equation gives two roots — Not every algebraic root fits the situation; a negative breadth has no physical meaning.
neither value, since a breadth this large is unrealistic — $10$ metres is a perfectly ordinary breadth for a hall; the only real basis for rejection is a negative value.
✓ $x=10$ metres, since a breadth cannot be negative — (D) Only $x=10$ is a physically valid breadth; $x=-15$ is an algebraic root with no meaning here.
The garden method also applies to a different pair of linked quantities. A rectangular field’s perimeter is $46$ metres and its area is $120$ square metres. Taking the length as $x$ metres, the breadth is $(23-x)$ metres from the perimeter condition. Which equation does $x$ satisfy?
$x^2+23x-120=0$
$x(23-x)=120x$
$x^2-46x+120=0$
$x^2-23x+120=0$
Check your answer
$x^2+23x-120=0$ — Rearranging $23x-x^2=120$ correctly gives $x^2-23x+120=0$; flipping both signs at once is not a valid step.
$x(23-x)=120x$ — The area condition is simply $x(23-x)=120$ — there is no reason to multiply the right side by $x$ again.
$x^2-46x+120=0$ — The breadth is $(23-x)$, half the perimeter minus the length — using $46$ directly skips that step.
✓ $x^2-23x+120=0$ — (D) The area condition $x(23-x)=120$ expands to $23x-x^2=120$, which rearranges to $x^2-23x+120=0$.
solve $x^2 - 7 x + 12 = 0$ Let us split the middle term first.
$(-3) + (-4) = -7$ and $(-3) \cdot (-4) = 12$ Find two numbers that multiply to $12$ and add to $-7$.
$x^2 - 3 x - 4 x + 12 = x (x - 3) - 4 (x - 3) = (x - 3)(x - 4)$ Split the middle term using those numbers. Then group in pairs, and factor each pair.
$x = 3$ or $x = 4$ Set each factor to $0$, in turn.
now solve $x^2 - 6 x + 9 = 0$ the same way Now a second equation, solved the same way.
$(-3) + (-3) = -6$ and $(-3) \cdot (-3) = 9$ The split this time lands on the same number, twice.
$(x - 3)(x - 3) = (x - 3)^2 = 0$ The same factor appears twice: not two different factors.
$x = 3$, a repeated root One factor set to $0$ gives one root, counted twice.
$3^2 - 7 \cdot 3 + 12 = 9 - 21 + 12 = 0$ and $4^2 - 7 \cdot 4 + 12 = 16 - 28 + 12 = 0$ *Check both roots: substitute them back into the original equation. Both give $0$, so $3$ and $4$ really are the roots.* Try the same check yourself, on the repeated root $x = 3$ of the second equation.
How do you solve a quadratic by factorising?
Write in standard form Get the equation as $a x^2 + b x + c = 0$ first.
Find the split Find two numbers that multiply to $a c$ and add to $b$.
Split and group Split the middle term using those two numbers. Group the four terms in pairs.
Factor each pair Pull the common factor out of each pair. The same bracket should appear in both.
Set each factor to zero Each factor gives one root. Two factors give two roots, unless they are the same factor twice.
Check yourself
Using the same splitting method as $x^2-7x+12=(x-3)(x-4)$, how does $x^2-9x+20$ factorise?
$(x-2)(x-10)$
$(x+4)(x+5)$
$(x-4)(x+5)$
$(x-4)(x-5)$
Check your answer
$(x-2)(x-10)$ — $-2$ and $-10$ multiply to $20$, but their sum is $-12$, not the required $-9$.
$(x+4)(x+5)$ — For the sum to be $-9$ with a positive product $20$, both numbers must be negative, not positive.
$(x-4)(x+5)$ — $(-4)(5)=-20$, not $20$ — the signs on the two factors do not match the required positive product.
✓ $(x-4)(x-5)$ — (D) $-4$ and $-5$ multiply to $20$ and add to $-9$, matching $a c=20$ and $b=-9$.
Using the same method, does $x^2-10x+25$ factorise to a repeated root, and if so what is it?
$(x-5)^2=0$, giving the repeated root $x=5$
$(x-5)(x+5)=0$, giving roots $5$ and $-5$
no repeated root exists, since $25$ is not a perfect square of $10$
$(x-25)(x-1)$, giving roots $25$ and $1$
Check your answer
✓ $(x-5)^2=0$, giving the repeated root $x=5$ — (A) $-5$ and $-5$ multiply to $25$ and add to $-10$, so the factorisation is $(x-5)(x-5)$.
$(x-5)(x+5)=0$, giving roots $5$ and $-5$ — $(x-5)(x+5)$ expands to $x^2-25$, a different equation; here both split numbers must be $-5$, not opposite signs.
no repeated root exists, since $25$ is not a perfect square of $10$ — The correct check is the discriminant, $100-100=0$, or the factor pair $(-5,-5)$ — comparing $25$ to $10$ is not the actual test.
$(x-25)(x-1)$, giving roots $25$ and $1$ — $-25$ and $-1$ multiply to $25$, but their sum is $-26$, not $-10$ — the sum condition was never checked.
A student factorises $x^2-x-12$ as $(x-3)(x+4)$. Expanding this gives $x^2+x-12$. Is the factorisation correct for $x^2-x-12$?
Yes — the constant term $-12$ matches, so the factorisation is correct
No — because $3$ and $4$ do not multiply to $-12$
No — the middle term’s sign is wrong
Yes — $(x-3)(x+4)$ and $(x-4)(x+3)$ give the same equation
Check your answer
Yes — the constant term $-12$ matches, so the factorisation is correct — The constant term matching is not enough — the middle term must also match, and here it does not.
No — because $3$ and $4$ do not multiply to $-12$ — $(-3)(4)=-12$ does match correctly — the actual fault is that this pairing gives $+x$ in the middle, not $-x$.
✓ No — the middle term’s sign is wrong — (C) The expansion produces $+x$ in the middle, not the required $-x$, so this factorisation does not match the original equation.
Yes — $(x-3)(x+4)$ and $(x-4)(x+3)$ give the same equation — $(x-3)(x+4)$ gives $+x$ in the middle while $(x-4)(x+3)$ gives $-x$ — they are genuinely different equations.
The worked examples show both a two-distinct-root case and a repeated-root case from factorisation. Which feature of the factorisation tells you which case has occurred?
whether the constant term is positive or negative
whether the two factors match
whether the coefficient of $x$ is even or odd
whether both roots are positive
Check your answer
whether the constant term is positive or negative — The sign of the constant term does not decide this — both cases in the worked examples can occur with either sign of $c$.
✓ whether the two factors match — (B) Two different factors give two distinct roots; two identical factors give one repeated root.
whether the coefficient of $x$ is even or odd — Whether $b$ is even or odd has no bearing on whether the factorisation produces a repeated root.
whether both roots are positive — A repeated root can be positive, negative, or zero — its sign says nothing about whether it is repeated.
Whatever the equation, its two roots are mirror images of each other in the axis of the parabola.
Worked example
Solving with the formula
solve $2 x^2 + 3 x - 5 = 0$: $a = 2$, $b = 3$, $c = -5$ Let us read the coefficients off the standard form, before touching the formula.
$D = b^2 - 4 a c = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$ Compute the discriminant first, as one number.
$\sqrt{49} = 7$ The discriminant is a perfect square here, so its square root is a whole number.
$x = (-3 \pm 7) / 4$ Substitute into the quadratic formula.
$x = 1$ or $x = -5/2$ Evaluate both signs of $\pm$, separately.
$2 (1)^2 + 3 (1) - 5 = 0$ and $2 (-5/2)^2 + 3 (-5/2) - 5 = 0$ Substitute each root back into the original equation, and confirm it works. This check is what the formula’s correctness rests on, not a separate proof. Try it yourself, on any equation you solve with the formula from here on.
How do you solve a quadratic with the formula?
Read off a, b, c Write the equation in standard form first. Then read $a$, $b$, $c$ straight off it.
Compute the discriminant Work out $b^2 - 4 a c$ as one number, before anything else.
Find its square root Take the square root of the discriminant. Simplify it if you can.
Substitute into the formula Put $a$, $b$ and the square root into $x = (-b \pm \sqrt{b^2 - 4 a c}) / (2 a)$.
Evaluate both signs Work out the plus case and the minus case separately. Each gives you one root.
Check yourself
Using the same formula method as $2x^2+3x-5=0$, what is the discriminant of $3x^2-2x-8=0$?
$4$
$-92$
$192$
$100$
Check your answer
$4$ — $b^2=4$ alone is not the discriminant; $4 a c=-96$ must still be subtracted, giving $100$.
$-92$ — Since $c=-8$ is negative, $-4 a c=+96$ must be ADDED to $b^2$, not combined with the opposite sign.
$192$ — $4 a c=-96$ should be subtracted once; counting it twice gives an inflated, incorrect value.
✓ $100$ — (D) $b^2-4 a c = (-2)^2 - 4(3)(-8) = 4 + 96 = 100$.
Continuing with $3x^2-2x-8=0$, whose discriminant is $100$, what are the roots?
$x=-2$ and $x=4/3$
$x=2$ and $x=-4/3$
$x=12$ and $x=-8$
$x=2$ and $x=4/3$
Check your answer
$x=-2$ and $x=4/3$ — $-b=2$ here, not $-2$; using the wrong sign flips both resulting roots.
✓ $x=2$ and $x=-4/3$ — (B) $x = (2 \pm 10)/6$ gives $x=12/6=2$ and $x=-8/6=-4/3$.
$x=12$ and $x=-8$ — The numerator $2 \pm 10$ must still be divided by $2a=6$, which was skipped here.
$x=2$ and $x=4/3$ — The second numerator is $2-10=-8$, not $8$; dropping the sign gives the wrong second root.
Using the same formula method, what are the roots of $x^2+2x-4=0$?
$x = -2 \pm \sqrt{5}$
$x = -1 \pm \sqrt{5}$
$x = -1 \pm \sqrt{20}$
$x = -1 \pm 5$
Check your answer
$x = -2 \pm \sqrt{5}$ — $-b=-2$ must also be divided by $2a=2$, giving $-1$, not left as $-2$.
✓ $x = -1 \pm \sqrt{5}$ — (B) $b^2-4 a c=4+16=20$, so $x=(-2 \pm \sqrt{20})/2 = -1 \pm \sqrt{5}$ after simplifying.
$x = -1 \pm \sqrt{20}$ — $\sqrt{20}/2$ simplifies to $\sqrt{5}$ once the common factor is taken out — leaving it as $\sqrt{20}$ is inconsistent.
$x = -1 \pm 5$ — $\sqrt{20}$ simplifies to $2\sqrt{5}$, not to the whole number $5$ — the square root cannot be dropped.
$x^2 + 4 x + 5 = 0$: $b^2 - 4 a c = 16 - 20 = -4$ Let us compute the discriminant first.
$D < 0$, so there is no real root A negative discriminant rules out any real root.
$x^2 - 4 x + 4 = 0$: $b^2 - 4 a c = 16 - 16 = 0$ Now a second equation, discriminant computed the same way.
$D = 0$, so there is one repeated real root, $x = 2$ A zero discriminant means the equation factors as one factor, squared.
$x^2 - 4 x + 3 = 0$: $b^2 - 4 a c = 16 - 12 = 4$ A third equation.
$D > 0$, so there are two distinct real roots, $x = 1$ and $x = 3$ A positive discriminant means two different real roots.
$2^2 - 4 \cdot 2 + 4 = 4 - 8 + 4 = 0$, and $1^2 - 4 \cdot 1 + 3 = 0$, $3^2 - 4 \cdot 3 + 3 = 9 - 12 + 3 = 0$ Check the roots from the second and third equations: substitute them back. Every one gives $0$, confirming the discriminant predicted the right number of roots, each time.
Three parabolas, three discriminants, three verdicts on how many times each meets the axis, all known without finding a single root.
How do you tell how many roots an equation has, without solving it?
Read off a, b, c Write the equation in standard form, then read $a$, $b$, $c$ off it.
Compute the discriminant Work out $b^2 - 4 a c$ as one number.
Read its sign A positive discriminant means two distinct roots. A zero discriminant means one repeated root. A negative discriminant means no real root.
Check your answer If you go on to solve the equation, the number of roots you find should match what the sign predicted.
Check yourself
Following the same three-case method, what is the nature of the roots of $x^2+6x+9=0$?
one repeated real root
two distinct real roots
no real root
two complex roots, since the equation cannot be factorised
Check your answer
✓ one repeated real root — (A) $b^2-4 a c=36-36=0$, exactly the boundary case that gives a repeated root, $x=-3$.
two distinct real roots — Two distinct roots need a positive discriminant; here $b^2-4 a c=0$ exactly.
no real root — No real root needs a negative discriminant; here $b^2-4 a c=0$ exactly.
two complex roots, since the equation cannot be factorised — $x^2+6x+9$ factorises as $(x+3)^2$; a repeated real root is a normal, solvable case, not a sign of failure.
What is the nature of the roots of $x^2+x+1=0$?
two distinct real roots
one repeated real root
two real roots, both negative
no real root
Check your answer
two distinct real roots — The discriminant is $-3$, negative — that rules out any real roots, distinct or otherwise.
one repeated real root — A repeated root needs a discriminant of exactly $0$; here it is $-3$.
two real roots, both negative — No real roots exist here at all, since the discriminant is negative — there is nothing to sign.
✓ no real root — (D) $b^2-4 a c=1-4=-3$, which is negative, so no real root exists.
What is the nature of the roots of $2x^2-7x+3=0$?
one repeated real root
no real root
two distinct real roots
cannot be decided without factorising first
Check your answer
one repeated real root — $25$ is a perfect square but is not $0$ — a repeated root needs the discriminant to equal exactly $0$.
no real root — Computing $b^2-4 a c$ correctly gives $49-24=25$, positive — reversing the subtraction order gives the wrong sign.
✓ two distinct real roots — (C) $b^2-4 a c=49-24=25$, a positive perfect square, giving two distinct real (and rational) roots.
cannot be decided without factorising first — The discriminant alone already fixes the nature of the roots — no factorising is needed to know this.
Worked: checking the discriminant before factoring
Worked example
Checking the discriminant before factoring
solve $x^2 - 2 x - 2 = 0$ Let us check $D$ before attempting to factorise.
$D = b^2 - 4 a c = (-2)^2 - 4 \cdot 1 \cdot (-2) = 4 + 8 = 12$ Compute the discriminant from $a = 1$, $b = -2$, $c = -2$.
$12$ is positive but not a perfect square A real root exists, since $D > 0$. But no integer or simple rational split of the middle term will be found: the roots are irrational.
$x = (2 \pm \sqrt{12}) / 2 = 1 \pm \sqrt{3}$ The quadratic formula reaches the roots directly. No factorisation attempt is needed.
$(1 + \sqrt{3})^2 - 2 (1 + \sqrt{3}) - 2 = 4 + 2 \sqrt{3} - 2 - 2 \sqrt{3} - 2 = 0$ Check the root: substitute it back into the original equation. The square-root terms cancel, leaving $0$, so the irrational root really does work. Check the other root, $1 - \sqrt{3}$, the same way yourself.
Check yourself
Following the same check-before-factoring method, what does the discriminant of $x^2-6x+2=0$ tell you before attempting to factorise?
$b^2-4 a c=28$, so the equation has no real root
positive, not a perfect square — formula needed
$b^2-4 a c=28$, so factorisation with integers must eventually be found by trying more pairs
$b^2-4 a c=28$, so the equation has a repeated root
Check your answer
$b^2-4 a c=28$, so the equation has no real root — $28$ is positive, so two real roots exist — ‘not a perfect square’ only means those roots are irrational.
✓ positive, not a perfect square — formula needed — (B) A positive, non-perfect-square discriminant confirms two real, irrational roots that factorisation with integers cannot reach.
$b^2-4 a c=28$, so factorisation with integers must eventually be found by trying more pairs — No integer pair can multiply to $2$ and add to $-6$ — the search would never succeed, however long it continued.
$b^2-4 a c=28$, so the equation has a repeated root — A repeated root requires the discriminant to be exactly $0$; here it is $28$.
Having confirmed $D=28$ for $x^2-6x+2=0$ is not a perfect square, what are the roots?
$x=3 \pm \sqrt{7}$
$x=6 \pm \sqrt{7}$
$x=3 \pm \sqrt{28}$
$x=3 \pm 7$
Check your answer
✓ $x=3 \pm \sqrt{7}$ — (A) $x=(6 \pm \sqrt{28})/2$ simplifies to $3 \pm \sqrt{7}$ once the common factor is taken out.
$x=6 \pm \sqrt{7}$ — $-b=6$ must also be divided by $2a=2$, giving $3$, not left as $6$.
$x=3 \pm \sqrt{28}$ — $\sqrt{28}/2$ simplifies to $\sqrt{7}$ once the common factor is taken out — leaving it as $\sqrt{28}$ is inconsistent.
$x=3 \pm 7$ — $\sqrt{28}$ simplifies to $2\sqrt{7}$, not to the whole number $7$ — the square root cannot be dropped.
Why check the discriminant before attempting to factorise, rather than after a failed attempt?
it makes the factorisation method unnecessary in every case
it is required only when the equation has a negative constant term
it replaces the need to write the equation in standard form first
it rules out impossible splits in advance
Check your answer
it makes the factorisation method unnecessary in every case — When the discriminant is a perfect square, factorisation is still the faster, valid method — the check does not replace it entirely.
it is required only when the equation has a negative constant term — The discriminant check applies to every quadratic equation, whatever the sign of $c$ happens to be.
it replaces the need to write the equation in standard form first — Standard form is still needed first, to identify $a$, $b$, $c$ — the discriminant cannot be computed without it.
✓ it rules out impossible splits in advance — (D) Checking first replaces trial and error with a direct, one-step test of whether factorisation is even possible.
Whichever lens you pick, the same roots come out; use the discriminant first to know how many to look for.
STANDARD FORM: $a x^2 + b x + c = 0$, $a \neq 0$. For $2 x^2 + 3 x - 5 = 0$: $a = 2$, $b = 3$, $c = -5$.
DISCRIMINANT: $D = b^2 - 4 a c$. Here $D = 3^2 - 4 \cdot 2 \cdot (-5) = 49$.
NATURE OF ROOTS: $D > 0$ gives two distinct roots, $D = 0$ gives one repeated root, $D < 0$ gives no real root. Here $D = 49 > 0$, so two distinct real roots.
FACTORISATION: split $b x$ into parts multiplying to $a c$ and adding to $b$. For $x^2 - 7 x + 12 = 0$, the split is $-3$ and $-4$, giving roots $3$ and $4$.
Let us look back over what we have covered. Every quadratic equation $a x^2 + b x + c = 0$, $a \neq 0$, has at most two real roots, never more. Two routes reach them. One factorises the expression into two linear pieces. The other applies the formula, and that works whether or not a factorisation is easy to spot.
One number decides which case you are in, before either method starts. It does this before you find a single root. Skip that check, and you can spend a long time hunting for a factorisation that does not exist.
A situational problem is no different in kind. You solve it by naming the unknown, writing the one stated condition as an equation, and solving it by the same two methods as any other quadratic. The only extra step a real situation adds is checking your answer against the situation itself. A length or a count can never come out negative.
Check yourself
Which single quantity, computed from $a$, $b$, $c$ alone, decides which solving method fits a given quadratic equation before either method is tried?
the sum of the coefficients, $a+b+c$
the value of $c$ alone
the standard form of the equation
the discriminant, $b^2-4 a c$
Check your answer
the sum of the coefficients, $a+b+c$ — $a+b+c$ has no established role in deciding which method fits — it is the discriminant that carries that job.
the value of $c$ alone — $a$ and $b$ both affect the discriminant just as much as $c$ does — the constant alone cannot decide this.
the standard form of the equation — Standard form only arranges the equation so $a$, $b$, $c$ can be read off — it is the discriminant, computed from those, that decides the method.
✓ the discriminant, $b^2-4 a c$ — (D) The discriminant is exactly the chapter’s tool for deciding the root count and, together with the perfect-square check, the right solving route.
A situational problem is turned into a quadratic equation and solved. What must be checked at the very end, that the discriminant and the formula alone cannot confirm?
whether each root fits the real situation
whether the discriminant was computed correctly
whether the equation was quadratic in the first place
whether a repeated root exists
Check your answer
✓ whether each root fits the real situation — (A) The discriminant and formula confirm the roots ARE roots, but only the situation itself can confirm which root is physically meaningful.
whether the discriminant was computed correctly — Checking the discriminant’s computation happens earlier, while solving — the final step is judging the roots against the real situation.
whether the equation was quadratic in the first place — Whether the equation is quadratic is confirmed at the very start, before any solving — not as a final check.
whether a repeated root exists — Whether a root is repeated is already known from the discriminant, long before the final situational check.
Two students solve the same quadratic equation, one by factorisation and one by the quadratic formula, and reach the same two roots. What does this confirm about the discriminant?
the discriminant must have been $0$, since both methods agree
the discriminant is irrelevant once both methods agree
the discriminant was a positive perfect square
the discriminant must have been negative, since two roots exist
Check your answer
the discriminant must have been $0$, since both methods agree — A discriminant of $0$ gives ONE repeated root — here two roots were reached, so the discriminant was positive, not $0$.
the discriminant is irrelevant once both methods agree — The agreement between the two methods is a direct consequence of the discriminant being a positive perfect square, not a reason to ignore it.
✓ the discriminant was a positive perfect square — (C) Factorisation over integers only succeeds when the discriminant is a positive perfect square, giving rational roots that the formula reaches too.
the discriminant must have been negative, since two roots exist — A negative discriminant rules out any real root — factorisation could never have succeeded in that case.
Check yourself: the whole chapter
A quadratic equation $a x^2 + b x + c = 0$ ($a \neq 0$) has at most two real roots. What tells you, before solving, whether it has two distinct roots, one repeated root, or none?
the sign of the constant term $c$ alone
the sign of the discriminant $b^2 - 4 a c$
whether $a$ is positive or negative
the number of terms written in the equation
Check your answer
the sign of the constant term $c$ alone — The sign of $c$ alone does not decide the root count — it is only one of three numbers that feed into the discriminant.
✓ the sign of the discriminant $b^2 - 4 a c$ — (B) The discriminant $b^2 - 4 a c$ says, before solving, whether the equation has two distinct real roots, one repeated real root, or no real root, from its sign alone.
whether $a$ is positive or negative — The sign of $a$ decides whether the parabola opens upward or downward, not how many real roots the equation has.
the number of terms written in the equation — Every quadratic equation in standard form has the same three terms — counting them tells you nothing about the roots.
$x^2 - x - 12$ factorises as $(x - 4)(x + 3)$. What are the roots of $x^2 - x - 12 = 0$?
$x = 4$ and $x = 3$
$x = -4$ and $x = -3$
$x = 4$ and $x = -3$
$x = -4$ and $x = 3$
Check your answer
$x = 4$ and $x = 3$ — Setting $x + 3 = 0$ gives $x = -3$, not $x = 3$ — carry the sign across when solving each factor.
$x = -4$ and $x = -3$ — Setting $x - 4 = 0$ gives $x = 4$ (not $-4$), since it is $x$ minus $4$ equalling zero — check each factor on its own.
✓ $x = 4$ and $x = -3$ — (C) Setting each factor equal to $0$ gives $x - 4 = 0$, so $x = 4$, and $x + 3 = 0$, so $x = -3$ — the two roots are $4$ and $-3$.
$x = -4$ and $x = 3$ — Each root comes from its own factor — $x - 4 = 0$ gives $x = 4$, and $x + 3 = 0$ gives $x = -3$ — not the other way round.
Using the quadratic formula $x = (-b \pm \sqrt{b^2 - 4 a c}) / (2 a)$ on $x^2 - 5x + 6 = 0$ (so $a = 1$, $b = -5$, $c = 6$), what are the roots?
$x = 5$ and $x = 6$
$x = -3$ and $x = -2$
$x = 1$ and $x = 6$
$x = 3$ and $x = 2$
Check your answer
$x = 5$ and $x = 6$ — The formula must actually be applied — $b$ and $c$’s raw values, $5$ and $6$, are not the roots.
$x = -3$ and $x = -2$ — With $b = -5$, $-b = 5$, not $-5$ — the sign flips because $b$ itself is negative; check $-b$ carefully before dividing.
$x = 1$ and $x = 6$ — $\sqrt{b^2 - 4 a c} = \sqrt{1} = 1$ is part of the formula, added to and subtracted from $-b = 5$ — it is not a root on its own.
✓ $x = 3$ and $x = 2$ — (D) $-b = 5$ and $b^2 - 4 a c = 25 - 24 = 1$, so $\sqrt{1} = 1$, giving $x = (5 \pm 1)/2$, which is $x = 3$ or $x = 2$.
For the equation $3x^2 - 2x + 5 = 0$, what is the discriminant?
$-4 - 60 = -64$
$4 - 60 = -56$
$4 - 4(3+5) = -28$
$4 - (3)(5) = -11$
Check your answer
$-4 - 60 = -64$ — $(-2)^2 = 4$, not $-4$ — squaring a negative number always gives a positive result.
$4 - 4(3+5) = -28$ — $4 a c$ needs $a$ and $c$ multiplied together, then multiplied by $4$ — $4(3)(5) = 60$, not $4(3+5)$.
$4 - (3)(5) = -11$ — The formula is $b^2 - 4 a c$, with the $4$ included — $4(3)(5) = 60$, not $(3)(5) = 15$ alone.
An equation has discriminant $b^2 - 4 a c = 0$. What does this tell you about its roots?
it has no real root
it has two distinct real roots
it has one repeated real root
it cannot be determined without solving the equation
Check your answer
it has no real root — No real root happens when the discriminant is negative — a discriminant of exactly $0$ is a different, in-between case.
it has two distinct real roots — Two distinct real roots happen when the discriminant is positive — a discriminant of exactly $0$ gives only one root, repeated.
✓ it has one repeated real root — (C) A discriminant of exactly $0$ means the equation has one repeated real root.
it cannot be determined without solving the equation — The discriminant’s sign is exactly what predicts the nature of the roots, from $a$, $b$, $c$ alone, before solving anything.
You will meet quadratics whenever one fact fixes two unknowns together, like a length and a width. Here are seven of those places.
Sizing a photo print. A photo print is $3$ cm longer than it is wide. Its area is $108$ square cm. The maths: name the width as $x$, then translate the one condition, area $= 108$, into a quadratic equation. Width $x$, length $x + 3$: $x(x + 3) = 108$, so $x^2 + 3x - 108 = 0$. Width $9$ cm, length $12$ cm.
Tiling a courtyard grid. A courtyard is tiled in rows and columns of square tiles. It has $5$ fewer rows than columns, and $150$ tiles in all. The maths: splitting the middle term factors the quadratic. Each factor then gives one root. Columns $x$: $x(x - 5) = 150$, so $x^2 - 5x - 150 = 0$. Split $-5$ into $-15$ and $10$: $(x - 15)(x + 10) = 0$, so $x = 15$ columns.
Checking a friend’s answer about a trip. A trip is $240$ km. Going $20$ km/h faster would save $1$ hour. A friend says the usual speed was $60$ km/h. The maths: a root must make the equation true when you put it back in. Here the equation is $x^2 + 20 x - 4800 = 0$. $60^2 + 20 \cdot 60 - 4800 = 3600 + 1200 - 4800 = 0$. And $240 / 60 - 240 / 80 = 4 - 3 = 1$ hour. It checks out.
Testing if a plot is possible. A rectangular plot has a fixed perimeter of $40$ m. You want its area to be exactly $120$ square m. The maths: a negative discriminant means the equation has no real root, so no real plot can give that area. Breadth $x$, length $20 - x$: $x^2 - 20x + 120 = 0$. The discriminant is $(-20)^2 - 4 \cdot 1 \cdot 120 = -80$.
A quadratic has a top, and a target above that top gives an equation with no real root at all.
Getting the most from a fixed fence. You have the same $40$ m of fence. This time you want an area of exactly $100$ square m. The maths: the equation factorises into the same factor twice. So it has one repeated root, and only one shape works. Breadth $x$, length $20 - x$: $x^2 - 20 x + 100 = 0$, which is $(x - 10)^2 = 0$. Only the $10$ m by $10$ m square works.
Sizing a rectangular poster. A rectangular poster has a diagonal of $13$ cm and an area of $60$ square cm. The maths: the quadratic formula finds both sides directly, without guessing a factorisation first. Sides add to $\sqrt{13^2 + 2 \cdot 60} = \sqrt{289} = 17$ cm. The formula gives $t = (17 \pm \sqrt{289 - 240}) / 2$: $12$ and $5$ cm.
Finding a deposit’s yearly rate. You put $₹10000$ in a deposit that adds interest once a year. After $2$ years it has grown to $₹12100$. The maths: name the yearly growth factor $x$. Two years of growth make the quadratic equation $10000 x^2 = 12100$. $x^2 = 1.21$, so $x = 1.1$. The other root, $-1.1$, makes no sense here. So the rate is $10%$ a year.
Your turn. A rectangular rug is $2$ m longer than it is wide. Its area is $24$ square m. How wide is it? Answer: Width $x$, length $x + 2$: $x^2 + 2x - 24 = 0$. Since $6 \cdot (-4) = -24$ and $6 + (-4) = 2$, the factors are $(x + 6)(x - 4) = 0$, so $x = 4$ m.
practice Check whether $x (x + 5) = 2 (x - 3)$ is a quadratic equation, after full simplification. (Worked in full below — read it, then do the next two the same way.)
practice Check whether $(x + 2)^2 = x^2 + 9$ is a quadratic equation, after full simplification. (Start the same way — expand both sides and collect everything on one side before looking at the coefficient of $x^2$.)
practice A rectangular room is $2$ metres longer than it is wide, and its area is $48$ square metres. Taking the width as $x$ metres, write the quadratic equation that $x$ must satisfy. (Name the width first, then write the length in terms of it.)
practice Check whether $2 x^2 - 3 x + 1 = (x - 1)(x + 2)$ is a quadratic equation, after full simplification.
practice Check whether $(x + 3)^2 = x^2 - 4$ is a quadratic equation, after full simplification.
practice A rectangular plot’s length is $3$ metres more than twice its breadth, and its area is $90$ square metres. Taking the breadth as $x$ metres, write the quadratic equation the breadth satisfies.
practice The product of two consecutive positive integers is $240$. Taking the smaller integer as $x$, write the quadratic equation it satisfies.
practice A train travels a distance of $300$ km at a uniform speed. If the speed had been $5$ km/h more, it would have taken $2$ hours less for the same journey. Taking the speed as $x$ km/h, write the quadratic equation $x$ satisfies.
practice The sum of the ages of two friends is $24$ years. Four years ago, the product of their ages in years was $45$. Taking the age of one friend as $x$ years, form the quadratic equation this problem gives.
Answers
Yes. Expanding gives $x^2 + 5 x = 2 x - 6$, so $x^2 + 3 x + 6 = 0$, and the coefficient of $x^2$ is $1 \neq 0$.
No. Expanding gives $x^2 + 4 x + 4 = x^2 + 9$, and the $x^2$ terms cancel, leaving $4 x - 5 = 0$ — a linear equation.
$x (x + 2) = 48$, that is $x^2 + 2 x - 48 = 0$.
Yes; expanding and collecting gives $x^2 - 4 x + 3 = 0$, coefficient of $x^2$ is $1 \neq 0$.
No — the $x^2$ terms cancel; expanding and collecting gives the linear equation $6 x + 13 = 0$.
$2 x^2 + 3 x - 90 = 0$.
$x^2 + x - 240 = 0$.
$x^2 + 5 x - 750 = 0$.
$x^2 - 24 x + 125 = 0$.
Exercise 4.1 — further practice
practice Check whether $(x - 4)(x + 2) = x^2 - 3 x + 6$ is a quadratic equation, after full simplification.
practice Check whether $3 x (x - 1) = 5 (x + 2)(x - 2) - 4 x$ is a quadratic equation, after full simplification.
practice Which of these equations, once fully simplified, is not a quadratic equation?
$(x + 2)^2 = 3 x + 10$
$x (x - 4) = (x - 1)^2 - 2$
$2 x^2 - 3 x = (x + 1)^2 - 5$
$3 x^2 = (x - 2)(x + 5)$
practice A rectangular hall’s carpet area is $126$ square metres, and its length is $5$ metres more than twice its breadth. Taking the breadth as $x$ metres, write the quadratic equation the breadth satisfies.
practice A cottage industry produces a certain number of pens in a day. The cost of production of each pen, in rupees, is $8$ more than the number of pens produced that day. If the total cost of production that day was ₹240, taking the number of pens produced as $x$, write the quadratic equation $x$ satisfies.
practice A woman is $3$ years older than her son. Five years from now, the product of their ages in years will be $150$. Taking the son’s present age as $x$ years, form the quadratic equation this problem gives.
practice The hypotenuse of a right triangle is $2$ metres more than twice its shorter leg, and its longer leg is $1$ metre less than twice the shorter leg. Taking the shorter leg as $x$ metres, write the quadratic equation $x$ satisfies, using the Pythagoras theorem.
Answers
No — the $x^2$ terms cancel; expanding and collecting gives the linear equation $x - 14 = 0$.
Yes; expanding and collecting gives $2 x^2 - x - 20 = 0$, coefficient of $x^2$ is $2 \neq 0$.
B — it simplifies to $-2 x + 1 = 0$, a linear equation.
practice Solve $x^2 - 9 x + 20 = 0$ by factorisation. (Worked in full below — read it, then do the next two the same way.)
practice Solve $x^2 + 2 x - 15 = 0$ by factorisation. (Start the same way — find two numbers that multiply to $-15$ and add to $2$. One of them is negative.)
practice Find two numbers whose sum is $15$ and whose product is $56$. (Call one of them $x$; the other is then $15 - x$.)
practice Solve $x^2 - 5 x - 24 = 0$ by factorisation.
practice Solve $4 x^2 - 12 x + 9 = 0$ by factorisation.
practice Find two numbers whose sum is $27$ and product is $182$.
practice The product of two consecutive positive odd integers is $143$. Find the integers.
practice The hypotenuse of a right triangle is $13$ cm, and one leg is $7$ cm longer than the other. Taking the shorter leg as $x$ cm, find the two legs by factorisation.
practice A shopkeeper buys a number of books for ₹720. If the price per book had been ₹4 more, the shopkeeper would have bought $6$ fewer books for the same amount. Taking the number of books bought as $x$, find $x$ by factorisation.
Answers
$x = 4$ and $x = 5$.
$x = -5$ and $x = 3$.
The numbers are $7$ and $8$.
$x = 8$ or $x = -3$.
A repeated root, $x = 3/2$; $4 x^2 - 12 x + 9 = (2 x - 3)^2$.
The two numbers are $14$ and $13$.
$11$ and $13$.
The legs are $5$ cm and $12$ cm ($x = 5$; $x = -12$ is rejected as a negative length).
$x = 36$ books, originally priced at ₹20 each ($x = -30$ is rejected).
Exercise 4.2 — further practice
practice Solve $x^2 - 2 x - 35 = 0$ by factorisation.
practice Solve $6 x^2 + x - 2 = 0$ by factorisation.
practice Which of these quadratic equations has a repeated root?
$x^2 - 5 x + 6 = 0$
$x^2 - 6 x + 9 = 0$
$x^2 - x - 6 = 0$
$x^2 + x - 6 = 0$
practice Find two positive numbers whose difference is $5$ and whose product is $204$.
practice Find two consecutive positive even integers whose product is $624$.
practice The sum of the squares of two consecutive positive integers is $181$. Find the integers.
practice A rectangular park’s length is $4$ metres more than its breadth, and its area is $192$ square metres. Taking the breadth as $x$ metres, find the length and breadth by factorisation.
practice The hypotenuse of a right triangle is $25$ m, and one leg is $17$ m longer than the other. Taking the shorter leg as $x$ m, find the two legs by factorisation.
practice A cottage industry produces a number of table lamps in a day. The cost of production of each lamp, in rupees, is $7$ more than twice the number of lamps produced that day. If the total cost of production that day was ₹114, taking the number of lamps produced as $x$, find $x$ by factorisation.
Answers
$x = 7$ or $x = -5$.
$x = -2/3$ or $x = 1/2$.
B — $x^2 - 6 x + 9 = 0$.
The numbers are $12$ and $17$ ($x = -17$ is rejected).
The integers are $24$ and $26$ ($x = -26$ is rejected).
The integers are $9$ and $10$ ($x = -10$ is rejected).
Breadth $= 12$ m, length $= 16$ m ($x = -16$ is rejected).
The legs are $7$ m and $24$ m ($x = -24$ is rejected).
$6$ lamps were produced, at ₹19 each ($x = -19/2$ is rejected).
A right triangle: legs of five and twelve centimetres, hypotenuse thirteen, with the shorter leg named x.
practice Find the discriminant of $3 x^2 + 5 x - 2 = 0$ and state the nature of its roots. (Worked in full below — read it, then do the next two the same way.)
practice Find the discriminant of $x^2 + 6 x + 9 = 0$ and state the nature of its roots. (Start the same way — write $a$, $b$ and $c$ down with their signs first.)
practice Find the value of $k$ for which $x^2 + k x + 9 = 0$ has equal roots. (Equal roots means the discriminant is zero.)
practice Find the discriminant of $2 x^2 - 3 x + 5 = 0$ and state the nature of its roots.
practice Find the discriminant of $5 x^2 - 6 x - 2 = 0$ and state the nature of its roots.
practice Solve $3 x^2 - 5 x - 2 = 0$ using the quadratic formula.
practice Find the value of $k$ for which the equation $k x^2 + 4 x + 1 = 0$ has equal roots.
practice Find the value of $p$ for which the equation $2 x^2 + p x + 8 = 0$ has real and equal roots.
practice A community wants to design a rectangular park with a perimeter of $80$ metres and an area of $500$ square metres. Taking the length as $x$ metres, check using the discriminant whether such a park is possible, before attempting to solve for $x$.
Answers
$D = 49$, which is positive, so the equation has two distinct real roots.
$D = 0$, so the equation has one repeated real root, $x = -3$.
$k = 6$ or $k = -6$.
$D = 9 - 40 = -31 < 0$; no real root.
$D = 36 + 40 = 76 > 0$; two distinct real roots (irrational, since $76$ is not a perfect square).
$x = 2$ or $x = -1/3$.
$k = 4$.
$p = 8$ or $p = -8$.
Breadth $= (40 - x)$, so $x^2 - 40 x + 500 = 0$ and $D = 1600 - 2000 = -400 < 0$: no real value of $x$ exists, so a park with these dimensions is not possible.
Exercise 4.3 — further practice
practice Find the discriminant of $x^2 - 8 x + 16 = 0$ and state the nature of its roots.
practice Find the discriminant of $4 x^2 + 3 x + 2 = 0$ and state the nature of its roots.
practice Which of these quadratic equations has two distinct real roots?
$x^2 + 2 x + 5 = 0$
$x^2 - 4 x + 4 = 0$
$2 x^2 - 3 x - 1 = 0$
$x^2 + x + 1 = 0$
practice Solve $2 x^2 - 7 x + 3 = 0$ using the quadratic formula.
practice Solve $x^2 - 4 x - 1 = 0$ using the quadratic formula.
practice Find the value of $k$ for which the equation $9 x^2 - 24 x + k = 0$ has equal roots.
practice Find the value of $n$ for which the equation $n x^2 + n x + 4 = 0$ (with $n \neq 0$) has equal roots.
practice A community wants to design a rectangular hall whose perimeter is $60$ metres and whose area is $250$ square metres. Taking the length as $x$ metres, check using the discriminant whether such a hall is possible, before attempting to solve for $x$.
Answers
$D = 0$; one repeated real root ($x = 4$).
$D = -23 < 0$; no real root.
C — $2 x^2 - 3 x - 1 = 0$.
$x = 3$ or $x = 1/2$.
$x = 2 + \sqrt{5}$ or $x = 2 - \sqrt{5}$.
$k = 16$.
$n = 16$ ($n = 0$ is rejected).
$D = -100 < 0$: no real value of $x$ exists, so a hall with these dimensions is not possible.
