Quadratics

Syllabus coverage

Carry out the process of completing the square for a quadratic polynomial $a x^{2} + b x + c$ and use a completed square form
Find the discriminant of a quadratic polynomial $a x^{2} + b x + c$ and use it, for example to locate the vertex of the graph of $y = a x^{2} + b x + c$ , to sketch the graph, or to determine the number of real roots of $a x^{2} + b x + c = 0$
Know and use the term 'repeated root'
Solve quadratic equations, and quadratic inequalities, in one unknown - by factorising, by completing the square and by using the formula
Solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic
Recognise and solve equations in $x$ which are quadratic in some function of $x$ , for example $x^{4} - 5 x^{2} + 4 = 0$ or $\tan^{2} x = 1 + \tan x$

Part of the AS & A Level Maths syllabus 9709.

The curve

y = x^{2} - 2 x - 3

, with the roots at

(- 1, 0)

and

(3, 0)

, the y-intercept at

(0, - 3)

and the minimum at

(1, - 4)

. The factorised form

y = (x + 1) (x - 3)

gives the roots; the completed square

y = (x - 1)^{2} - 4

gives the turning point.

The three forms of a quadratic, and what each one tells you

A quadratic is any expression of the form $a x^{2} + b x + c$ with $a \neq 0$ . If $a$ were zero the $x^{2}$ term would vanish and you would be looking at a straight line, so every method on this page assumes $a \neq 0$ . The graph of a quadratic is a parabola: a U shape opening upwards when $a > 0$ , and an upside-down U when $a < 0$ .

Cambridge asks you to move between three ways of writing the same quadratic, and each one is useful for a different job. The general form $a x^{2} + b x + c$ gives you the y-intercept immediately, because putting $x = 0$ leaves $y = c$ . The factorised form $a (x - p) (x - q)$ gives you the roots: the curve crosses the x-axis at $x = p$ and $x = q$ . The completed square form $a (x + p)^{2} + q$ gives you the turning point, which sits at $(- p, q)$ .

A question almost always tells you which form it wants. 'Express in the form $(x + p)^{2} + q$ ' wants the completed square; 'factorise' wants the product of brackets; 'find the values of $x$ for which the curve meets the x-axis' wants the roots, which you can get from either of the other two forms.

Form	Written as	What it tells you at a glance
General	$y = a x^{2} + b x + c$	the y-intercept $c$ , and the direction of the curve from the sign of $a$
Factorised	$y = a (x - p) (x - q)$	the roots $x = p$ and $x = q$ , where the curve meets the x-axis
Completed square	$y = a (x + p)^{2} + q$	the turning point $(- p, q)$ , the line of symmetry $x = - p$ , and the least or greatest value $q$

If a question says 'express in the form

(x + p)^{2} + q

', that form is the answer. Expanding it back to

x^{2} + b x + c

at the end throws away the marks you have just earned.

Completing the square

Completing the square rewrites $x^{2} + b x + c$ as $(x + \frac{b}{2})^{2} + c - \frac{b^{2}}{4}$ . You do not need to memorise that line; you need to be able to build the square that matches the first two terms and then put the constant right. For $x^{2} + 6 x + 5$ : half of 6 is 3, so start with $(x + 3)^{2} = x^{2} + 6 x + 9$ , which is 4 too big, so $x^{2} + 6 x + 5 = (x + 3)^{2} - 4$ .

When the coefficient of $x^{2}$ is not 1, take that coefficient out of the first two terms first and keep it outside the bracket. For $2 x^{2} - 12 x + 11$ : $2 x^{2} - 12 x + 11 = 2 (x^{2} - 6 x) + 11 = 2 [(x - 3)^{2} - 9] + 11 = 2 (x - 3)^{2} - 18 + 11 = 2 (x - 3)^{2} - 7$ . The number you subtract inside the bracket is multiplied by 2 when the bracket is expanded, which is why it becomes 18 rather than 9.

If the coefficient is negative, take out $- 1$ and treat the bracket as usual. For $5 - 2 x - x^{2}$ : $5 - 2 x - x^{2} = - (x^{2} + 2 x) + 5 = - [(x + 1)^{2} - 1] + 5 = - (x + 1)^{2} + 6$ . The square is now subtracted, so the turning point of this curve is a maximum.

$x^{2} + 6 x + 5 = (x + 3)^{2} - 4$
$x^{2} - 10 x + 3 = (x - 5)^{2} - 22$
$3 x^{2} + 6 x + 1 = 3 (x + 1)^{2} - 2$
$5 - 2 x - x^{2} = - (x + 1)^{2} + 6$

Check every completed square by expanding it. It takes ten seconds, and it catches the commonest error in the topic, which is subtracting

b^{2}

instead of

\frac{b^{2}}{4}

The completed square gives you the turning point

Once a quadratic is written as $a (x + p)^{2} + q$ you can read off its whole shape without drawing it. A square is never negative, so $(x + p)^{2} \geq 0$ for every value of $x$ , and it is zero only when $x = - p$ . If $a > 0$ , the smallest value of the expression is therefore $q$ , reached at $x = - p$ , and $(- p, q)$ is a minimum. If $a < 0$ , the same point is a maximum, because the negative $a$ turns the square upside down.

This is how you answer questions that ask for the least or greatest value of something, and questions that ask you to show a quadratic is always positive. Since $x^{2} - 4 x + 7 = (x - 2)^{2} + 3$ and $(x - 2)^{2} \geq 0$ , the expression is at least 3, so $x^{2} - 4 x + 7 > 0$ for every $x$ .

The line of symmetry of the parabola is $x = - p$ , which is the same line as $x = - \frac{b}{2 a}$ in the general form. That is where the turning point's x-coordinate comes from, and it is also the midpoint of the two roots whenever there are two of them.

$x^{2} - 4 x + 7 = (x - 2)^{2} + 3$ : least value 3, at $x = 2$
$5 - 2 x - x^{2} = - (x + 1)^{2} + 6$ : greatest value 6, at $x = - 1$
$2 x^{2} + 8 x + 11 = 2 (x + 2)^{2} + 3$ : least value 3, at $x = - 2$

The discriminant decides how many roots there are

For the equation $a x^{2} + b x + c = 0$ the quantity $b^{2} - 4 a c$ is called the discriminant. It is the expression under the square root in the quadratic formula, so its sign decides whether the formula hands you two numbers, one number, or no real number at all.

If $b^{2} - 4 a c > 0$ there are two distinct real roots and the curve cuts the x-axis twice. If $b^{2} - 4 a c = 0$ there is exactly one real root, called a repeated root, and the curve touches the x-axis at its turning point. If $b^{2} - 4 a c < 0$ there are no real roots and the curve lies entirely above or entirely below the axis.

Questions usually use the discriminant the other way round: they give you an equation with an unknown constant and tell you how many roots it has. For $x^{2} + k x + 9 = 0$ to have a repeated root, $k^{2} - 36 = 0$ , so $k = 6$ or $k = - 6$ . If you were told instead that it has two distinct real roots, you would solve the inequality $k^{2} - 36 > 0$ , which is the kind of work the next block covers.

Discriminant	Number of real roots	What the graph does
$b^{2} - 4 a c > 0$	two distinct real roots	cuts the x-axis at two points
$b^{2} - 4 a c = 0$	one repeated root	touches the x-axis at its turning point
$b^{2} - 4 a c < 0$	no real roots	lies wholly above or wholly below the x-axis

A zero discriminant and a perfect square are the same statement.

b^{2} - 4 a c = 0

happens exactly when the quadratic is

a (x - r)^{2}

, and then the

q

in the completed square form is zero.

Solving a quadratic equation: three methods, one decision

Factorising is the fastest method when the numbers are friendly, and it is worth trying first. For $3 x^{2} - 5 x - 2 = 0$ the product $3 \times (- 2) = - 6$ needs two factors that add to $- 5$ , namely $- 6$ and $1$ , so the middle term splits: $3 x^{2} - 6 x + x - 2 = 3 x (x - 2) + (x - 2) = (3 x + 1) (x - 2)$ . Setting each factor to zero gives $x = 2$ or $x = - \frac{1}{3}$ .

Completing the square solves the same equation without any guessing, and it is the method to use when a question has already put you in that form. From $(x - 3)^{2} = 11$ you get $x = 3 \pm \sqrt{11}$ , and the answer is exact until somebody asks you to round it.

The formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ always works, and it is the right choice when the answer will not be a neat fraction. Read it as a whole: the entire numerator $- b \pm \sqrt{b^{2} - 4 a c}$ is divided by $2 a$ , and $b$ keeps its own sign. If a question says 'giving your answers correct to 3 significant figures', the formula is almost certainly what it wants.

If the discriminant comes out negative, stop and say so. An equation with no real roots has no solution to give, and inventing one by ignoring the minus sign under the square root is an expensive error.

Quadratic inequalities

A quadratic inequality such as $x^{2} - 3 x - 10 > 0$ is solved in two steps: solve the equation $x^{2} - 3 x - 10 = 0$ to find the critical values, then decide which side of those values the inequality selects. Factorising gives $(x - 5) (x + 2) = 0$ , so the critical values are $x = 5$ and $x = - 2$ .

Now picture the parabola $y = (x - 5) (x + 2)$ . It is a U shape crossing the axis at $- 2$ and $5$ , so it is below the axis between those values and above the axis outside them. The inequality asks where the curve is above the axis, so the answer is $x < - 2$ or $x > 5$ . Had the inequality been $x^{2} - 3 x - 10 < 0$ , the answer would have been the inside interval $- 2 < x < 5$ .

Two details decide marks. First, if the coefficient of $x^{2}$ is negative, multiply the whole inequality by $- 1$ and reverse the inequality sign before you start, so that you are always reasoning about a U-shaped curve. Second, never write $- 2 < x < 5$ for the answer $x < - 2$ or $x > 5$ : the two describe completely different sets of numbers, and the inside interval is the answer to the opposite inequality.

Inequality, with $a > 0$	Solution
$(x - p) (x - q) > 0$ with $p < q$	$x < p$ or $x > q$
$(x - p) (x - q) < 0$ with $p < q$	$p < x < q$
$x^{2} < 9$	$- 3 < x < 3$
$x^{2} \geq 9$	$x \leq - 3$ or $x \geq 3$

If you are unsure which side to take, test a number. Put

x = 0

into

x^{2} - 3 x - 10 > 0

and you get

- 10 > 0

, which is false, so 0 is not in the solution. Since 0 lies between

- 2

and

5

, the solution must be the outside.

Simultaneous equations: one linear, one quadratic

When one equation is linear and the other is quadratic, the linear equation is your tool for removing a variable. Rearrange it to give $y$ in terms of $x$ (or $x$ in terms of $y$ , whichever is easier) and substitute that expression into the quadratic equation. What is left is an ordinary quadratic in one variable.

For $y = x + 1$ and $x^{2} + y^{2} = 25$ : substituting gives $x^{2} + (x + 1)^{2} = 25$ , so $2 x^{2} + 2 x + 1 = 25$ and therefore $x^{2} + x - 12 = 0$ . Factorising gives $(x + 4) (x - 3) = 0$ , so $x = 3$ or $x = - 4$ . Each of those needs its own value of $y$ : from $y = x + 1$ , $x = 3$ gives $y = 4$ , and $x = - 4$ gives $y = - 3$ .

The two answers are the two points where the line meets the curve, so they must be given as pairs. The discriminant tells you how many pairs to expect: two distinct real roots of the substituted quadratic means the line cuts the curve twice, a repeated root means the line is a tangent to the curve, and no real roots means the line misses it altogether.

Never mix the two solutions up. Writing 'x = 3, y = -3' takes correct work and turns it into a wrong answer, because

(3, - 3)

is not on the line

y = x + 1

. Pair each x with the y that came from it.

Equations that are quadratic in some function of x

Some equations look far worse than they are. If you can see a repeated block - $x^{2}$ inside an $x^{4}$ , $\sqrt{x}$ inside an $x$ , $2^{x}$ inside a $2^{2 x}$ , or $\tan x$ inside a $\tan^{2} x$ - then a substitution turns the equation into an ordinary quadratic. A question usually hands you the substitution; if it does not, choose one yourself and write it down.

Take $x^{4} - 13 x^{2} + 36 = 0$ . Put $u = x^{2}$ . Then $x^{4} = u^{2}$ and the equation becomes $u^{2} - 13 u + 36 = 0$ , which factorises as $(u - 4) (u - 9) = 0$ , so $u = 4$ or $u = 9$ . Now go back: $x^{2} = 4$ gives $x = \pm 2$ , and $x^{2} = 9$ gives $x = \pm 3$ . Four solutions, all real.

The substitution can carry a condition, and that is where marks are won and lost. If $u = \sqrt{x}$ then $u \geq 0$ , so a negative root of the quadratic in $u$ has to be rejected. If $u = 2^{x}$ then $u > 0$ for the same reason. Solve $x - 6 \sqrt{x} + 8 = 0$ : with $u = \sqrt{x}$ , $u^{2} - 6 u + 8 = 0$ , so $u = 2$ or $u = 4$ ; both are positive, so $x = 4$ and $x = 16$ both count. Solve $2^{2 x} - 9 (2^{x}) + 8 = 0$ : with $u = 2^{x}$ , $u^{2} - 9 u + 8 = 0$ gives $u = 1$ or $u = 8$ , so $2^{x} = 1$ gives $x = 0$ and $2^{x} = 8$ gives $x = 3$ .

Do not stop at the value of

u

. The question asked for

x

, so the last line must convert back - and it must give every value of

x

, which often means two answers from a single root of the quadratic, because

x^{2} = 4

has two solutions.

Key terms you must be able to define

Quadratic: An expression of the form $a x^{2} + b x + c$ with $a \neq 0$ . Its graph is a parabola.

Parabola: The U-shaped curve that is the graph of a quadratic. It opens upwards when $a > 0$ and downwards when $a < 0$ .

Root of an equation: A value of $x$ that satisfies the equation. The roots of $a x^{2} + b x + c = 0$ are where the curve crosses the x-axis.

Completed square form: A quadratic written as $a (x + p)^{2} + q$ , which shows the turning point $(- p, q)$ and the line of symmetry $x = - p$ .

Turning point: The minimum or maximum point of a parabola. It always has x-coordinate $- \frac{b}{2 a}$ .

Line of symmetry: The vertical line $x = - \frac{b}{2 a}$ through the turning point, about which the two halves of the parabola are mirror images.

Discriminant: The quantity $b^{2} - 4 a c$ for $a x^{2} + b x + c = 0$ . Its sign tells you how many real roots the equation has.

Repeated root: The single root that occurs when $b^{2} - 4 a c = 0$ . The curve touches the x-axis, and the quadratic is a perfect square.

Critical values: The values of $x$ where a quadratic expression equals zero. They split the number line into the intervals a quadratic inequality has to choose between.

Factorising: Writing an expression as a product of factors, for example $x^{2} - 7 x + 12 = (x - 3) (x - 4)$ .

Quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , which solves $a x^{2} + b x + c = 0$ for every $a \neq 0$ .

Real roots: Roots that are real numbers. An equation has no real roots when its discriminant is negative.

Simultaneous equations: Two equations in two unknowns, solved by using one equation to eliminate a variable from the other.

Substitution: Replacing a variable, or an expression in a variable, by another expression. It is the method that solves a linear-and-quadratic pair.

Quadratic in disguise: An equation such as $x^{4} - 13 x^{2} + 36 = 0$ that becomes a quadratic after a substitution such as $u = x^{2}$ .

Minimum value: The least value a quadratic expression takes. It is $q$ in the form $a (x + p)^{2} + q$ when $a > 0$ .

Maximum value: The greatest value a quadratic expression takes. It is $q$ in the form $a (x + p)^{2} + q$ when $a < 0$ .

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Greatest height

A ball is kicked and its height in metres after $t$ seconds is $h = 20 t - 5 t^{2}$ . Completing the square gives $h = - 5 (t^{2} - 4 t) = - 5 [(t - 2)^{2} - 4] = - 5 (t - 2)^{2} + 20$ , so the greatest height is 20 m, reached after 2 seconds.

Tangency

The line $y = 2 x + k$ touches the circle $x^{2} + y^{2} = 5$ . Substituting gives $x^{2} + (2 x + k)^{2} = 5$ , so $5 x^{2} + 4 k x + k^{2} - 5 = 0$ . Touching means a repeated root, so the discriminant is zero: $16 k^{2} - 20 (k^{2} - 5) = 0$ , which gives $k^{2} = 25$ , so $k = 5$ or $k = - 5$ .

Profit inequality

A workshop makes a profit of $P = - x^{2} + 60 x - 500$ pounds on a day when it produces $x$ items. It makes at least 300 pounds when $- x^{2} + 60 x - 500 \geq 300$ , that is $x^{2} - 60 x + 800 \leq 0$ , which factorises as $(x - 20) (x - 40) \leq 0$ . So the daily output must be between 20 and 40 items.

Perimeter and area

A rectangle has perimeter 26 cm and area 36 cm squared. From $2 x + 2 y = 26$ , $y = 13 - x$ , and $x y = 36$ becomes $x (13 - x) = 36$ , so $x^{2} - 13 x + 36 = 0$ . Factorising gives $(x - 4) (x - 9) = 0$ , so the sides are 4 cm and 9 cm.

Showing a quadratic is always positive

To show that $x^{2} - 6 x + 11$ is positive for every value of $x$ , complete the square: $x^{2} - 6 x + 11 = (x - 3)^{2} + 2$ . A square is never negative, so the expression is at least 2, and therefore greater than 0 for every $x$ .

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What does the completed square form

a (x + p)^{2} + q

tell you?

The turning point is at

(- p, q)

, and the line of symmetry is

x = - p

Q2Complete the square:

x^{2} - 8 x + 3

(x - 4)^{2} - 13

. Half of

- 8

- 4

, and

(- 4)^{2} = 16

is taken off the constant:

3 - 16 = - 13

Q3What is the discriminant of

a x^{2} + b x + c

b^{2} - 4 a c

. It is the expression under the square root in the quadratic formula.

b^{2} - 4 a c < 0

: how many real roots are there?

None. The curve lies wholly above or wholly below the x-axis.

Q5What is a repeated root?

The single root that occurs when

b^{2} - 4 a c = 0

. The curve touches the x-axis at its turning point.

Q6Solve

x^{2} - 3 x - 10 > 0

x < - 2

x > 5

. The parabola is above the axis outside its roots.

Q7Solve

x^{2} < 9

- 3 < x < 3

. The parabola is below the axis between its roots.

Q8How do you solve a pair of simultaneous equations when one is linear and one is quadratic?

Rearrange the linear equation to give one variable in terms of the other, substitute into the quadratic, solve it, then find the matching value for each solution.

Q9How many times can a line meet a parabola?

Twice, once (the line is a tangent), or not at all - decided by the sign of the discriminant of the substituted quadratic.

Q10Which substitution solves

x^{4} - 13 x^{2} + 36 = 0

u = x^{2}

. Then

u^{2} - 13 u + 36 = 0

, so

u = 4

u = 9

, giving

x = \pm 2

x = \pm 3

Q11What condition comes with the substitution

u = \sqrt{x}

u \geq 0

, so a negative root of the quadratic in

u

must be rejected.

Q12What is the least value of

x^{2} + 4 x + 7

, and where does it occur?

3, at

x = - 2

, because

x^{2} + 4 x + 7 = (x + 2)^{2} + 3

and a square is never negative.

Q13Write down the quadratic formula.

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

for

a x^{2} + b x + c = 0

. The whole numerator is divided by

2 a

Q14How do you show that a quadratic is always positive?

Complete the square and show that the constant outside the bracket is positive. For

x^{2} - 6 x + 11

(x - 3)^{2} + 2

, and 2 is positive.

Q15What is the line of symmetry of

y = a x^{2} + b x + c

x = - \frac{b}{2 a}

, the vertical line through the turning point. It is also the midpoint of the two roots.

0 of 15 marked as known

Exam technique for this topic

Read the form the question asks for before you start. 'Express in the form $(x + p)^{2} + q$ ' and 'express in the form $a (x + p)^{2} + q$ ' both want the completed square left as it is, not expanded back out.
Learn the discriminant wording. 'Two distinct real roots' means $b^{2} - 4 a c > 0$ , 'a repeated root' or 'the curve touches the axis' means $b^{2} - 4 a c = 0$ , and 'no real roots' means $b^{2} - 4 a c < 0$ . Write $b^{2} - 4 a c$ down first and substitute $a$ , $b$ and $c$ with their signs.
For an inequality, sketch the parabola and mark the critical values on the x-axis. The sketch takes ten seconds and it stops you writing the inside interval when the question wants the outside.
In a simultaneous-equations question, finish with the points as pairs and check each pair in the linear equation. An answer written as two x-values and two y-values with no pairing scores the method marks and loses the accuracy mark.
When you substitute $u = \sqrt{x}$ or $u = 2^{x}$ , write the condition $u \geq 0$ or $u > 0$ beside the substitution. It shows why a root is rejected, and rejecting the impossible root is often a mark on its own.
A 'Show (that)' question has no answer box and no automatic mark: every line is ticked by you, so write the completing-the-square or substitution steps in full rather than jumping to the given result.

Why students lose marks here

A student wrote

(x + 3)^{2} = x^{2} + 9

Why it loses marksSquaring a bracket is not the same as squaring each term.

(x + 3)^{2} = x^{2} + 6 x + 9

, and that missing

6 x

is the whole reason completing the square works.

The answer that scores

(x + 3)^{2} = x^{2} + 6 x + 9

. Expand whenever you write a completed square, and the error cannot survive.

A student wrote

x^{2} + 6 x + 1 = (x + 3)^{2} + 1

Why it loses marksThe bracket contributes 9, not 0. Half of 6 is 3, and

3^{2} = 9

has to be taken back off the constant.

The answer that scores

x^{2} + 6 x + 1 = (x + 3)^{2} - 9 + 1 = (x + 3)^{2} - 8

A student wrote

x^{2} - 3 x - 10 > 0

, so

- 2 < x < 5

Why it loses marksThat is the solution of the opposite inequality. The parabola is above the axis outside its roots, so

> 0

selects the outside.

The answer that scores

x < - 2

x > 5

. Test

x = 0

0 - 0 - 10 = - 10 > 0

is false, and 0 lies between

- 2

and

5

, so the middle is not in the solution.

A student wrote

x^{2} + k x + 9 = 0

has a repeated root when

k^{2} - 9 = 0

, so

k = 3

k = - 3

Why it loses marksThe discriminant is

b^{2} - 4 a c

, and both the 4 and the

c

matter. With

a = 1

and

c = 9

the discriminant is

k^{2} - 36

, not

k^{2} - 9

The answer that scores

k^{2} - 4 (1) (9) = k^{2} - 36 = 0

, so

k = 6

k = - 6

A student wrote

x - 6 \sqrt{x} + 8 = 0

gives

\sqrt{x} = 2

\sqrt{x} = 4

, so

x = 2

x = 4

Why it loses marksThe substitution solves for

\sqrt{x}

, not for

x

. Each value has to be squared, and both are positive so neither is rejected.

The answer that scores

\sqrt{x} = 2

gives

x = 4

, and

\sqrt{x} = 4

gives

x = 16

A student wrote

y = x + 1

with

x^{2} + y^{2} = 25

gives

x = 3

x = - 4

y = 4

y = - 3

Why it loses marksThe four numbers are right but they are not paired, and the pairs are what the question asks for. The point

(3, - 3)

is on neither the line nor the circle.

The answer that scoresThe solutions are

(3, 4)

and

(- 4, - 3)

Syllabus coverage

The three forms of a quadratic, and what each one tells you

Completing the square

The completed square gives you the turning point

The discriminant decides how many roots there are

Solving a quadratic equation: three methods, one decision

Quadratic inequalities

Simultaneous equations: one linear, one quadratic

Equations that are quadratic in some function of x

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here