Completing the square and solving quadratics

AS · P1 · 15 min

A quadratic is any expression $ax^2 + bx + c$ with $a \neq 0$ , and it is the most used piece of algebra in Paper 1. Completing the square tells you where the graph turns and what its least or greatest value is; solving tells you where it crosses the $x$ -axis. Almost every other P1 topic, from ranges of functions to tangents to circles, eventually reduces to one of these two skills, so they need to be automatic.

The language of quadratics

Definition

A quadratic polynomial in $x$ is an expression $ax^2 + bx + c$ where $a$ , $b$ , $c$ are constants and $a \neq 0$ . The numbers $a$ , $b$ , $c$ are its coefficients ( $c$ is also called the constant term). A quadratic equation is $ax^2 + bx + c = 0$ , and its solutions are called its roots.

Two roots that happen to be equal form a repeated root. For example $x^2 - 6x + 9 = (x - 3)^2 = 0$ has the repeated root $x = 3$ : the graph touches the $x$ -axis at one point instead of crossing it twice. Which of these happens is decided by the discriminant.

Completing the square

Why it works

Expand a perfect square:

(x + p)^2 = x^2 + 2px + p^2

The coefficient of $x$ is $2p$ , so $p$ is half the coefficient of $x$ . That means any $x^2 + bx$ can be rebuilt from a square:

x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2

You subtract $\left(\tfrac{b}{2}\right)^2$ because squaring the bracket creates that number, and it was not in the original expression. For example $x^2 + 6x = (x + 3)^2 - 9$ , because $(x + 3)^2 = x^2 + 6x + 9$ has a $9$ too many.

Why it is useful

In $(x + 3)^2 - 7$ the bracket is a square, so it is never negative, and it equals $0$ only when $x = -3$ . So the whole expression is at least $-7$ , with the least value $-7$ reached at $x = -3$ . Completing the square turns a quadratic into a form where its smallest (or largest) value, and where that happens, can simply be read off.

Key result

a(x + p)^2 + q

The vertex (turning point) of $y = a(x + p)^2 + q$ is $(-p,\ q)$ .
The line of symmetry is $x = -p$ .
If $a > 0$ the graph is $\cup$ -shaped and $q$ is the minimum value.
If $a < 0$ the graph is $\cap$ -shaped and $q$ is the maximum value.

The general result, for reference:

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}

Watch the sign: $(x - 3)^2 + 5$ has vertex $(3, 5)$ , not $(-3, 5)$ . The $x$ -coordinate of the vertex is the value that makes the bracket zero.

Completing the square

If $a \neq 1$ , take $a$ out of the $x^2$ and $x$ terms only: $ax^2 + bx + c = a\left(x^2 + \tfrac{b}{a}x\right) + c$ .
Inside the bracket, halve the coefficient of $x$ . Write $\left(x + \text{half}\right)^2 - \text{half}^2$ .
Multiply the subtracted number by $a$ as you remove the outer bracket.
Collect the constants.
Check by expanding mentally: the $x^2$ and $x$ coefficients must match the original.

y = 2(x - 3)^2 - 11 x = 3 (3, -11) (0, 7)

The graph above is $y = 2x^2 - 12x + 7 = 2(x - 3)^2 - 11$ : vertex $(3, -11)$ , line of symmetry $x = 3$ , $y$ -intercept $(0, 7)$ .

Completing the square when a = 1

Express $x^2 - 8x + 21$ in the form $(x + a)^2 + b$ , and hence state the coordinates of the vertex of $y = x^2 - 8x + 21$ .

Solution

Half of $-8$ is $-4$ , and $(x - 4)^2 = x^2 - 8x + 16$ :

x^2 - 8x + 21 = (x - 4)^2 - 16 + 21 = (x - 4)^2 + 5

So $a = -4$ , $b = 5$ . The bracket is zero when $x = 4$ , so the vertex is $(4, 5)$ . Because $5 > 0$ and the graph is $\cup$ -shaped, this also shows $x^2 - 8x + 21 > 0$ for every real $x$ .

Completing the square when a is not 1

Express $2x^2 - 12x + 7$ in the form $a(x + b)^2 + c$ , and state the minimum value of $2x^2 - 12x + 7$ .

Solution

Take out $2$ from the first two terms only:

\begin{aligned} 2x^2 - 12x + 7 &= 2\left(x^2 - 6x\right) + 7 \\ &= 2\left[(x - 3)^2 - 9\right] + 7 \\ &= 2(x - 3)^2 - 18 + 7 \\ &= 2(x - 3)^2 - 11 \end{aligned}

So $a = 2$ , $b = -3$ , $c = -11$ . The minimum value is $-11$ , when $x = 3$ .

Check: $2(x - 3)^2 = 2x^2 - 12x + 18$ , and $18 - 11 = 7$ . Correct.

A negative x-squared term

Express $16 - 6x - x^2$ in the form $a - (x + b)^2$ . Hence state the greatest value of $16 - 6x - x^2$ and the value of $x$ at which it occurs.

Solution

Take out $-1$ from the $x$ terms:

\begin{aligned} 16 - 6x - x^2 &= 16 - \left(x^2 + 6x\right) \\ &= 16 - \left[(x + 3)^2 - 9\right] \\ &= 25 - (x + 3)^2 \end{aligned}

So $a = 25$ , $b = 3$ . Since $(x + 3)^2 \ge 0$ , the expression is at most $25$ . The greatest value is $25$ , when $x = -3$ .

Using a completed square to find a greatest value

(a) Express $2x^2 - 12x + 23$ in the form $a(x + b)^2 + c$ .

(b) Hence find the greatest value of $\dfrac{1}{2x^2 - 12x + 23}$ and the value of $x$ at which it occurs.

Solution

(a)

2x^2 - 12x + 23 = 2\left[(x - 3)^2 - 9\right] + 23 = 2(x - 3)^2 + 5

(b) The denominator $2(x - 3)^2 + 5$ is always at least $5$ , with least value $5$ when $x = 3$ . A fraction with numerator $1$ and positive denominator is largest when the denominator is smallest. So the greatest value is

\frac{1}{5}, \quad \text{when } x = 3

The word hence is the clue: the examiner wants the completed square from (a), not calculus.

Watch out

Taking $a$ out of the constant too. $2x^2 - 12x + 7 = 2(x^2 - 6x + 7)$ is wrong, because it doubles the $7$ . Factor $a$ out of the $x^2$ and $x$ terms only, or, if you do factor it out of everything, write $2\left(x^2 - 6x + \tfrac{7}{2}\right)$ .

Forgetting to multiply the subtracted square by $a$ . In $2\left[(x - 3)^2 - 9\right] + 7$ the $-9$ becomes $-18$ , not $-9$ .

Reading the vertex sign wrongly. $(x + 3)^2 - 7$ has vertex $(-3, -7)$ .

Solving quadratic equations

There are three methods. All three give the same roots; choose the quickest.

Method	Use it when
Factorising	the roots are whole numbers or simple fractions
Completing the square	the question has just asked you to complete the square, or says hence
The quadratic formula	anything else, especially when exact (surd) answers are asked for

Factorising

Rewrite $ax^2 + bx + c$ as a product of two linear factors, then use the fact that if a product is zero, one of the factors is zero.

When $a \neq 1$ , find two numbers that multiply to $ac$ and add to $b$ , split the middle term with them, and factorise in pairs. For $6x^2 - x - 12$ : $ac = -72$ , and $-9 + 8 = -1$ , so

6x^2 - 9x + 8x - 12 = 3x(2x - 3) + 4(2x - 3) = (3x + 4)(2x - 3)

Completing the square

Once the equation is in completed square form, isolate the square and take square roots, remembering both signs.

The quadratic formula

Key result

The roots of $ax^2 + bx + c = 0$ are

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

This formula is printed in the list of formulae (MF19), but you should know it.

The formula is just completing the square done once and for all.

Deriving the formula

Divide by $a$ and complete the square:

\begin{aligned} x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \\ \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} &= 0 \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{aligned}

Exact answers and surds

"Exact" or "in surd form" means leave the square root in and simplify it. Use $\sqrt{ab} = \sqrt{a}\sqrt{b}$ to pull out square factors: $\sqrt{40} = \sqrt{4}\sqrt{10} = 2\sqrt{10}$ . Then cancel any common factor across the whole numerator and the denominator.

Factorising with a not equal to 1

Solve $6x^2 - x - 12 = 0$ .

Solution

$ac = 6 \times (-12) = -72$ . Two numbers with product $-72$ and sum $-1$ are $-9$ and $8$ .

\begin{aligned} 6x^2 - 9x + 8x - 12 &= 0 \\ 3x(2x - 3) + 4(2x - 3) &= 0 \\ (3x + 4)(2x - 3) &= 0 \end{aligned}

So $x = -\dfrac{4}{3}$ or $x = \dfrac{3}{2}$ .

Exact roots from the formula

Solve $3x^2 + 4x - 2 = 0$ , giving your answers in the form $\dfrac{p \pm \sqrt{q}}{r}$ .

Solution

$a = 3$ , $b = 4$ , $c = -2$ , so $b^2 - 4ac = 16 + 24 = 40$ .

x = \frac{-4 \pm \sqrt{40}}{6} = \frac{-4 \pm 2\sqrt{10}}{6} = \frac{-2 \pm \sqrt{10}}{3}

Every term in the numerator was divisible by $2$ , so the $2$ cancels with the $6$ .

Solving by completing the square

(a) Express $x^2 - 8x + 3$ in the form $(x - a)^2 - b$ .

(b) Hence solve $x^2 - 8x + 3 = 0$ , giving exact answers.

Solution

(a) $x^2 - 8x + 3 = (x - 4)^2 - 16 + 3 = (x - 4)^2 - 13$ .

(b)

(x - 4)^2 = 13 \quad\Rightarrow\quad x - 4 = \pm\sqrt{13} \quad\Rightarrow\quad x = 4 \pm \sqrt{13}

An equation that has to be rearranged first

Solve $\dfrac{1}{x} + \dfrac{1}{x + 2} = \dfrac{3}{4}$ .

Solution

Multiply every term by the common denominator $4x(x + 2)$ , which is non-zero for the solutions we want ( $x \neq 0$ , $x \neq -2$ ):

\begin{aligned} 4(x + 2) + 4x &= 3x(x + 2) \\ 8x + 8 &= 3x^2 + 6x \\ 0 &= 3x^2 - 2x - 8 \\ 0 &= (3x + 4)(x - 2) \end{aligned}

So $x = 2$ or $x = -\dfrac{4}{3}$ . Neither makes a denominator zero, so both are valid.

Check $x = 2$ : $\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}$ . Correct.

Roots in terms of a constant

Show that the roots of $x^2 + 2kx + k^2 - 9 = 0$ differ by $6$ for every value of the constant $k$ .

Solution

The first two terms are the start of $(x + k)^2 = x^2 + 2kx + k^2$ , so

x^2 + 2kx + k^2 - 9 = (x + k)^2 - 9

Setting this to zero, $(x + k)^2 = 9$ , so $x + k = \pm 3$ and

x = -k + 3 \quad\text{or}\quad x = -k - 3

The difference is $(-k + 3) - (-k - 3) = 6$ , which does not depend on $k$ .

Watch out

Dividing by $x$ loses a root. From $x^2 = 5x$ , dividing by $x$ gives only $x = 5$ . Instead write $x^2 - 5x = 0$ , so $x(x - 5) = 0$ , and $x = 0$ or $x = 5$ .

Forgetting $\pm$ . $(x - 4)^2 = 13$ has two solutions, $4 + \sqrt{13}$ and $4 - \sqrt{13}$ .

Cancelling only part of the numerator. $\dfrac{-4 \pm 2\sqrt{10}}{6}$ is not $\dfrac{-4 \pm \sqrt{10}}{3}$ . Divide every term of the numerator by the same number.

Not setting the equation to zero. $x(x - 3) = 10$ does not mean $x = 10$ or $x - 3 = 10$ . Expand and rearrange to $x^2 - 3x - 10 = 0$ first.

Choosing and checking

A quadratic with integer coefficients factorises nicely exactly when $b^2 - 4ac$ is a perfect square. If you cannot see the factors within a few seconds, use the formula: it never fails.

Always check roots by substituting into the original equation, or check that the sum of the roots is $-\dfrac{b}{a}$ and their product is $\dfrac{c}{a}$ . For $6x^2 - x - 12 = 0$ : $-\tfrac{4}{3} + \tfrac{3}{2} = \tfrac{1}{6} = -\tfrac{-1}{6}$ and $-\tfrac{4}{3} \times \tfrac{3}{2} = -2 = \tfrac{-12}{6}$ . Both agree.

Tip

The sum and product of roots are a quick check, not a P1 topic you will be asked to prove. Use them silently on your calculator work.

Exam tip

Command words. "Express in the form" wants the completed square with the constants identified. "Hence" means use the previous part; a different method may score nothing. "Exact" means no decimals.
Calculators. Your calculator can solve quadratics, but answers from a calculator with no working may earn no marks. Show the factorisation or the formula with numbers substituted.
Marks. For completing the square, a typical 3-mark question awards one mark for each of $a$ , $b$ and $c$ correct. A slip in one constant loses only one mark if the method is visible.
Accuracy. If decimals are acceptable, give them to 3 significant figures unless told otherwise.

Summary

Completing the square: $x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2$ .
For $a \neq 1$ , factor $a$ out of the $x^2$ and $x$ terms only, and multiply the subtracted square by $a$ .
$a(x + p)^2 + q$ has vertex $(-p, q)$ , line of symmetry $x = -p$ , and least value $q$ if $a > 0$ (greatest if $a < 0$ ).
Solve by factorising, completing the square, or $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Rearrange to $= 0$ before factorising; never divide by $x$ .
Exact answers: simplify the surd and cancel across the whole numerator.
"Hence" means use the completed square you have just found.

Practice questions

Question

Express $x^2 - 10x + 30$ in the form $(x + a)^2 + b$ and state the minimum value of $x^2 - 10x + 30$ .
Express $3x^2 + 12x - 1$ in the form $a(x + b)^2 + c$ and state the coordinates of the vertex of $y = 3x^2 + 12x - 1$ .
(a) Express $8 + 2x - x^2$ in the form $a - (x - b)^2$ . (b) Hence solve $8 + 2x - x^2 = 0$ .
Solve $5x^2 - 7x - 6 = 0$ .
Solve $2x^2 + 6x - 3 = 0$ , giving your answers in exact form.
Solve $x(2x + 1) = 3(x + 4)$ .
(a) Express $4x^2 - 12x + 13$ in the form $(2x + a)^2 + b$ . (b) Hence explain why $4x^2 - 12x + 13 > 0$ for all $x$ , and find the greatest value of $\dfrac{6}{4x^2 - 12x + 13}$ .
Solve $\dfrac{2}{x - 1} - \dfrac{1}{x + 1} = \dfrac{1}{2}$ .
A quadratic curve $y = ax^2 + bx + c$ has its vertex at $(2, -3)$ and passes through $(0, 5)$ . Find $a$ , $b$ and $c$ .
Show that, for every value of the constant $k$ , the roots of $x^2 - 2kx + k^2 - 4 = 0$ are $k + 2$ and $k - 2$ . Hence find the value of $k$ for which one root is three times the other and both are positive.

Answers

$x^2 - 10x + 30 = (x - 5)^2 - 25 + 30 = (x - 5)^2 + 5$ . So $a = -5$ , $b = 5$ , and the minimum value is $5$ (at $x = 5$ ).
$3x^2 + 12x - 1 = 3\left(x^2 + 4x\right) - 1 = 3\left[(x + 2)^2 - 4\right] - 1 = 3(x + 2)^2 - 13$ . Vertex $(-2, -13)$ .
(a) $8 + 2x - x^2 = 8 - \left(x^2 - 2x\right) = 8 - \left[(x - 1)^2 - 1\right] = 9 - (x - 1)^2$ , so $a = 9$ , $b = 1$ . (b) $(x - 1)^2 = 9$ , so $x - 1 = \pm 3$ , giving $x = 4$ or $x = -2$ .
$ac = -30$ ; $-10 + 3 = -7$ . $5x^2 - 10x + 3x - 6 = 5x(x - 2) + 3(x - 2) = (5x + 3)(x - 2) = 0$ . So $x = 2$ or $x = -\tfrac{3}{5}$ .
$b^2 - 4ac = 36 + 24 = 60$ . $x = \dfrac{-6 \pm \sqrt{60}}{4} = \dfrac{-6 \pm 2\sqrt{15}}{4} = \dfrac{-3 \pm \sqrt{15}}{2}$ .
$2x^2 + x = 3x + 12$ , so $2x^2 - 2x - 12 = 0$ , i.e. $x^2 - x - 6 = 0$ , $(x - 3)(x + 2) = 0$ . So $x = 3$ or $x = -2$ .
(a) $(2x - 3)^2 = 4x^2 - 12x + 9$ , so $4x^2 - 12x + 13 = (2x - 3)^2 + 4$ . $a = -3$ , $b = 4$ . (b) $(2x - 3)^2 \ge 0$ , so $4x^2 - 12x + 13 \ge 4 > 0$ . The least value of the denominator is $4$ (at $x = \tfrac{3}{2}$ ), so the greatest value of the fraction is $\tfrac{6}{4} = \tfrac{3}{2}$ .
Multiply by $2(x - 1)(x + 1)$ : $4(x + 1) - 2(x - 1) = (x - 1)(x + 1)$ , so $2x + 6 = x^2 - 1$ , giving $x^2 - 2x - 7 = 0$ . Then $x = \dfrac{2 \pm \sqrt{4 + 28}}{2} = \dfrac{2 \pm 4\sqrt{2}}{2} = 1 \pm 2\sqrt{2}$ . Neither is $\pm 1$ , so both are valid.
Vertex $(2, -3)$ gives $y = a(x - 2)^2 - 3$ . At $(0, 5)$ : $5 = 4a - 3$ , so $a = 2$ . Then $y = 2(x - 2)^2 - 3 = 2x^2 - 8x + 5$ . So $a = 2$ , $b = -8$ , $c = 5$ .
$x^2 - 2kx + k^2 - 4 = (x - k)^2 - 4$ . Setting this to zero, $x - k = \pm 2$ , so the roots are $k + 2$ and $k - 2$ . The larger root is $k + 2$ , so we need $k + 2 = 3(k - 2)$ , giving $k + 2 = 3k - 6$ and $k = 4$ . The roots are then $6$ and $2$ , both positive. (The other ordering, $k - 2 = 3(k + 2)$ , gives $k = -4$ and roots $-2$ and $-6$ , which are not positive.)

The language of quadratics

Completing the square

Why it works

Why it is useful

Solving quadratic equations

Factorising

Completing the square

The quadratic formula

Exact answers and surds

Choosing and checking

Practice questions

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