Completing the Square

Q: Can I use completing the square in the exam?

Yes. In the IEB exam you may solve a quadratic by any valid method, including completing the square. It is also needed for turning points and for finding the centre and radius of a circle.

Q: Why learn completing the square if the formula exists?

It shows where the formula comes from and is the quickest way to find a parabola's turning point by hand.

Q: What do I add to complete the square?

Half the coefficient of $x$, squared, after making the coefficient of $x^2$ equal to 1.

Completing the square rewrites a quadratic so the variable appears only once. It is taught in Grade 11 to solve equations and to derive the quadratic formula. It stays useful in Grade 12 for turning points and circle equations.

Solving quadratics by completing the square

Make the coefficient of $x^{2}$ equal to 1 (divide by $a$ ).
Move the constant to the right.
Add $(\frac{b}{2})^{2}$ to both sides.
Write the left side as a square and take the square root of both sides (with $\pm$ ).

Example. $x^{2} + 6 x - 7 = 0$

$x^{2} + 6 x + 9 = 7 + 9 \Rightarrow (x + 3)^{2} = 16 \Rightarrow x + 3 = \pm 4$

$x = 1 or x = - 7$

Example. $2 x^{2} - 8 x + 3 = 0$

$x^{2} - 4 x = - \frac{3}{2} \Rightarrow (x - 2)^{2} = \frac{5}{2} \Rightarrow x = 2 \pm \frac{5}{2}$

$x \approx 3, 58$ or $x \approx 0, 42$ .

Deriving the quadratic formula

Complete the square on $a x^{2} + b x + c = 0$ :

$x^{2} + \frac{b}{a} x = - \frac{c}{a}$

$x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a}$

$(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}$

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

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

This derivation can be asked as a question in Grade 11.

Finding the turning point of a parabola

Completing the square changes $y = a x^{2} + b x + c$ into $y = a (x + p)^{2} + q$ , which shows the turning point $(- p; q)$ .

Example. $y = - 2 x^{2} + 12 x - 10$

$y = - 2 (x^{2} - 6 x) - 10 = - 2 [(x - 3)^{2} - 9] - 10 = - 2 (x - 3)^{2} + 8$

Turning point $(3; 8)$ . Since $a < 0$ , this is a maximum, and the range is $y \leq 8$ .

Take out $a$ from the $x$ terms only, not from the constant.

Key terms

Completing the square: Adding $(\frac{b}{2})^{2}$ to make a perfect square trinomial.
Perfect square: An expression like $(x + 3)^{2} = x^{2} + 6 x + 9$ .
Turning point form: $y = a (x + p)^{2} + q$ .

Common mistakes

Adding $(\frac{b}{2})^{2}$ to one side only.
Forgetting to divide by $a$ first.
Forgetting $\pm$ when taking the square root.
Multiplying the subtracted square by the wrong factor when $a \neq = 1$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Solve $x^{2} + 6 x - 7 = 0$ by completing the square.

Worked solution

Hidden - try the question first

Solution

$x^{2} + 6 x + 9 = 7 + 9$
$(x + 3)^{2} = 16$
$x + 3 = \pm 4$
$x = 1$ or $x = - 7$

Question 2

5 marks total

routine

Solve $2 x^{2} - 8 x + 3 = 0$ by completing the square. Give answers to two decimal places.

Worked solution

Hidden - try the question first

Solution

$x^{2} + 6 x + 9 = 7 + 9$
$x^{2} - 4 x + 4 = - \frac{3}{2} + 4$
$(x - 2)^{2} = \frac{5}{2}$
$x = 2 \pm \frac{5}{2}$
$x \approx 3, 58$ or $x \approx 0, 42$

Question 3

6 marks total

complex

Derive the quadratic formula by completing the square on $a x^{2} + b x + c = 0$ , $a \neq = 0$ .

Worked solution

Hidden - try the question first

Solution

$x^{2} + 6 x + 9 = 7 + 9$
$x^{2} + \frac{b}{a} x + \frac{b ^{2}}{4 a ^{2}} = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a}$
$(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}$
$x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a}$
$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Question 4

8 marks total

problem-solving

Given $y = - 2 x^{2} + 12 x - 10$ .

(a) Write the equation in the form $y = a (x + p)^{2} + q$ and write down the turning point. (4)

(b) Determine the $x$ -intercepts using your answer to (a). (2)

Worked solution

Hidden - try the question first

Solution

(a) $y = - 2 (x^{2} - 6 x) - 10$
$= - 2 [(x - 3)^{2} - 9] - 10$
$= - 2 (x - 3)^{2} + 8$
Turning point $(3; 8)$
(b) $(x - 3)^{2} = 4 \Rightarrow x - 3 = \pm 2$
$x = 1$ or $x = 5$
(c) $a < 0$ , so the turning point is a maximum
$y \leq 8$

FAQs

Can I use completing the square in the exam?

Yes. In the IEB exam you may solve a quadratic by any valid method, including completing the square. It is also needed for turning points and for finding the centre and radius of a circle.

Why learn completing the square if the formula exists?

It shows where the formula comes from and is the quickest way to find a parabola's turning point by hand.

What do I add to complete the square?

Half the coefficient of $x$ , squared, after making the coefficient of $x^{2}$ equal to 1.

Solving quadratics by completing the square

Deriving the quadratic formula

Finding the turning point of a parabola

Key terms

Common mistakes

Practice questions

FAQs

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