The Quadratic Formula and Completing the Square

Completing the square (including a number or a minus in front of x²), turning points, and solving with the quadratic formula.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Edexcel Further Pure Maths 4PM1, Cambridge O Level Maths 4024, IGCSE Additional Maths 0606 · 73 real past-paper questions below

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Two Tools for Quadratics

Some quadratics do not factorise. You have two tools for them.

Completing the square

(x + p)^{2} + q

Rewrites the quadratic. Gives the turning point and exact answers.

The formula

a x^{2} + b x + c

Solves the equation when it equals 0. Best for decimal answers.

Key idea

Read the question: “write in the form …” means completing the square; “give your answers to 2 decimal places” means the formula.

Completing the Square

Completing the square means rewriting a quadratic so all the x's sit inside one squared bracket. Watch it one small move at a time.

Worked example

1
$x^{2} + 6 x + 2$ — The question. Look at the $x^{2}$ and the 6x first. The +2 waits at the end.
2
$(x + 3)^{2}$ — Halve the 6 to get 3. Why half? Because $(x + 3)^{2}$ gives 3x twice, and 3x + 3x = 6x.
3
$(x + 3)^{2} = x^{2} + 6 x + 9$ — Check by expanding. We get the $x^{2}$ and the 6x we wanted, plus an extra 9 we did not ask for.
4
$x^{2} + 6 x = (x + 3)^{2} - 9$ — So take the 9 away again. Now both sides are equal.
5
$x^{2} + 6 x + 2 = (x + 3)^{2} - 9 + 2$ — Bring back the +2 that was waiting.
6
$= (x + 3)^{2} - 7$ — −9 + 2 = −7. This is the form $(x + p)^{2} + q$ , with p = 3 and q = −7.

Memory trick

Halve it, square it, take it away.

Common mistake

Adding the 9 instead of taking it away. The bracket already makes the 9, so you must remove it.

A Number in Front of x²

The halving trick only works when $x^{2}$ stands alone. If it has a number in front, take that number out of the x terms first. Then do the same trick inside the bracket.

Worked example

1
$3 x^{2} + 12 x + 5$ — The question. $x^{2}$ has a 3 in front, so we cannot halve straight away.
2
$3 (x^{2} + 4 x) + 5$ — Take the 3 out of the first two terms only: $3 x^{2} \div 3 = x^{2}$ and $12 x \div 3 = 4 x$ . The +5 stays outside.
3
$x^{2} + 4 x = (x + 2)^{2} - 4$ — Now work inside the bracket, like the last card. Half of 4 is 2, and $(x + 2)^{2}$ makes an extra 4, so take 4 away.
4
$3 ((x + 2)^{2} - 4) + 5$ — Put that back in place of $x^{2} + 4 x$ .
5
$3 (x + 2)^{2} - 12 + 5$ — Open the big bracket. The 3 multiplies both parts, so 3 × (−4) = −12.
6
$3 (x + 2)^{2} - 7$ — −12 + 5 = −7. Done.

Key idea

Check by expanding back: $3 (x^{2} + 4 x + 4) - 7 = 3 x^{2} + 12 x + 12 - 7 = 3 x^{2} + 12 x + 5$ . It matches, so the answer is right.

Common mistake

Writing 3(x + 2)² − 4 + 5. The −4 is inside the 3's bracket, so it becomes −12.

A Negative x² (Extended/Higher)

Questions like “write $5 + 8 x - x^{2}$ in the form $a - (x - b)^{2}$ ” work the same way. The only new move: take out the minus sign first.

Worked example

1
$5 + 8 x - x^{2}$ — The question. The x² has a minus, so we cannot halve straight away.
2
$5 - (x^{2} - 8 x)$ — Take the minus out of the x terms. It flips each sign inside: $- x^{2} + 8 x$ becomes $- (x^{2} - 8 x)$ .
3
$x^{2} - 8 x = (x - 4)^{2} - 16$ — Complete the square inside. Half of −8 is −4, and $(x - 4)^{2}$ makes an extra 16, so take 16 away.
4
$5 - ((x - 4)^{2} - 16)$ — Put that back in place of $x^{2} - 8 x$ .
5
$5 + 16 - (x - 4)^{2}$ — Open the outer bracket. The minus in front turns −16 into +16.
6
$21 - (x - 4)^{2}$ — 5 + 16 = 21. So a = 21 and b = 4.

Common mistake

Writing 5 − 16 = −11. The minus outside the bracket flips the −16 to +16.

With −3 in front. Exams also write $a - b (x - c)^{2}$ , with a number bigger than 1 in front of the minus. Take out the whole −3.

Worked example

1
$7 + 12 x - 3 x^{2}$ — The question. The x² has −3 in front. Aim for the form $a - b (x - c)^{2}$ .
2
$7 - 3 (x^{2} - 4 x)$ — Take −3 out of the x terms only: $- 3 x^{2} \div (- 3) = x^{2}$ and $12 x \div (- 3) = - 4 x$ . The 7 stays outside.
3
$x^{2} - 4 x = (x - 2)^{2} - 4$ — Complete the square inside. Half of −4 is −2, and $(x - 2)^{2}$ makes an extra 4, so take 4 away.
4
$7 - 3 ((x - 2)^{2} - 4)$ — Put that back in place of $x^{2} - 4 x$ .
5
$7 + 12 - 3 (x - 2)^{2}$ — Open the big bracket. The −3 multiplies the −4: (−3) × (−4) = +12.
6
$19 - 3 (x - 2)^{2}$ — 7 + 12 = 19. So a = 19, b = 3 and c = 2.

Common mistake

Writing 7 − 12 = −5. It is −3 times −4, and minus times minus is plus, so +12.

Shortcut: One Formula for Any Quadratic

If you do the last two cards with letters instead of numbers, you get one formula that does the whole job for any $a x^{2} + b x + c$ :

Formula

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

Read $a$ , $b$ and $c$ from the question, with their signs. Then work out two numbers: $\frac{b}{2 a}$ goes inside the bracket, and $c - \frac{b ^{2}}{4 a}$ goes on the end.

Common mistake

Watch the letters. Here a, b and c are the numbers in the question. Exam questions often use a, b and c for the answer instead, e.g. “write in the form a(x + b)² + c”. For $3 x^{2} + 12 x + 5 = 3 (x + 2)^{2} - 7$ that answer's a, b, c are 3, 2 and −7, not 3, 12 and 5. Read the letters from the question.

Worked example

1
$3 x^{2} + 12 x + 5$ — The example from two cards ago. a = 3, b = 12, c = 5.
2
$\frac{b}{2 a} = \frac{12}{6} = 2$ — The number inside the bracket: x + 2.
3
$c - \frac{b ^{2}}{4 a} = 5 - \frac{144}{12} = 5 - 12 = - 7$ — The number on the end. Square b first: $1 2^{2} = 144$ .
4
$3 (x + 2)^{2} - 7$ — Put a, the bracket and the end number together. Same answer as before.

Worked example

1
$- x^{2} + 8 x + 5$ — The last card's 5 + 8x − x², with x² written first. a = −1, b = 8, c = 5. Keep the minus on a.
2
$\frac{b}{2 a} = \frac{8}{- 2} = - 4$ — So the bracket is x − 4.
3
$c - \frac{b ^{2}}{4 a} = 5 - \frac{64}{- 4} = 5 + 16 = 21$ — Dividing by a negative flips the sign: $\frac{64}{- 4} = - 16$ , and taking away −16 adds 16.
4
$- (x - 4)^{2} + 21$ — The same as $21 - (x - 4)^{2}$ from the last card.

Key idea

It also gives the turning point straight away: $(- \frac{b}{2 a}, c - \frac{b ^{2}}{4 a})$ . For $3 x^{2} + 12 x + 5$ that is (−2, −7).

Common mistake

Forgetting the minus when a is negative. Always write a with its sign before you divide.

A letter in front. The formula still works when a coefficient is a letter, like q. Treat q as a number and keep it in the answer.

Worked example

1
$q + 12 x - q x^{2}$ — The question. Written as $- q x^{2} + 12 x + q$ , it has a = −q, b = 12, c = q.
2
$\frac{b}{2 a} = \frac{12}{- 2 q} = - \frac{6}{q}$ — The number inside the bracket, so the bracket is $x - \frac{6}{q}$ . Dividing by −2q makes it negative.
3
$\frac{b ^{2}}{4 a} = \frac{144}{- 4 q} = - \frac{36}{q}$ — Work out the fraction on the end first: $1 2^{2} = 144$ , and dividing by −4q makes it −36/q.
4
$c - \frac{b ^{2}}{4 a} = q - (- \frac{36}{q}) = q + \frac{36}{q}$ — Taking away a negative adds it.
5
$q + \frac{36}{q} - q (x - \frac{6}{q})^{2}$ — Put them together: the end number, then a = −q times the bracket squared.

When Half Is a Fraction

When the x number is odd, half of it is a fraction. The method is the same; keep the fraction exact.

Worked example

1
$x^{2} + 5 x + 1$ — The question. 5 is odd, so half of it is not a whole number.
2
$(x + \frac{5}{2})^{2}$ — Half of 5 is $\frac{5}{2}$ . Keep it as a fraction, not 2.5, so the next line stays exact.
3
$(\frac{5}{2})^{2} = \frac{25}{4}$ — The bracket makes an extra 25/4. Square the top and the bottom.
4
$(x + \frac{5}{2})^{2} - \frac{25}{4} + 1$ — Take the extra away, and bring back the +1.
5
$- \frac{25}{4} + \frac{4}{4} = - \frac{21}{4}$ — Write 1 as 4/4 so the bottoms match.
6
$(x + \frac{5}{2})^{2} - \frac{21}{4}$ — Done. p = 5/2 and q = −21/4.

Key idea

A surd works the same way. For $x^{2} + 62 x - 1$ , half of $62$ is $32$ , and $(32)^{2} = 9 \times 2 = 18$ . So it is $(x + 32)^{2} - 19$ .

Common mistake

Squaring 5/2 as 25/2. Square the bottom too: 2² = 4, so it is 25/4.

Using the Completed Square

Why bother completing the square? Because the finished form answers two more exam questions almost for free.

1. Find the turning point of $y = (x + 3)^{2} - 7$

Worked example

1
$(x + 3)^{2} \geq 0$ — A squared number is never negative. The smallest it can ever be is 0.
2
$x + 3 = 0, so x = - 3$ — The bracket is 0 when x = −3.
3
$y = 0 - 7 = - 7$ — Put x = −3 in. The bracket disappears, leaving y = −7.
4
$turning point (- 3, - 7)$ — This is the lowest point of the curve, so the least value is −7. Notice: the 3 changes sign, the −7 does not.

Key idea

If the x² is negative, the curve is upside down, so the turning point is the highest point. For $21 - (x - 4)^{2}$ the turning point is (4, 21), and 21 is the greatest value.

2. Solve $(x + 3)^{2} - 7 = 0$

Worked example

1
$(x + 3)^{2} = 7$ — Add 7 to both sides, so the squared bracket is alone.
2
$x + 3 = \pm 7$ — Square root both sides. Both $7$ and $- 7$ square to 7, so keep the ±.
3
$x = - 3 \pm 7$ — Take 3 from both sides. This is the exact answer, and ± means two solutions.
4
$x = - 0.354 or x = - 5.65$ — Only if the question asks for decimals: −3 + 2.646 and −3 − 2.646, to 3 s.f.

Key idea

One completed square, three answers: the turning point, the least value, and the solutions.

3. “Hence solve” with a bracket in place of x

Worked example

1
$(y + 3)^{2} + 6 (y + 3) + 2 = 0$ — The question. It is $x^{2} + 6 x + 2 = 0$ with (y + 3) wherever x was.
2
$x = y + 3$ — So call the bracket x. Now it is the equation we just solved.
3
$y + 3 = - 3 \pm 7$ — We know $x = - 3 \pm 7$ , so y + 3 equals those answers.
4
$y = - 6 \pm 7$ — Take 3 from both sides: −3 − 3 = −6.
5
$y = - 3.35 or y = - 8.65$ — Only if asked for decimals: −6 + 2.646 and −6 − 2.646, to 3 s.f.

Common mistake

Expanding the brackets. “Hence” means use the answer you already have: spot the repeated bracket and shift the answers.

The Quadratic Formula

When a quadratic equals 0 and will not factorise, this formula always works:

Formula

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

$a$ is the number in front of $x^{2}$ , $b$ the number in front of $x$ , and $c$ the number on its own. The ± means you work it out twice: once with +, once with −.

Worked example

1
$2 x^{2} + 5 x - 4 = 0$ — The question. It already equals 0. Read off a = 2, b = 5, c = −4, and keep the minus on c.
2
$b^{2} - 4 a c = 5^{2} - 4 \times 2 \times (- 4)$ — Work out the part under the square root first. Put c in brackets so its minus is not lost.
3
$= 25 + 32 = 57$ — −4 × 2 × (−4) = +32, because minus times minus is plus.
4
$x = \frac{- 5 \pm 57}{2 \times 2} = \frac{- 5 \pm 57}{4}$ — Put it all in the formula: −b is −5, and the bottom is 2a = 4. Keep √57 exact for now.
5
$x = \frac{- 5 + 57}{4} = 0.637$ — The + version: √57 = 7.5498…, so (−5 + 7.5498) ÷ 4 = 0.637 to 3 s.f.
6
$x = \frac{- 5 - 57}{4} = - 3.14$ — The − version: (−5 − 7.5498) ÷ 4 = −3.137…, which is −3.14 to 3 s.f.

Common mistake

Losing the minus in c. Put c in brackets: −4 × 2 × (−4).

When b is negative. Exams often give the x term a minus. Here is how to handle it.

Worked example

1
$2 x^{2} - 7 x - 3 = 0$ — The question. Read off a = 2, b = −7, c = −3. This time b has a minus too.
2
$- b = - (- 7) = 7$ — The top of the formula starts with −b. Minus a minus is plus, so −b = 7.
3
$b^{2} = (- 7)^{2} = 49$ — A negative squared is positive. Keep the brackets.
4
$b^{2} - 4 a c = 49 - 4 \times 2 \times (- 3)$ — Put c in brackets, as before.
5
$= 49 + 24 = 73$ — −4 × 2 × (−3) = +24, because minus times minus is plus.
6
$x = \frac{7 \pm 73}{2 \times 2} = \frac{7 \pm 73}{4}$ — Put it all in: 7 on top, 2a = 4 on the bottom.
7
$x = \frac{7 + 73}{4} = 3.89$ — The + version: √73 = 8.544…, so (7 + 8.544) ÷ 4 = 3.886…, which is 3.89 to 2 d.p.
8
$x = \frac{7 - 73}{4} = - 0.39$ — The − version: (7 − 8.544) ÷ 4 = −0.386…, which is −0.39 to 2 d.p.

Common mistake

Typing −7² on a calculator. It gives −49, because the calculator squares 7 first. Type (−7)², which is 49.

Key idea

Write the formula with your numbers in before you press anything. That line earns marks even if a later slip spoils the answer.

Key idea

In 628, the Indian mathematician Brahmagupta wrote a rule, in words, for solving quadratic equations. It follows the same steps as the formula you use today, about 1,000 years before the letters and symbols existed.

Which Method? Read the Question

The wording tells you which tool.

You see	You do
“In the form (x + p)² + q”	Complete the square
“In the form a − (x − b)²”	Take out the minus, then complete
“Turning point” or “least value”	Complete the square, read it off
“To 2 decimal places”	Use the formula
“In the form p ± √q” or “exact”	Complete the square, then solve. Keep the √
It factorises	See Solving Quadratics by Factorising

Key idea

On a formula question, write a, b and c down first, with their signs.

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