Extended. Rewrite a quadratic as a squared bracket plus a number. The completed square form solves equations exactly, and it shows the turning point of the graph at a glance.
Halve the coefficient of \(x\): \(x^{2} + bx = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}\). If \(a \ne 1\), take \(a\) out of the \(x^{2}\) and \(x\) terms first. The graph of \(y = (x + p)^{2} + q\) has its minimum point at \((-p,\ q)\).
Fig. 1: x² + 6x is a square of side x with two strips 3 by x. Moved into an L, they almost make a square of side (x + 3), but a 3 by 3 corner is missing, so x² + 6x = (x + 3)² - 9
Section 01
Half of b, then square.
In the lab set \(b\) to 6 and press Make the square. Watch the \(6x\) strip split into two halves of \(3x\) that wrap around the \(x^{2}\) square, leaving a 3 by 3 corner empty (Fig. 1). Change \(b\) and \(c\) and predict the completed square before you press.
Worked example Positive b
Write \(x^{2} + 6x + 5\) in the form \((x + p)^{2} + q\).
Half of 6 is 3, so \(x^{2} + 6x = (x + 3)^{2} - 9\).
\(x^{2} + 6x + 5 = (x + 3)^{2} - 9 + 5\).
Answer: \((x + 3)^{2} - 4\)
Worked example Negative b
Write \(x^{2} - 10x + 7\) in the form \((x + p)^{2} + q\).
Half of \(-10\) is \(-5\): \(x^{2} - 10x = (x - 5)^{2} - 25\).
\((x - 5)^{2} - 25 + 7\).
Answer: \((x - 5)^{2} - 18\)
Worked example Odd b
Write \(x^{2} + 5x - 2\) in the form \((x + p)^{2} + q\).
Half of 5 is \(\dfrac{5}{2}\), and \(\left(\dfrac{5}{2}\right)^{2} = \dfrac{25}{4}\).
The number you subtract is \(\left(\dfrac{b}{2}\right)^{2}\), and it is always subtracted, even when \(b\) is negative: \(x^{2} - 10x = (x - 5)^{2} - 25\), not \(+ 25\).
Section 02
When a is not 1.
Take \(a\) out as a factor of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out the number you subtracted.
Worked example Factor out a
Write \(2x^{2} + 12x + 7\) in the form \(a(x + p)^{2} + q\).
In \(2[(x + 3)^{2} - 9]\) the 2 multiplies the \(-9\) too, giving \(-18\). Subtracting only 9 gives the wrong constant \(-2\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Write \(x^{2} + 8x\) in completed square form.
Quick check+10 XP
Write \(x^{2} - 6x + 10\) in the form \((x + p)^{2} + q\).
Quick check+10 XP
Solve \((x - 2)^{2} = 9\).
Quick check+10 XP
What is the minimum point of \(y = (x + 1)^{2} - 4\)?
Section 03
Solving and turning points.
Once a quadratic is in completed square form, the unknown appears only once, so you can solve by undoing: add, square root (both signs), subtract. The same form gives the turning point of the graph.
Worked example Exact solutions
Solve \(x^{2} - 6x + 4 = 0\) by completing the square. Give your answers in surd form.
\((x - 2)^{2}\) is never negative, and is smallest (zero) when \(x = 2\).
Then \(y = 0 - 3 = -3\). It is a minimum because the \(x^{2}\) term is positive.
Answer: Minimum point \((2,\ -3)\)
Your turn
Solve \(x^{2} + 8x + 3 = 0\) by completing the square, giving exact answers and then answers correct to 2 decimal places.
\((x + 4)^{2} - 16 + 3 = 0\), so \((x + 4)^{2} = 13\), \(x = -4 \pm \sqrt{13}\). That is \(x = -0.39\) or \(x = -7.61\).
Examiner trap
Square rooting gives two values: \(x - 3 = +\sqrt{5}\) or \(-\sqrt{5}\). And the turning point of \((x - 2)^{2} - 3\) is at \(x = +2\), the opposite sign to the number in the bracket.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Paper 2 and Paper 4 standard.
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Exam style+20 XP
Write \(x^{2} - 10x + 7\) in the form \((x + p)^{2} + q\).
Exam style+20 XP
Write \(2x^{2} + 12x + 7\) in the form \(a(x + p)^{2} + q\).
Exam style+20 XP
Solve \(x^{2} - 6x + 4 = 0\) by completing the square.
Exam style+20 XP
Find the coordinates of the turning point of \(y = x^{2} + 5x + 1\).
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