Completing the Square

Completing the square rewrites a quadratic \(ax^2+bx+c\) in the form \(a(x-h)^2+k\). When \(a=1\) it rests on one identity: \(x^2+bx=\left(x+\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2\).

Why: \(x^2+6x\) is a square of side \(x\) with two strips of \(3x\), half of \(6x\) on each side. Adding the missing corner \(3^2=9\) makes a square of side \(x+3\), so \(x^2+6x=(x+3)^2-9\). Once \(a\) has been taken out (here \(a=1\)), half of the coefficient of \(x\) is the number inside the bracket.

Method

1. If \(a\ne1\), take \(a\) out of the \(x^2\) and \(x\) terms only: \(ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x\right)+c\).
2. Halve the coefficient of \(x\) inside the bracket; add and subtract the square of that half.
3. Write the first three terms inside the bracket as a perfect square.
4. Multiply the subtracted square by \(a\) and collect the constants.

Example

\(\begin{aligned} x^2-10x+7&=(x-5)^2-25+7 \\ &=(x-5)^2-18 \end{aligned}\)

\(\begin{aligned} 3x^2+12x-2&=3(x^2+4x)-2 \\ &=3\left[(x+2)^2-4\right]-2 \\ &=3(x+2)^2-14 \end{aligned}\)

With a negative \(a\), take out the minus sign with it: \(-2x^2+4x+1=-2(x^2-2x)+1=-2\left[(x-1)^2-1\right]+1=-2(x-1)^2+3\).

Exam check Expand your answer to check it: \(3(x+2)^2-14=3x^2+12x+12-14=3x^2+12x-2\). The commonest slip is not multiplying the subtracted square by \(a\): \(3\left[(x+2)^2-4\right]\) is \(3(x+2)^2-12\), not \(3(x+2)^2-4\).