Completing the SquareEdexcel IGCSE Maths: Revision notes

Completing the square rewrites a quadratic expression $ax^2+bx+c$ in the form $a(x+p)^2+q$, where $p$ and $q$ are constants. This form is useful because it shows the vertex (turning point) of the parabola directly, and lets you solve equations without factorising.

For $a=1$: $x^2+bx+c = \left(x+\dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c$

The key idea: half the coefficient of $x$, square it, then adjust with a correction term so the expansion still matches the original.

Step-by-step method for $x^2+bx+c$:

1. Write $x^2+bx$ as $\left(x+\dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2$
2. Add the constant $c$ from the original expression.
3. Simplify the two constant terms together.

Example: $x^2-8x+3$

Half of $-8$ is $-4$, and $(-4)^2=16$.

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

First factor $a$ out of the $x^2$ and $x$ terms only, then complete the square inside the bracket.

Example: $2x^2+12x+7$

Factor out 2 from the first two terms: $2(x^2+6x)+7$

Complete the square inside: $x^2+6x = (x+3)^2-9$

Substitute back: $2\left[(x+3)^2-9\right]+7 = 2(x+3)^2-18+7 = 2(x+3)^2-11$

Notice the $-9$ gets multiplied by 2 when the bracket is expanded out — a common source of errors.

Once in the form $a(x+p)^2+q$, the turning point is at $(-p, q)$.

If $a>0$, the parabola opens upwards and $q$ is the minimum value of the function, occurring when $x=-p$.

If $a<0$, the parabola opens downwards and $q$ is the maximum value.

Example: $f(x) = (x-4)^2-13$ has a minimum value of $-13$ at $x=4$.

Example: $g(x) = 2(x+3)^2-11$ has a minimum value of $-11$ at $x=-3$ (since $a=2>0$).

Once written as $a(x+p)^2+q=0$, rearrange to solve for $x$.

Example: solve $x^2+6x+5=0$

Completed square form: $(x+3)^2-4=0$

Rearrange: $(x+3)^2=4$

Square root both sides (remember $\pm$): $x+3=\pm2$

So $x=-3+2=-1$ or $x=-3-2=-5$

This method always works, even when the quadratic does not factorise nicely, and gives exact (surd) answers rather than decimals.