Square Completion

$x^2+6x-4=(x+3)^2-13.$

The two forms are equivalent: they give the same value for every value of $x$. Nothing is being solved because there is no equation here; the expression is simply being rewritten. This form is useful because the squared part is easy to recognise and work with.

Recognising a perfect square

Expanding a squared bracket reveals the pattern behind the method:

$(x+p)^2=(x+p)(x+p)=x^2+2px+p^2.$

Similarly,

$(x-p)^2=x^2-2px+p^2.$

The middle coefficient is twice the number inside the bracket. Therefore, to turn $x^2+bx$ into the beginning of a perfect square, the number placed beside $x$ in the bracket must be half of $b$.

Completing the square when the coefficient of $x^2$ is 1

Consider

$x^2+6x-4.$

Half of $6$ is $3$, so the required squared bracket is $(x+3)^2$. Expanding it gives

$(x+3)^2=x^2+6x+9.$

This contains an extra $9$. Subtracting the same $9$ keeps the expression unchanged:

This is the central idea: add the number needed to make a perfect square, then subtract it again.

For a negative middle term, its sign is retained when it is halved. In

$x^2-8x+7,$

half of $-8$ is $-4$, so

The whole procedure is summarised by

$x^2+bx+c=\left(x+\frac b2\right)^2+c-\frac{b^2}{4}.$

This formula is not a separate trick: it records the same process of halving $b$, squaring the result and compensating for the added term.

When the coefficient of $x^2$ is not 1

A squared bracket $(x+p)^2$ has coefficient $1$ on $x^2$. If the original expression begins with $ax^2$, first factor out $a$ from the terms involving $x$—or, more conveniently, from the whole expression.

For example,

$2x^2+5x+1=2\left(x^2+\frac52x+\frac12\right).$

Inside the bracket, half of $\frac52$ is $\frac54$. Its square is $\frac{25}{16}$, so add and subtract that value:

The outer factor multiplies every term inside the bracket, including the remaining constant. This is why the final constant is $-\frac{17}{8}$ rather than $-\frac{17}{16}$.

More generally, for $a\ne0$,

$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.$

Checking the result

Expansion provides a reliable check. For instance,

Since this matches the original expression exactly, the square has been completed correctly.