Completing the square

Completing the square rewrites a quadratic so that its shape and turning point are easy to see.

For a quadratic beginning $x^2+bx$, half the coefficient of $x$ appears inside the bracket:

$x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2.$

The basic method

Consider $x^2+6x+5$. Start by building the square:

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

This contains $4$ more than the original expression, so subtract $4$:

$x^2+6x+5=(x+3)^2-4.$

Expanding the result is a reliable way to check it.

What the form tells you

The graph of $y=(x+3)^2-4$ is the graph of $y=x^2$ translated three units left and four units down. Its minimum point is therefore

$(-3,-4).$

This makes completed-square form useful for sketching graphs, finding ranges and solving optimization problems.

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

Factor it from the first two terms before completing the square:

$2x^2+12x+7=2(x^2+6x)+7.$

Now complete the square inside the bracket:

$2\big((x+3)^2-9\big)+7=2(x+3)^2-11.$

The factor $2$ multiplies both terms inside the bracket. Forgetting this is a common source of incorrect constants.

Solving a quadratic

To solve $x^2+6x+5=0$, use the completed form:

$(x+3)^2-4=0$ $(x+3)^2=4$ $x+3=\pm2.$

So $x=-1$ or $x=-5$. The $\pm$ is essential because both $2^2$ and $(-2)^2$ equal $4$.

Common mistakes

• The number inside the bracket is half the coefficient of $x$, not the full coefficient.
• The turning point of $(x+a)^2+b$ is $(-a,b)$; the horizontal sign reverses.
• When taking a square root during solving, include both the positive and negative roots.
• If you factor out a leading coefficient, apply it to the correction term too.