Completing the Square: Practice

Take half of the coefficient of $x$, then square it. Half of $6$ is $3$.

$\left(\frac{6}{2}\right)^2 = 3^2 = 9$

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

Halve the coefficient of $x$, then square the result. Half of $4$ is $2$.

$\left(\frac{4}{2}\right)^2 = 2^2 = 4$

So $x^2 + 4x + 4 = (x + 2)^2$.

The missing constant is the square of half the linear coefficient. Half of $10$ is $5$.

$\left(\frac{10}{2}\right)^2 = 5^2 = 25$

Then $x^2 + 10x + 25 = (x + 5)^2$.

The rule uses half of the coefficient of $x$, and squaring makes the constant positive. Half of $-2$ is $-1$.

$\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1$

So $x^2 - 2x + 1 = (x - 1)^2$.

Half of the middle coefficient $6$ is $3$, and $3^2 = 9$ matches the constant, so the trinomial is a perfect square.

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

The binomial uses half of $6$, not $6$ itself.

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

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

The sign inside the binomial follows the sign of the middle term.

Half of $4$ is $2$, and $2^2 = 4$. Add and subtract it: $x^2 + 4x + 1 = (x^2 + 4x + 4) - 4 + 1$.

$x^2 + 4x + 1 = (x + 2)^2 - 3$

The leftover constant is $1 - 4 = -3$.

Half of $-6$ is $-3$, and $(-3)^2 = 9$. Add and subtract it: $x^2 - 6x + 2 = (x^2 - 6x + 9) - 9 + 2$.

$x^2 - 6x + 2 = (x - 3)^2 - 7$

The leftover constant is $2 - 9 = -7$.

Take the square root of both sides with a $\pm$: $x + 1 = \pm 2$.

$x = -1 + 2 = 1 \quad\text{or}\quad x = -1 - 2 = -3$

Subtract $1$ from each case.

The constant is the square of half the linear coefficient. Half of $-12$ is $-6$.

$\left(\frac{-12}{2}\right)^2 = (-6)^2 = 36$

Then $x^2 - 12x + 36 = (x - 6)^2$.

Take the square root of both sides with a $\pm$: $x + 4 = \pm 1$.

$x = -4 + 1 = -3 \quad\text{or}\quad x = -4 - 1 = -5$

Subtract $4$ from each case.

Half of $8$ is $4$, and $4^2 = 16$. Add and subtract it: $x^2 + 8x + 20 = (x^2 + 8x + 16) - 16 + 20$.

$x^2 + 8x + 20 = (x + 4)^2 + 4$

The leftover constant is $20 - 16 = 4$.