Completing the Square: Practice

Move the constant across and add $\left(\frac{6}{2}\right)^2 = 9$ to both sides.

$x^2 + 6x + 9 = -5 + 9 = 4$

Then $(x + 3)^2 = 4$, so $x + 3 = \pm 2$ and $x = -1$ or $x = -5$.

Move the constant across and add $\left(\frac{-4}{2}\right)^2 = 4$ to both sides.

$x^2 - 4x + 4 = 5 + 4 = 9$

Then $(x - 2)^2 = 9$, so $x - 2 = \pm 3$ and $x = 5$ or $x = -1$.

Half of the odd coefficient $5$ is the fraction $\frac{5}{2}$, and squaring it gives a fraction over $4$.

$\left(\frac{5}{2}\right)^2 = \frac{25}{4}$

Then $x^2 + 5x + \frac{25}{4} = \left(x + \frac{5}{2}\right)^2$.

Move the constant across and add $\left(\frac{4}{2}\right)^2 = 4$ to both sides.

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

Then $(x + 2)^2 = 3$, so $x + 2 = \pm\sqrt{3}$ and $x = -2 \pm \sqrt{3}$.

Move the constant across and add $\left(\frac{8}{2}\right)^2 = 16$ to both sides.

$x^2 + 8x + 16 = -12 + 16 = 4$

Then $(x + 4)^2 = 4$, so $x + 4 = \pm 2$ and $x = -2$ or $x = -6$.

The left side is already a perfect square, since $\left(\frac{-8}{2}\right)^2 = 16$ matches the constant.

$(x - 4)^2 = 0 \quad\Rightarrow\quad x = 4$

The right side is $0$, so the root $4$ is repeated.

Move the constant across and add $\left(\frac{2}{2}\right)^2 = 1$ to both sides.

$x^2 + 2x + 1 = 3 + 1 = 4$

Then $(x + 1)^2 = 4$, so $x + 1 = \pm 2$ and $x = 1$ or $x = -3$.

Half of $5$ is $\frac{5}{2}$, and $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$. The leftover constant is $4 - \frac{25}{4} = \frac{16}{4} - \frac{25}{4}$.

$x^2 + 5x + 4 = \left(x + \frac{5}{2}\right)^2 - \frac{9}{4}$

The fraction $-\frac{9}{4}$ is what remains after borrowing $\frac{25}{4}$ to build the square.

Divide every term by the leading coefficient $2$ first: $x^2 - 2x - 3 = 0$. Then add $\left(\frac{-2}{2}\right)^2 = 1$ to both sides.

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

Then $(x - 1)^2 = 4$, so $x - 1 = \pm 2$ and $x = 3$ or $x = -1$.

Complete the square: move the constant across and add $\left(\frac{6}{2}\right)^2 = 9$ to both sides.

$(x + 3)^2 = -13 + 9 = -4$

The right side is negative, so there is no real square root. The equation has no real solutions; its two roots, $-3 \pm 2i$, are complex.

Move the constant across and add $\left(\frac{-10}{2}\right)^2 = 25$ to both sides.

$x^2 - 10x + 25 = -16 + 25 = 9$

Then $(x - 5)^2 = 9$, so $x - 5 = \pm 3$ and $x = 8$ or $x = 2$.

Move the constant across and add $\left(\frac{4}{2}\right)^2 = 4$ to both sides.

$x^2 + 4x + 4 = -13 + 4 = -9$

Then $(x + 2)^2 = -9$, so $x + 2 = \pm\sqrt{-9} = \pm 3i$ and $x = -2 \pm 3i$.