Completing the Square: Practice

Add $\left(\frac{2}{2}\right)^2 = 1$ to both sides after moving the constant across.

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

Since $\sqrt{-2} = i\sqrt{2}$, we get $x + 1 = \pm i\sqrt{2}$, so $x = -1 \pm i\sqrt{2}$.

Add $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$ to both sides. On the right, $-5 = -\frac{20}{4}$.

$\left(x + \frac{5}{2}\right)^2 = -\frac{20}{4} + \frac{25}{4} = \frac{5}{4}$

Then $x + \frac{5}{2} = \pm\frac{\sqrt{5}}{2}$, so $x = \frac{-5 \pm \sqrt{5}}{2}$.

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

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

Since $\sqrt{-2} = i\sqrt{2}$, we get $x = -1 \pm i\sqrt{2}$.

Divide every term by $2$ first: $x^2 - 4x + 5 = 0$. Then add $\left(\frac{-4}{2}\right)^2 = 4$ to both sides.

$(x - 2)^2 = -5 + 4 = -1$

Since $\sqrt{-1} = i$, we get $x - 2 = \pm i$, so $x = 2 \pm i$.

Add $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ to both sides. On the right, $-1 = -\frac{4}{4}$.

$\left(x + \frac{1}{2}\right)^2 = -\frac{4}{4} + \frac{1}{4} = -\frac{3}{4}$

Since $\sqrt{-\frac{3}{4}} = \frac{i\sqrt{3}}{2}$, we get $x = \frac{-1 \pm i\sqrt{3}}{2}$.

Move the constant across and add $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ to both sides. On the right, $6 + \frac{1}{4} = \frac{25}{4}$.

$\left(x - \frac{1}{2}\right)^2 = \frac{25}{4}$

Then $x - \frac{1}{2} = \pm\frac{5}{2}$, so $x = \frac{1 \pm 5}{2}$, giving $x = 3$ or $x = -2$.

For a perfect square, the constant must equal $\left(\frac{b}{2}\right)^2$, so $\left(\frac{b}{2}\right)^2 = 9$.

$\frac{b}{2} = \pm 3 \quad\Rightarrow\quad b = \pm 6$

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

Add $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ to both sides. On the right, $-3 = -\frac{12}{4}$.

$\left(x + \frac{3}{2}\right)^2 = -\frac{12}{4} + \frac{9}{4} = -\frac{3}{4}$

Since $\sqrt{-\frac{3}{4}} = \frac{i\sqrt{3}}{2}$, we get $x = \frac{-3 \pm i\sqrt{3}}{2}$.

Divide every term by $2$ first: $x^2 + 6x + 2 = 0$. Then add $\left(\frac{6}{2}\right)^2 = 9$ to both sides.

$(x + 3)^2 = -2 + 9 = 7$

Then $x + 3 = \pm\sqrt{7}$, so $x = -3 \pm \sqrt{7}$.

Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides after moving the constant across.

$(x + 3)^2 = -11 + 9 = -2$

Since $\sqrt{-2} = i\sqrt{2}$, we get $x + 3 = \pm i\sqrt{2}$, so $x = -3 \pm i\sqrt{2}$.

Add $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$ to both sides. On the right, $-5 = -\frac{20}{4}$.

$\left(x - \frac{5}{2}\right)^2 = -\frac{20}{4} + \frac{25}{4} = \frac{5}{4}$

Then $x - \frac{5}{2} = \pm\frac{\sqrt{5}}{2}$, so $x = \frac{5 \pm \sqrt{5}}{2}$.

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

$(x - 1)^2 = \frac{3}{2} + 1 = \frac{5}{2}$

Since $\sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$, we get $x = 1 \pm \frac{\sqrt{10}}{2}$.