Completing the Square: Additional Practice (Set 1)

Half of $-7$ is $-\frac{7}{2}$, and squaring makes it positive.

$\left(-\frac{7}{2}\right)^2 = \frac{49}{4}$

Then $x^2 - 7x + \frac{49}{4} = \left(x - \frac{7}{2}\right)^2$.

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

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

Then $x - 1 = \pm\sqrt{5}$, so $x = 1 \pm \sqrt{5}$.

Divide every term by $3$ first: $x^2 - 4x + 3 = 0$. Then add $\left(\frac{-4}{2}\right)^2 = 4$ to both sides.

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

Then $x - 2 = \pm 1$, so $x = 3$ or $x = 1$.

Add $\left(\frac{-4}{2}\right)^2 = 4$ to both sides after moving the constant across.

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

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

Half of $-3$ is $-\frac{3}{2}$, and $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$. The leftover constant is $2 - \frac{9}{4} = \frac{8}{4} - \frac{9}{4}$.

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

The remaining constant is $-\frac{1}{4}$.

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

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

Then $x + 1 = \pm\sqrt{2}$, so $x = -1 \pm \sqrt{2}$.

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

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

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

The left side is already a perfect square, since $\left(\frac{-12}{2}\right)^2 = 36$ matches the constant.

$(x - 6)^2 = 0 \quad\Rightarrow\quad x = 6$

The right side is $0$, so the root $6$ is repeated.

Divide every term by $2$ first: $x^2 + 4x + 10 = 0$. Then add $\left(\frac{4}{2}\right)^2 = 4$ to both sides.

$(x + 2)^2 = -10 + 4 = -6$

Since $\sqrt{-6} = i\sqrt{6}$, we get $x + 2 = \pm i\sqrt{6}$, so $x = -2 \pm i\sqrt{6}$.

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

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

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

Half of $-2$ is $-1$, and $(-1)^2 = 1$. Add and subtract it: $x^2 - 2x + 7 = (x^2 - 2x + 1) - 1 + 7$.

$x^2 - 2x + 7 = (x - 1)^2 + 6$

The leftover constant is $7 - 1 = 6$.

Divide every term by $4$ first: $x^2 + 2x - 3 = 0$. Move the constant across and add $\left(\frac{2}{2}\right)^2 = 1$ to both sides.

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

Then $x + 1 = \pm 2$, so $x = 1$ or $x = -3$.