Completing the Square: Additional Practice (Set 2)

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

$(x - 2)^2 = -7 + 4 = -3$

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

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

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

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

Half of $-7$ is $-\frac{7}{2}$, and $\left(\frac{7}{2}\right)^2 = \frac{49}{4}$. The leftover constant is $3 - \frac{49}{4} = \frac{12}{4} - \frac{49}{4}$.

$x^2 - 7x + 3 = \left(x - \frac{7}{2}\right)^2 - \frac{37}{4}$

The remaining constant is $-\frac{37}{4}$.

Half of $12$ is $6$, and $6^2 = 36$. Add and subtract it: $x^2 + 12x + 40 = (x^2 + 12x + 36) - 36 + 40$.

$x^2 + 12x + 40 = (x + 6)^2 + 4$

The leftover constant is $40 - 36 = 4$.

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

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

Then $x - 3 = \pm\sqrt{2}$, so $x = 3 \pm \sqrt{2}$.

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

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

Then $x - 2 = \pm\sqrt{3}$, so $x = 2 \pm \sqrt{3}$.

Completing the square needs a leading coefficient of $1$, but here it is $2$. Divide every term by $2$ first.

$2x^2 + 4x + 5 = 0 \quad\Rightarrow\quad x^2 + 2x + \frac{5}{2} = 0$

Only after that does adding $\left(\frac{b}{2}\right)^2$ build a genuine perfect square.

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 $x = \frac{1 \pm i\sqrt{3}}{2}$.

Add $\left(\frac{10}{2}\right)^2 = 25$ to both sides after moving the constant across.

$(x + 5)^2 = -29 + 25 = -4$

Since $\sqrt{-4} = 2i$, we get $x + 5 = \pm 2i$, so $x = -5 \pm 2i$.

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

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

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

Moving the constant gives $x^2 + 6x = -2$. Adding $\left(\frac{6}{2}\right)^2 = 9$ must happen on both sides.

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

The right side should be $7$, not $-2$. The student added $9$ only on the left, breaking the balance.

Add $\left(\frac{8}{2}\right)^2 = 16$ to both sides after moving the constant across.

$(x + 4)^2 = -20 + 16 = -4$

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