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Solve for $q$ in $(q+2)^2-12=0\text{.}$

Enter multiple answers in the form q=1 or q=-1. If you need to use the square root symbol, as in $\sqrt{17}\text{,}$ type it as sqrt(17).

Solution

It’s important here to suppress any urge you may have to expand the squared binomial.

$\begin{aligned} (q+2)^2-12\amp=0\\ (q+2)^2\amp=12 \end{aligned}$

$\begin{aligned} (q+2)\amp=-\sqrt{12}\amp\text{or}\amp\amp(q+2)\amp=\sqrt{12}\\ q+2\amp=-2\sqrt{3}\amp\text{or}\amp\amp q+2\amp=2\sqrt{3}\\ q\amp=-2\sqrt{3}-2\amp\text{or}\amp\amp q\amp=2\sqrt{3}-2 \end{aligned}$

The solution set is $\left\{-2\sqrt{3}-2,2\sqrt{3}-2\right\}\text{.}$

To check the solution, we would replace $q$ with each of $-2\sqrt{3}-2$ and $2\sqrt{3}-2$ in the original equation, as shown here:

$\begin{aligned} \left(\left(\substitute{-2\sqrt{3}-2}\right)+2\right)^2-12\amp\stackrel{?}{=}0\amp\amp\amp\left(\left(\substitute{2\sqrt{3}-2}\right)+2\right)^2-12\amp\stackrel{?}{=}0\\ \left(-2\sqrt{3}\right)^2-12\amp\stackrel{?}{=}0\amp\amp\amp \left(2\sqrt{3}\right)^2-12\amp\stackrel{?}{=}0\\ (-2)^2\left(\sqrt{3}\right)^2-12\amp\stackrel{?}{=}0\amp\amp\amp (2)^2\left(\sqrt{3}\right)^2-12\amp\stackrel{?}{=}0\\ (4)(3)-12\amp\stackrel{?}{=}0\amp\amp\amp (4)(3)-12\amp\stackrel{?}{=}0\\ 12-12\amp\stackrel{\checkmark}{=}0\amp\amp\amp 12-12\amp\stackrel{\checkmark}{=}0 \end{aligned}$

Note that these simplifications relied on exponent rules and the multiplicative property of square roots.