Q.9.17 - Intermediate Algebra

Question

Solve ${(3r+4)}^2=-8$

Step-by-Step Solution

Answer

The solutions are $r=-\frac{4}{3}+i\frac{2\sqrt{2}}{3}, r=-\frac{4}{3}-i\frac{2\sqrt{2}}{3}$

Step 1. Given information

Consider the quadratic equation ${(3r+4)}^2=-8$

Use the Square Root Property and simplify the radical.

If $x^2=k$, then $x=\pm \sqrt{k}$

$$\begin{gathered} (3r+4)=\pm \sqrt{-8} \\ (3r+4)=\pm \sqrt{8}i \\ (3r+4)=\pm \sqrt{4·2}i \\ (3r+4)=\pm 2\sqrt{2}i \end{gathered}$$

Subtract 4 from both sides.

$3r=\pm 2\sqrt{2}i-4$

Divide both sides by 3.

$r=\frac{\pm 2\sqrt{2}i-4}{3}$

Rewrite in standard form.

$r=\pm \frac{2\sqrt{2}}{3}i-\frac{4}{3}$

Rewrite to show two solutions.

$r=\frac{2\sqrt{2}}{3}i-\frac{4}{3}, r=-\frac{2\sqrt{2}}{3}i-\frac{4}{3}$

Step 2. Check for solutions.

Substitute $r=\frac{2\sqrt{2}}{3}i-\frac{4}{3}$in the original equation.

$$\begin{gathered} { (3r+4)}^2=-8 \\ {\left(3\left(\frac{2\sqrt{2}}{3}i-\frac{4}{3}\right)+4\right)}^2\stackrel{?}{=}-8 \\ {\left(-2\sqrt{2}-4+4\right)}^2\stackrel{?}{=}-8 \\ -4·2\stackrel{?}{=}-8 \\ -8=-8 \end{gathered}$$

Substitute $r=-\frac{2\sqrt{2}}{3}i-\frac{4}{3}$in the original equation.

$$\begin{gathered} { (3r+4)}^2=-8 \\ {\left(3\left(-\frac{2\sqrt{2}}{3}i-\frac{4}{3}\right)+4\right)}^2\stackrel{?}{=}-8 \\ {\left(-2\sqrt{2}-4+4\right)}^2\stackrel{?}{=}-8 \\ -4·2\stackrel{?}{=}-8 \\ -8=-8 \end{gathered}$$