Question

In the following exercises, solve using the Square Root Property.

$\frac{4}{3}{v}^2+4=28$

Short answer

The required answer is$v=3\sqrt{2}\mathrm{and}v=-3\sqrt{2}$.

Step-by-step solution

Step 1. Given information.

The given quadratic equation is:

$\frac{4}{3}{v}^2+4=28$

Step 2. Solve using the quadratic equation by using Square Root Property.

$\frac{4}{3}{v}^2+4=28$

Isolate the quadratic term and make its coefficient one by subtracting 4 from both sides,

$$\begin{gathered} \frac{4}{3}{v}^2+4-4=28-4 \\ \frac{4}{3}{v}^2=24 \end{gathered}$$

Multiply both sides by $\frac{3}{4}$,

$$\begin{gathered} \frac{4}{3}{v}^2\times \frac{3}{4}=24\times \frac{3}{4} \\ {v}^2=18 \end{gathered}$$

Use Square Root Property,

$v=\pm \sqrt{18}$

Simplify the radical,

$$\begin{gathered} v=\pm \sqrt{9}·\sqrt{2} \\ v=\pm 3\sqrt{2} \end{gathered}$$

Rewrite to show two solutions,

$v=3\sqrt{2}\mathrm{and}v=-3\sqrt{2}$