Q.9 - Intermediate Algebra

Question

Solve Quadratic Equations of the Form $a{x}^2=k$Using the Square Root Property.

In the following exercises, solve each equation $\frac{4}{3}{x}^2=48$

Step-by-Step Solution

Answer

The solution of the quadratic equation:  $x=6, x=-6$

Step 1. Given information

Consider the quadratic equation $\frac{4}{3}{x}^2=48$

The objective is to solve the quadratic equation by using the square root property. 

If $x^2=k$, then $x=\pm \sqrt{k}$

The quadratic term is isolated. Divide by $\frac{3}{4}$ to make its coefficient 1.

$$\begin{gathered} \frac{3}{4}\left(\frac{4}{3}{x}^2\right)=\frac{3}{4}\left(48\right) \\ x^2=36 \end{gathered}$$

Step 2. Apply square root property

So, $x=\pm \sqrt{36}$

Simplify the radical.    

$$\begin{gathered} x=\pm \sqrt{6·6} \\ x=\pm 6 \end{gathered}$$

Rewrite to show two solutions.  

$x=6, x=-6$

Step 3. Check the solutions

Substitute $x=6$ in the original equation.

$$\begin{gathered} \frac{4}{3}{x}^2=48 \\ \frac{4}{3}{\left(6\right)}^2\stackrel{?}{=}48 \\ 4\left(12\right)\stackrel{?}{=}48 \\ 48=48 \end{gathered}$$

Substitute $x=-6$ in the original equation.

$$\begin{gathered} \frac{4}{ 3}{x}^2=48 \\ \frac{4}{3}{(-6)}^2\stackrel{?}{=}48 \\ 4\left(12\right)\stackrel{?}{=}48 \\ 48=48 \end{gathered}$$