Q.10 - 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 $\frac{5}{3}{y}^2=60$.

Step-by-Step Solution

Answer

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

Step 1. Given information

Consider the quadratic equation $\frac{5}{3}{y}^2=60$

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 $\frac{3}{5}$ to make its coefficient 1.  

$$\begin{gathered} \frac{3}{5}\left(\frac{5}{3}{y}^2\right)=\frac{3}{5}\left(60\right) \\ y^2=36 \end{gathered}$$

Step 2. Apply square root property

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

Simplify the radical.     

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

Rewrite to show two solutions.  

$y=6, y=-6$

Step 3. Check the solutions

Substitute $y=6$ in the original equation.

$$\begin{gathered} \frac{5}{3}{y}^2=60 \\ \frac{5}{3}{\left(6\right)}^2\stackrel{?}{=}60 \\ \frac{5}{3}\left(36\right)\stackrel{?}{=}60 \\ 60=60 \end{gathered}$$

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

 $$\begin{gathered} \frac{5}{3}{y}^2=60 \\ \frac{5}{3}{(-6)}^2\stackrel{?}{=}60 \\ \frac{5}{3}\left(36\right)\stackrel{?}{=}60 \\ 60=60 \end{gathered}$$