Q 291 - Intermediate Algebra

Question

In the following exercises, solve:

$16y^2=81$

Step-by-Step Solution

Answer

On solving the equation, we get $y=-\frac{9}{4},\frac{9}{4}$.

Step 1. Given information.

Consider the given equation,

$16y^2=81$

Write the quadratic equation in standard form,

$16y^2-81=0$

Step 2. Factor the quadratic expression.

On factoring, we get,

$$\begin{gathered} {\left(4y\right)}^2-9^2=0 \\ \left(4y-9\right)\left(4y+9\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} 4y-9=0 \\ y=\frac{9}{4} \end{gathered}$$

Then,

$$\begin{gathered} 4y+9=0 \\ y=-\frac{9}{4} \end{gathered}$$

Step 3. Check the answers.

Substitute $y=\frac{9}{4}$ in the given equation,

$$\begin{gathered} 16{\left(\frac{9}{4}\right)}^2=81 \\ 16\times \frac{81}{16}=81 \\ 81=81 \end{gathered}$$

This is true.

Substitute $x=-\frac{9}{4}$ in the given equation,

$$\begin{gathered} 16{\left(-\frac{9}{4}\right)}^2=81 \\ 16\times \frac{81}{16}=81 \\ 81=81 \end{gathered}$$

This is also true.

Hence, the answer is $y=\frac{9}{4},-\frac{9}{4}$.