Q 293 - Intermediate Algebra

Question

In the following exercises, solve:

$121n^2=36$

Step-by-Step Solution

Answer

On solving the equation, we get $n=\frac{6}{11},-\frac{6}{11}$.

Step 1. Given information.

Consider the given equation,

$121n^2=36$

Write the quadratic equation in standard form,

$121n^2-36=0$

Step 2. Factor the quadratic expression.

On factoring, we get,

$$\begin{gathered} {\left(11n\right)}^2-6^2=0 \\ \left(11n+6\right)\left(11n-6\right)=0 \end{gathered}$$

Use the zero product property,$$\begin{gathered} 11n+6=0 \\ n=-\frac{6}{11} \end{gathered}$$

Then,

$$\begin{gathered} 11n-6=0 \\ n=\frac{6}{11} \end{gathered}$$

Step 3. Check the answers.

Substitute $n=\frac{6}{11}$ in the given equation,

$$\begin{gathered} 121{\left(\frac{6}{11}\right)}^2=36 \\ 121\times \frac{36}{121}=36 \\ 36=36 \end{gathered}$$

This is true.

Substitute $n=-\frac{6}{11}$ in the given equation,

$$\begin{gathered} 121{\left(-\frac{6}{11}\right)}^2=36 \\ 121\times \frac{36}{121}=36 \\ 36=36 \end{gathered}$$

This is also true.

Hence, the answer is $n=\frac{6}{11},-\frac{6}{11}$.