Q41. - Algebra 1

Question

Solve each equation. Check your solution.

$\sqrt{x+4}=x-8$

Step-by-Step Solution

Answer

The solution of the given equation is $x=12$.

 

Check:

When $x=12$,

Substitute 12 for x in the given equation $\sqrt{x+4}=x-8$.

$$\begin{gathered} \sqrt{x+4}=x-8 \\ \sqrt{12+4}\underset{¯}{\underset{¯}{?}}12-8 \\ \sqrt{16}\underset{¯}{\underset{¯}{?}}4 \\ \text{ 4}=4 \end{gathered}$$

Therefore, $x=12$ is the solution of the given equation $\sqrt{x+4}=x-8$.

 

When $x=5$,

Substitute 5 for x in the given equation $\sqrt{x+4}=x-8$.

$$\begin{gathered} \sqrt{x+4}=x-8 \\ \sqrt{5+4}\underset{¯}{\underset{¯}{?}}5-8 \\ \sqrt{9}\underset{¯}{\underset{¯}{?}}-3 \\ \text{ 3}\ne -3 \end{gathered}$$

Therefore, $x=5$ is not the solution of the given equation $\sqrt{x+4}=x-8$.

Step 1. Rewrite the given equation.

The given equation can be written as:

$$\begin{gathered} \sqrt{x+4}=x-8 \\ {\left(\sqrt{x+4}\right)}^2={\left(x-8\right)}^2\left(\text{squaring both sides}\right) \\ x+4={\left(x-8\right)}^2 \\ x+4=x^2-2\left(x\right)\left(8\right)+8^2\left(\because {\left(a-b\right)}^2=a^2-2ab+b^2\right) \\ x+4=x^2-16x+64 \end{gathered}$$

Step 2. Solve the given equation.

The solution of the given equation is:

$$\begin{gathered} x+4=x^2-16x+64 \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }0=x^2-16x-x+64-4 \\ 0=x^2-17x+60 \\ 0=x^2-12x-5x+60 \\ x^2-12x-5x+60=0 \\ x\left(x-12\right)-5\left(x-12\right)=0 \\ \left(x-12\right)\left(x-5\right)=0 \\ x-12=0\text{ or }x-5=0 \\ x=12\text{ or }x=5 \end{gathered}$$

Therefore, the values x are 12 and 5.

Step 3. Check the solution.

When $x=12$,

Substitute 12 for x in the given equation $\sqrt{x+4}=x-8$.

$$\begin{gathered} \sqrt{x+4}=x-8 \\ \sqrt{12+4}\underset{¯}{\underset{¯}{?}}12-8 \\ \sqrt{16}\underset{¯}{\underset{¯}{?}}4 \\ \text{ 4}=4 \end{gathered}$$

Therefore, $x=12$ is the solution of the given equation $\sqrt{x+4}=x-8$.

 

When $x=5$,

Substitute 5 for x in the given equation $\sqrt{x+4}=x-8$.

$$\begin{gathered} \sqrt{x+4}=x-8 \\ \sqrt{5+4}\underset{¯}{\underset{¯}{?}}5-8 \\ \sqrt{9}\underset{¯}{\underset{¯}{?}}-3 \\ \text{ 3}\ne -3 \end{gathered}$$

Therefore, $x=5$ is not the solution of the given equation $\sqrt{x+4}=x-8$.