Q41. - Algebra 1

Question

_{Solve each equation. Check your solution.}

_{$\sqrt{\mathbf{x}\mathbf{+}\mathbf{4}}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{8}$}

Step-by-Step Solution

Answer

The solution of the given equation is $\mathit{x}\mathbf{=}\mathbf{12}$.

Check:

When $\mathit{x}\mathbf{=}\mathbf{12}$,

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

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

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

When $\mathit{x}\mathbf{=}\mathbf{5}$,

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

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

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

Step 1. Rewrite the given equation.

The given equation can be written as:

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

Step 2. Solve the given equation.

The solution of the given equation is:

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

Therefore, the values $x$ are 12 and 5.

Step 3. Check the solution.

When $x=12$,

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

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

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

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

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