Q42. - Algebra 1

Question

Solve each equation. Check your solution.

$\sqrt{\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{14}}\mathbf{+}\mathit{x}\mathbf{=}\mathbf{6}$

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 data-custom-editor="chemistry" $\mathit{x}$ 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 $\mathit{x}$ in the given equation $\sqrt{\mathbf{x}\mathbf{+}\mathbf{4}}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{8}$.

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

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:

$\begin{array}{l}\sqrt{3x-14}+x=6\\ \sqrt{3x-14}=6-x\\ {\left(\sqrt{3x-14}\right)}^2={(6-x)}^2\left(\text{squaring both sides}\right)\\ 3x-14={(6-x)}^2\\ 3x-14=6^2-2\left(6\right)\left(x\right)+x^2(\because {(a-b)}^2=a^2-2ab+b^2)\\ 3x-14=36-12x+x^2\end{array}$

Step 2. Solve the given equation.

The solution of the given equation is:

$\begin{array}{l}3x-14=36-12x+x^2\\ x^2-12x-3x+36+14=0\\ x^2-15x+50=0\\ x^2-10x-5x+50=0\\ x(x-10)-5(x-10)=0\\ (x-10)(x-5)=0\\ x-10=0\text{ or }x-5=0\\ x=10\text{ or }x=5\end{array}$

Therefore, the values $x$ are 10 and 5.

Step 3. Check the solution.

When $x=10$,

Substitute 10 for $x$ in the given equation $\sqrt{3x-14}+x=6$.

$\begin{array}{l}\sqrt{3x-14}+x=6\\ \sqrt{3\left(10\right)-14}+10\underset{¯}{\underset{¯}{?}}6\\ \sqrt{30-14}+10\underset{¯}{\underset{¯}{?}}6\\ \sqrt{16}+10\underset{¯}{\underset{¯}{?}}6\\ 4+10\underset{¯}{\underset{¯}{?}}6\\ \text{14}\ne 6\end{array}$

Therefore, $x=10$ is not the solution of the given equation $\sqrt{3x-14}+x=6$.

When $x=5$,

Substitute 5 for $x$ in the given equation $\sqrt{3x-14}+x=6$.

$\begin{array}{l}\sqrt{3x-14}+x=6\\ \sqrt{3\left(5\right)-14}+5\underset{¯}{\underset{¯}{?}}6\\ \sqrt{15-14}+5\underset{¯}{\underset{¯}{?}}6\\ \sqrt{1}+5\underset{¯}{\underset{¯}{?}}6\\ \text{1}+5\underset{¯}{\underset{¯}{?}}6\\ \text{6}=6\end{array}$

Therefore, $x=5$ is the solution of the given equation $\sqrt{3x-14}+x=6$.