91Ó°ÊÓ

The given equation can be written as:

$$\begin{gathered} \sqrt{3x−14}+x=6 \\ \sqrt{3x−14}=6−x \\ {\left(\sqrt{3x−14}\right)}^2={\left(6−x\right)}^2\left(\text{squaringbothsides}\right) \\ 3x−14={\left(6−x\right)}^2 \\ 3x−14=6^2−2\left(6\right)\left(x\right)+x^2\left(âˆ\mu {\left(a−b\right)}^2=a^2−2ab+b^2\right) \\ 3x−14=36−12x+x^2 \end{gathered}$$

The solution of the given equation is:

$$\begin{gathered} 3x−14=36−12x+x^2 \\ {x}^2−12x−3x+36+14=0 \\ {x}^2−15x+50=0 \\ {x}^2−10x−5x+50=0 \\ x\left(x−10\right)−5\left(x−10\right)=0 \\ \left(x−10\right)\left(x−5\right)=0 \\ x−10=0\text{or}x−5=0 \\ x=10\text{or}x=5 \end{gathered}$$

Therefore, the values x are 10 and 5.

When $x=10$,

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

$$\begin{gathered} \sqrt{3x−14}+x=6 \\ \sqrt{3\left(10\right)−14}+10\underset{¯}{\underset{¯}{?}}6 \\ \sqrt{30−14}+10\underset{¯}{\underset{¯}{?}}6 \\ \sqrt{16}+10\underset{¯}{\underset{¯}{?}}6 \\ 4+10\underset{¯}{\underset{¯}{?}}6 \\ \text{14}≠6 \end{gathered}$$

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 $\sqrt{3x−14}+x=6$.

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

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