91Ó°ÊÓ

The given equation can be written as:

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

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.

When $x=10$,

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

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

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 $\sqrt{3x−14}+x=6$.