91Ó°ÊÓ

The given equation can be written as:

$$\begin{gathered} \sqrt{2x−1}=2x−7 \\ {\left(\sqrt{2x−1}\right)}^2={\left(2x−7\right)}^2\left(\text{squaringbothsides}\right) \\ 2x−1={\left(2x\right)}^2−2\left(2x\right)\left(7\right)+{\left(7\right)}^2\left(âˆ\mu {\left(a−b\right)}^2=a^2−2ab+b^2\right) \\ 2x−1=4x^2−28x+49 \end{gathered}$$

The solution of the given equation is:

$$\begin{gathered} 2x−1=4x^2−28x+49 \\ 4x^2−28x−2x+49+1=0 \\ 4x^2−30x+50=0 \\ 2\left(2x^2−15x+25\right)=0 \\ 2x^2−15x+25=0 \\ 2x^2−10x−5x+25=0 \\ 2x\left(x−5\right)−5\left(x−5\right)=0 \\ \left(x−5\right)\left(2x−5\right)=0 \\ x−5=0\text{or}2x−5=0 \\ x=5\text{or}x=\frac{5}{2} \end{gathered}$$

Therefore, the values of x are$x=5$ or $x=\frac{5}{2}$.

Check the solution $x=5$,

Substitute 5 for x in the given equation $\sqrt{2x−1}=2x−7$.

$$\begin{gathered} \sqrt{2x−1}=2x−7 \\ \sqrt{2\left(5\right)−1}\underset{¯}{\underset{¯}{?}}2\left(5\right)−7 \\ \sqrt{10−1}\underset{¯}{\underset{¯}{?}}10−7 \\ \sqrt{9}\underset{¯}{\underset{¯}{?}}3 \\ \text{3}=3 \end{gathered}$$

Therefore,$x=5$ is the solution of the given equation $\sqrt{2x−1}=2x−7$.

Check the solution $x=\frac{5}{2}$,

Substitute$\frac{5}{2}$ for x in the given equation $\sqrt{2x−1}=2x−7$.

$$\begin{gathered} \sqrt{2x−1}=2x−7 \\ \sqrt{2\left(\frac{5}{2}\right)−1}\underset{¯}{\underset{¯}{?}}2\left(\frac{5}{2}\right)−7 \\ \sqrt{5−1}\underset{¯}{\underset{¯}{?}}5−7 \\ \sqrt{4}\underset{¯}{\underset{¯}{?}}−2 \\ \text{2}≠−2 \end{gathered}$$

Therefore,$x=\frac{5}{2}$ is not the solution of the given equation $\sqrt{2x−1}=2x−7$.

Therefore, the only solution of the given equation$\sqrt{2x−1}=2x−7$ is $x=5$.