Answer
The solution set of equation ${{\left( 3x-1 \right)}^{2}}=75$ is $\left\{ \left. \frac{1-5\sqrt{3}}{3},\frac{1+5\sqrt{3}}{3} \right\} \right.$.
Work Step by Step
Consider the equation: ${{\left( 3x-1 \right)}^{2}}=75$
${{\left( 3x-1 \right)}^{2}}=75$
Take the square root on both sides
$\begin{align}
& 3x-1=\pm \sqrt{75} \\
& =\pm \sqrt{25\cdot 3} \\
& =\pm \sqrt{25}\sqrt{3} \\
& =\pm 5\sqrt{3}
\end{align}$
Now
$3x-1=\pm 5\sqrt{3}$
Add 1 on both sides,
$\begin{align}
& 3x-1+1=1\pm 5\sqrt{3} \\
& 3x=1\pm 5\sqrt{3}
\end{align}$
Divide by 3 on both sides,
$\begin{align}
& \frac{3x}{3}=\frac{1\pm 5\sqrt{3}}{3} \\
& x=\frac{1\pm 5\sqrt{3}}{3}
\end{align}$
Check:
For $x=\frac{1+5\sqrt{3}}{3}$
Put $x=\frac{1+5\sqrt{3}}{3}$ in ${{\left( 3x-1 \right)}^{2}}=75$
$\begin{align}
& {{\left( 3\left( \frac{1+5\sqrt{3}}{3} \right)-1 \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& {{\left( 1+5\sqrt{3}-1 \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& {{\left( 5\sqrt{3} \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& 75=75
\end{align}$
Which is true.
For $x=\frac{1-5\sqrt{3}}{3}$
Put $x=\frac{1-5\sqrt{3}}{3}$ in ${{\left( 3x-1 \right)}^{2}}=75$
$\begin{align}
& {{\left( 3\left( \frac{1-5\sqrt{3}}{3} \right)-1 \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& {{\left( 1-5\sqrt{3}-1 \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& {{\left( -5\sqrt{3} \right)}^{2}}\overset{?}{\mathop{=}}\,75 \\
& 75=75
\end{align}$
Which is true.
Therefore, the solution set of the equation ${{\left( 3x-1 \right)}^{2}}=75$ is $\left\{ \left. \frac{1-5\sqrt{3}}{3},\frac{1+5\sqrt{3}}{3} \right\} \right.$.
