Answer
$x=\dfrac{1}{2}$
Work Step by Step
Squaring both sides of the given equation, $
4x=\sqrt{6x+1}
,$ results to
\begin{align*}\require{cancel}
(4x)^2&=\left(\sqrt{6x+1}\right)^2
\\
16x^2&=6x+1
\\
16x^2-6x-1&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(8x+1)(2x-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
8x+1=0 & 2x-1=0
\\
8x=-1 & 2x=1
\\\\
x=-\dfrac{1}{8} & x=\dfrac{1}{2}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-\dfrac{1}{8}: & \text{If }x=\dfrac{1}{2}:
\\\\
4\left(-\dfrac{1}{8}\right)\overset{?}=\sqrt{6\left(-\dfrac{1}{8}\right)+1} &
4\left(\dfrac{1}{2}\right)\overset{?}=\sqrt{6\left(\dfrac{1}{2}\right)+1}
\\\\
\cancelto14\left(-\dfrac{1}{\cancelto28}\right)\overset{?}=\sqrt{\cancelto36\left(-\dfrac{1}{\cancelto48}\right)+1} &
\cancelto24\left(\dfrac{1}{\cancelto12}\right)\overset{?}=\sqrt{\cancelto36\left(\dfrac{1}{\cancelto12}\right)+1}
\\\\
-\dfrac{1}{2}\overset{?}=\sqrt{-\dfrac{3}{4}+1} &
2\overset{?}=\sqrt{3+1}
\\\\
-\dfrac{1}{2}\overset{?}=\sqrt{-\dfrac{3}{4}+\dfrac{4}{4}} &
2\overset{?}=\sqrt{4}
\\\\
-\dfrac{1}{2}\overset{?}=\sqrt{\dfrac{1}{4}} &
2\overset{\checkmark}=2
\\\\
-\dfrac{1}{2}\ne\dfrac{1}{2}
.\end{array}
Since $x=-\dfrac{1}{8}$ does not satisfy the original equation, then the only solution of the equation $
4x=\sqrt{6x+1}
$ is $x=\dfrac{1}{2}$.
