Answer
$\{4\}$
Work Step by Step
Isolating the radical expression on one side, the given equation, $
2x-\sqrt{x}=6
,$ is equivalent to
\begin{align*}\require{cancel}
2x-6&=\sqrt{x}
.\end{align*}
Squaring both sides, the equation above is equivalent to
\begin{align*}
(2x-6)^2&=\left(\sqrt{x}\right)^2
\\
(2x)^2+2(2x)(-6)+(-6)^2&=x
\\
4x^2-24x+36&=x
&(\text{use }(a+b)^2=a^2+2ab+b^2)
.\end{align*}
In the form $ax^2+bx+c=0$, the equation above is equivalent to
\begin{align*}
4x^2+(-24x-x)+36&=0
\\
4x^2-25x+36&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x-4)(4x-9)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving the variable, then
\begin{array}{l|r}
x-4=0 & 4x-9=0
\\
x=4 & 4x=9
\\\\
& x=\dfrac{9}{4}
.\end{array}
Since both sides of the original equation were raised to the second power, then checking of solutions is a must. Substituting the solutions in the original equation results to
\begin{array}{l|r}
\text{If }x=4: & \text{If }x=\dfrac{9}{4}:
\\\\
2(4)-\sqrt{4}\overset{?}=6 & 2\left(\dfrac{9}{4}\right)-\sqrt{\left(\dfrac{9}{4}\right)}\overset{?}=6
\\\\
8-2\overset{?}=6 & \dfrac{9}{2}-\dfrac{9}{2}\overset{?}=6
\\\\
6\overset{\checkmark}=6 & 0\ne6
.\end{array}
Since $x=\dfrac{9}{4}$ does not satisfy the original equation, then the only solution is $x=4$. Hence, the solution set of the equation $
2x-\sqrt{x}=6
$ is $\{4\}$.
