#### Answer

$x=1$

#### Work Step by Step

$3x=\sqrt{8x+1}$
Square both sides of the equation:
$(3x)^{2}=(\sqrt{8x+1})^{2}$
$9x^{2}=8x+1$
Take all terms to the left side of the equation:
$9x^{2}-8x-1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=9$, $b=-8$ and $c=-1$.
Substitute:
$x=\dfrac{-(-8)\pm\sqrt{(-8)^{2}-4(9)(-1)}}{2(9)}=\dfrac{8\pm\sqrt{64+36}}{18}=...$
$...=\dfrac{8\pm\sqrt{100}}{18}=\dfrac{8\pm10}{18}$
The two solutions are:
$x=\dfrac{8+10}{18}=1$
$x=\dfrac{8-10}{18}=-\dfrac{1}{9}$
Verify the solutions:
$3(1)=\sqrt{8(1)+1}$
$3=3$ True
$3\Big(-\dfrac{1}{9}\Big)=\sqrt{8\Big(-\dfrac{1}{9}\Big)+1}$
$-\dfrac{1}{3}=\dfrac{1}{3}$ False
The answer is $x=1$