Answer
$x=\dfrac{-3+3\sqrt{5}}{2}$
Work Step by Step
$x-\sqrt{9-3x}=0$
$x=\sqrt{9-3x}$
Square the both sides $\Rightarrow(x^2=\sqrt{9-3x})^2$
Expand $\Rightarrow x^2=9-3x$
$\Rightarrow x^2+3x-9=0$
Use the quadratic formula, where $a=1, b=3$ and $c=-9$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(3)\pm\sqrt{(3)^2-4(1)(-9)}}{2(1)}$
$\Rightarrow x=\dfrac{-3\pm\sqrt{9+36}}{2}=\dfrac{-3\pm\sqrt{45}}{2}=\dfrac{-3\pm3\sqrt{5}}{2}$
$x=\dfrac{-3+3\sqrt{5}}{2}$
or $x=\dfrac{-3-3\sqrt{5}}{2}$
Check the answer:
$x=\dfrac{-3+3\sqrt{5}}{2}$
LHS $\dfrac{-3+3\sqrt{5}}{2}-\sqrt{9-3(\dfrac{-3+3\sqrt{5}}{2})}$
$=3(\dfrac{-1+\sqrt{5}}{2})-3\sqrt{1-(\dfrac{-1+\sqrt{5}}{2})}$
RHS $0$
LHS $\ne$ RHS
$x=0 \color{red}{reject}$
$x=\dfrac{-3+3\sqrt{5}}{2}$
LHS $\dfrac{-3+3\sqrt{5}}{2}-\sqrt{9-3(\dfrac{-3+3\sqrt{5}}{2})}=0$
RHS $0$
LHS $=$ RHS
$x=\dfrac{-3+3\sqrt{5}}{2}$
