Answer
The solution is $x=1$.
Work Step by Step
The given equation is
$\Rightarrow \sqrt{4-3x}=x$
Square each side of the equation.
$\Rightarrow (\sqrt{4-3x})^2=x^2$
Simplify.
$\Rightarrow 4-3x=x^2$
Add $3x-4$ to each side.
$\Rightarrow 4-3x+3x-4=x^2+3x-4$
Simplify.
$\Rightarrow 0=x^2+3x-4$
Factor.
$\Rightarrow 0=(x+4)(x-1)$
Use zero-product property.
$\Rightarrow x+4=0$ or $x-1=0$
Solve for $x$.
$\Rightarrow x=-4$ or $x=1$
Check $x=-4$.
$\Rightarrow \sqrt{4-3x}=x$
$\Rightarrow \sqrt{4-3(-4)}=-4$
$\Rightarrow \sqrt{4+12}=-4$
$\Rightarrow \sqrt{16}=-4$
$\Rightarrow 4=-4$
False.
Check $x=1$.
$\Rightarrow \sqrt{4-3x}=x$
$\Rightarrow \sqrt{4-3(1)}=1$
$\Rightarrow \sqrt{4-3}=1$
$\Rightarrow \sqrt{1}=1$
$\Rightarrow 1=1$
True.
Hence, the solution is $x=1$.
