<s>[INST] How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$. 
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$


*

*make exponent postive


$2x\frac{1}{\sqrt{4-x}} - 3\sqrt{4-x} = 0$


*

*square everything 


$4x^2\frac{1}{4-x} - 9(4-x) =0$


*

*combine into one fraction


$\frac{1-9(4-x)[4x^2(4-x)]}{4x^2(4-x)} = 0$


*

*remove like terms


$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$
 [/INST] If $$\frac{2x}{\sqrt{4-x}}-3\sqrt{4-x}=0$$
Then
$$\frac{2x}{\sqrt{4-x}}=\frac{3\sqrt{4-x}}{1}$$
If we cross multiply we get that 
$$2x=3(4-x)=12-3x$$
You can go from here....
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