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


$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$

This step is wrong 

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

Once you square everything you get (i.e. $a=b$, then $a^2=b^2$, then $a^2-b^2=0$)
$$4x^2\frac{1}{4-x} - 9(4-x) =0$$
Now multiply by $(4-x)$, we get
$$4x^2 - 9(4-x)^2 =0$$
or
$$(2x)^2 - [3(4-x)]^2 =0$$
which gives 
$$(2x -12 + 3x)(2x +12 - 3x)=0$$
Finally we get
$$(5x -12 )(12 - x)=0$$
Obviously $12=x$ is not accepted, but $x = \frac{12}{5}$ is.
