Question

What is the solution to the equation below?

Answer 1

Answer: OPTION C

Step-by-step explanation:

 Given the equation $\frac{\sqrt{3-2x} }{\sqrt{4x} } =2$, you need to solve for  the variable "x".

First, you need to multiply both sides of the equation by $\sqrt{4x}$:

$(\frac{\sqrt{3-2x}}{\sqrt{4x}})(\sqrt{4x} })=2(\sqrt{4x} })\\\\\sqrt{3-2x}=2\sqrt{4x}$

Now you need to square both sides of the equation:

$(\sqrt{3-2x})^2=(2\sqrt{4x})^2\\\\3-2x=4(4x)\\\\3-2x=16x$

Subtrac 3 and 16x from both sides:

$3-2x-(3)-(16x)=16x-(3)-(16x)\\\\-18x=-3$

Divide both sides by -18:

$\frac{-18x}{-18}=\frac{-3}{-18}\\\\x=\frac{1}{6}$