Answer: OPTION C
Step-by-step explanation:
Given the equation [tex]\frac{\sqrt{3-2x} }{\sqrt{4x} } =2[/tex], you need to solve for the variable "x".
First, you need to multiply both sides of the equation by [tex]\sqrt{4x}[/tex]:
[tex](\frac{\sqrt{3-2x}}{\sqrt{4x}})(\sqrt{4x} })=2(\sqrt{4x} })\\\\\sqrt{3-2x}=2\sqrt{4x}[/tex]
Now you need to square both sides of the equation:
[tex](\sqrt{3-2x})^2=(2\sqrt{4x})^2\\\\3-2x=4(4x)\\\\3-2x=16x[/tex]
Subtrac 3 and 16x from both sides:
[tex]3-2x-(3)-(16x)=16x-(3)-(16x)\\\\-18x=-3\\[/tex]
Divide both sides by -18:
[tex]\frac{-18x}{-18}=\frac{-3}{-18}\\\\x=\frac{1}{6}[/tex]
