Question

If $\sqrt{3-2x}=\sqrt{2x}+1$ then, $4x^2=$

• A. $1$
• B. $4$
• C. $2-2x$
• D. $4x-2$
• E. $6x-1$

Answer 1

The correct option is E $6x-1$

Work with the equation to create $4x^2$ on one side.
$\sqrt{3-2x}=\sqrt{2x}+1$
$(\sqrt{3-2x}{)}^2=(\sqrt{2x}+1{)}^2$ square both sides
$3-2x=2x+2\sqrt{2x+1}$
$2-4x=2\sqrt{2x}$ move all non-square-root terms to one side (i.e., subtract $2x$ and $1$)
$1-2x=\sqrt{2x}$ divide both sides by $2$
$(1-2x{)}^2=(\sqrt{2x}{)}^2$ square both sides
$1-4x+4x^2=2x$
$4x^2=6x-1$ isolate the $4x^2$ (add $4x$ and subtract $1$ from both sides)

So, the correct answer will be $6x-1$