Answer
No real number solutions
Work Step by Step
Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain:
Step 1: $(4x-1)^{2}=-2$
Step 2: $4x-1=\pm \sqrt {-2}$
Since $\sqrt {-2}$ does not have any real number solutions, $x$ does not have any real number solutions either.
