Answer
$x=\left\{ -\dfrac{\sqrt{2}}{2},0,\dfrac{\sqrt{2}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^2-2x^4=0
,$ use factoring. Then equate each factor to zero and solve for the value of the variable using the Square Root Principle.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=x^2,$ the factored form of the given expression is
\begin{array}{l}\require{cancel}
x^2(1-2x^2)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
x^2=0
\\\\\text{OR}\\\\
1-2x^2=0
.\end{array}
Isolating the squared variable results to
\begin{array}{l}\require{cancel}
x^2=0
\\\\\text{OR}\\\\
-2x^2=-1
\\\\
x^2=\dfrac{-1}{-2}
\\\\
x^2=\dfrac{1}{2}
.\end{array}
Taking the square root of both sides (Square Root Principle) results to
\begin{array}{l}\require{cancel}
x^2=0
\\\\
x=\pm\sqrt{0}
\\\\
x=0
\\\\\text{OR}\\\\
x^2=\dfrac{1}{2}
\\\\
x=\pm\sqrt{\dfrac{1}{2}}
\\\\
x=\pm\sqrt{\dfrac{1}{2}\cdot\dfrac{2}{2}}
\\\\
x=\pm\sqrt{\dfrac{1}{4}\cdot2}
\\\\
x=\pm\sqrt{\left( \dfrac{1}{2}\right)^2\cdot2}
\\\\
x=\pm\dfrac{1}{2}\sqrt{2}
\\\\
x=\pm\dfrac{\sqrt{2}}{2}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{\sqrt{2}}{2},0,\dfrac{\sqrt{2}}{2} \right\}
.$