Answer
$x=0$, $x=2$ and $x=-2$ satisfy the given conditions
Work Step by Step
$y_{1}=(x^{2}-1)^{2}$, $y_{2}=2(x^{2}-1)$, and $y_{1}$ exceeds $y_{2}$ by $3$
If $y_{1}$ exceeds $y_{2}$, then $y_{2}+3=y_{1}$
If $y_{2}+3=y_{1}$, then $y_{1}-y_{2}$ is equal to $3$ and $(x^{2}-1)^{2}-2(x^{2}-1)$ is also equal to $3$:
$(x^{2}-1)^{2}-2(x^{2}-1)=3$
Evaluate all operations indicated on the left side:
$x^{4}-2x^{2}+1-2x^{2}+2=3$
Take $3$ to the left side and simplify:
$x^{4}-2x^{2}+1-2x^{2}+2-3=0$
$x^{4}-4x^{2}=0$
Take out common factor $x^{2}$ from the left side:
$x^{2}(x^{2}-4)=0$
Factor $x^{2}-4$:
$x^{2}(x-2)(x+2)=0$
Set all factors equal to $0$ and solve each individual equation for $x$:
$x^{2}=0$
$x=0$
$x-2=0$
$x=2$
$x+2=0$
$x=-2$
$x=0$, $x=2$ and $x=-2$ satisfy the given conditions
