Substitute \(y=\frac{{x}^{2}}{2}\) into \(y=\frac{1}{1+{x}^{2}}\).
Start with the original equation.
\[y=\frac{1}{1+{x}^{2}}\]
Let \(y=\frac{{x}^{2}}{2}\).
\[\frac{{x}^{2}}{2}=\frac{1}{1+{x}^{2}}\]
\[\frac{{x}^{2}}{2}=\frac{1}{1+{x}^{2}}\]
Solve for \(x\) in \(\frac{{x}^{2}}{2}=\frac{1}{1+{x}^{2}}\).
Solve for \(x\).
\[\frac{{x}^{2}}{2}=\frac{1}{1+{x}^{2}}\]
Multiply both sides by \(2\).
\[{x}^{2}=\frac{2}{1+{x}^{2}}\]
Multiply both sides by \(1+{x}^{2}\).
\[{x}^{2}(1+{x}^{2})=2\]
Expand.
\[{x}^{2}+{x}^{4}=2\]
Move all terms to one side.
\[{x}^{2}+{x}^{4}-2=0\]
Factor \({x}^{2}+{x}^{4}-2\).
Ask: Which two numbers add up to \(1\) and multiply to \(-2\)?
Rewrite the expression using the above.
\[({x}^{2}-1)({x}^{2}+2)\]
\[({x}^{2}-1)({x}^{2}+2)=0\]
Solve for \(x\).
Ask: When will \(({x}^{2}-1)({x}^{2}+2)\) equal zero?
When \({x}^{2}-1=0\) or \({x}^{2}+2=0\)
Solve each of the 2 equations above.
\[x=\pm 1,\pm \sqrt{2}\imath \]
\[x=\pm 1,\pm \sqrt{2}\imath \]
\[x=\pm 1,\pm \sqrt{2}\imath \]
Substitute \(x=\pm 1,\pm \sqrt{2}\imath \) into \(y=\frac{{x}^{2}}{2}\).
Start with the original equation.
\[y=\frac{{x}^{2}}{2}\]
Let \(x=\pm 1,\pm \sqrt{2}\imath \).
\[y=\frac{{1}^{2}}{2},\frac{{(-1)}^{2}}{2},\frac{{(\sqrt{2}\imath )}^{2}}{2},\frac{{(-\sqrt{2}\imath )}^{2}}{2}\]
Simplify.
\[y=\frac{1}{2},\frac{1}{2},-1,-1\]
\[y=\frac{1}{2},\frac{1}{2},-1,-1\]
Therefore,
\[\begin{aligned}&x=1,-1,\sqrt{2}\imath ,-\sqrt{2}\imath \\&y=\frac{1}{2},-1\end{aligned}\]
x=1,-1,sqrt(2)*IM,-sqrt(2)*IM;y=1/2,1/2,-1,-1
