Solve for y
y=1
y=-1
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y^{4}+1-2y^{2}=0
Subtract 2y^{2} from both sides.
t^{2}-2t+1=0
Substitute t for y^{2}.
t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and 1 for c in the quadratic formula.
t=\frac{2±0}{2}
Do the calculations.
t=1
Solutions are the same.
y=-1 y=1
Since y=t^{2}, the solutions are obtained by evaluating y=±\sqrt{t} for positive t.
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