Solve x^2+1/x^2=2

$${ x }^{ 2 } + \frac{ 1 }{ { x }^{ 2 } } =2$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$x^{2}x^{2}+1=2x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$x^{4}+1=2x^{2}$$

Subtract 2x^{2} from both sides.

$$x^{4}+1-2x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-2t+1=0$$

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.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{2±0}{2}$$

Solutions are the same.

$$t=1$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=-1 x=1$$