Solve x^2+1/x^2=4

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

Solve for x

$$x=\frac{-\sqrt{2}-\sqrt{6}}{2}\approx -1.931851653$$

$$x = \frac{\sqrt{2} + \sqrt{6}}{2} \approx 1.931851653$$

$$x=\frac{\sqrt{6}-\sqrt{2}}{2}\approx 0.51763809$$

$$x=\frac{\sqrt{2}-\sqrt{6}}{2}\approx -0.51763809$$

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}.

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

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

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

Subtract 4x^{2} from both sides.

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

Substitute t for x^{2}.

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

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, -4 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{4±2\sqrt{3}}{2}$$

Solve the equation t=\frac{4±2\sqrt{3}}{2} when ± is plus and when ± is minus.

$$t=\sqrt{3}+2 t=2-\sqrt{3}$$

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

$$x=\frac{\sqrt{2}+\sqrt{6}}{2} x=-\frac{\sqrt{2}+\sqrt{6}}{2} x=-\frac{\sqrt{2}-\sqrt{6}}{2} x=\frac{\sqrt{2}-\sqrt{6}}{2}$$