Solve x^2+1/x^2=3

$$x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = 3$$

Solve for x

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

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}.

$$x^{2}x^{2}+1=3x^{2}$$

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

$$x^{4}+1=3x^{2}$$

Subtract 3x^{2} from both sides.

$$x^{4}+1-3x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-3t+1=0$$

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, -3 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{3±\sqrt{5}}{2}$$

Solve the equation t=\frac{3±\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}+1}{2} x=-\frac{\sqrt{5}+1}{2} x=-\frac{1-\sqrt{5}}{2} x=\frac{1-\sqrt{5}}{2}$$