Solve x^2+1/x^2=38

$${ x }^{ 2 } + \frac{ 1 }{ { x }^{ 2 } } =38$$

Solve for x

$$x=-\left(\sqrt{10}+3\right)\approx -6.16227766$$

$$x=\sqrt{10}+3\approx 6.16227766$$

$$x=\sqrt{10}-3\approx 0.16227766$$

$$x=3-\sqrt{10}\approx -0.16227766$$

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}.

$$x^{2}x^{2}+1=38x^{2}$$

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

$$x^{4}+1=38x^{2}$$

Subtract 38x^{2} from both sides.

$$x^{4}+1-38x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-38t+1=0$$

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, -38 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{38±12\sqrt{10}}{2}$$

Solve the equation t=\frac{38±12\sqrt{10}}{2} when ± is plus and when ± is minus.

$$t=6\sqrt{10}+19 t=19-6\sqrt{10}$$

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

$$x=\sqrt{10}+3 x=-\left(\sqrt{10}+3\right) x=-\left(3-\sqrt{10}\right) x=3-\sqrt{10}$$