Solve (x+1/x)^2+(x-1/x)^2=16

$$( x + \frac { 1 } { x } ) ^ { 2 } + ( x - \frac { 1 } { x } ) ^ { 2 } = 16$$

Solve for x

$$x=\frac{\sqrt{10}-\sqrt{6}}{2}\approx 0.356393959$$

$$x=\frac{\sqrt{6}-\sqrt{10}}{2}\approx -0.356393959$$

$$x=\frac{-\sqrt{6}-\sqrt{10}}{2}\approx -2.805883701$$

$$x = \frac{\sqrt{6} + \sqrt{10}}{2} \approx 2.805883701$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$\left(\frac{xx}{x}+\frac{1}{x}\right)^{2}+\left(x-\frac{1}{x}\right)^{2}=16$$

Since \frac{xx}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$\left(\frac{xx+1}{x}\right)^{2}+\left(x-\frac{1}{x}\right)^{2}=16$$

Do the multiplications in xx+1.

$$\left(\frac{x^{2}+1}{x}\right)^{2}+\left(x-\frac{1}{x}\right)^{2}=16$$

To raise \frac{x^{2}+1}{x} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x^{2}+1\right)^{2}}{x^{2}}+\left(x-\frac{1}{x}\right)^{2}=16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$\frac{\left(x^{2}+1\right)^{2}}{x^{2}}+\left(\frac{xx}{x}-\frac{1}{x}\right)^{2}=16$$

Since \frac{xx}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x^{2}+1\right)^{2}}{x^{2}}+\left(\frac{xx-1}{x}\right)^{2}=16$$

Do the multiplications in xx-1.

$$\frac{\left(x^{2}+1\right)^{2}}{x^{2}}+\left(\frac{x^{2}-1}{x}\right)^{2}=16$$

To raise \frac{x^{2}-1}{x} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(x^{2}+1\right)^{2}}{x^{2}}+\frac{\left(x^{2}-1\right)^{2}}{x^{2}}=16$$

Since \frac{\left(x^{2}+1\right)^{2}}{x^{2}} and \frac{\left(x^{2}-1\right)^{2}}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}+1\right)^{2}+\left(x^{2}-1\right)^{2}}{x^{2}}=16$$

Do the multiplications in \left(x^{2}+1\right)^{2}+\left(x^{2}-1\right)^{2}.

$$\frac{x^{4}+2x^{2}+1+x^{4}-2x^{2}+1}{x^{2}}=16$$

Combine like terms in x^{4}+2x^{2}+1+x^{4}-2x^{2}+1.

$$\frac{2x^{4}+2}{x^{2}}=16$$

Subtract 16 from both sides.

$$\frac{2x^{4}+2}{x^{2}}-16=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{x^{2}}{x^{2}}.

$$\frac{2x^{4}+2}{x^{2}}-\frac{16x^{2}}{x^{2}}=0$$

Since \frac{2x^{4}+2}{x^{2}} and \frac{16x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x^{4}+2-16x^{2}}{x^{2}}=0$$

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

$$2x^{4}+2-16x^{2}=0$$

Substitute t for x^{2}.

$$2t^{2}-16t+2=0$$

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 2 for a, -16 for b, and 2 for c in the quadratic formula.

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Do the calculations.

$$t=\frac{16±4\sqrt{15}}{4}$$

Solve the equation t=\frac{16±4\sqrt{15}}{4} when ± is plus and when ± is minus.

$$t=\sqrt{15}+4 t=4-\sqrt{15}$$

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

$$x=\frac{\sqrt{6}+\sqrt{10}}{2} x=-\frac{\sqrt{6}+\sqrt{10}}{2} x=-\frac{\sqrt{6}-\sqrt{10}}{2} x=\frac{\sqrt{6}-\sqrt{10}}{2}$$