Solve 3=(x-1/x)^2

$$3 = ( x - \frac { 1 } { x } ) ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{3}-\sqrt{7}}{2}\approx -0.456850252$$

$$x = \frac{\sqrt{3} + \sqrt{7}}{2} \approx 2.188901059$$

$$x=\frac{\sqrt{7}-\sqrt{3}}{2}\approx 0.456850252$$

$$x=\frac{-\sqrt{3}-\sqrt{7}}{2}\approx -2.188901059$$

Steps Using the Quadratic Formula

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

$$3=\left(\frac{xx}{x}-\frac{1}{x}\right)^{2}$$

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

$$3=\left(\frac{xx-1}{x}\right)^{2}$$

Do the multiplications in xx-1.

$$3=\left(\frac{x^{2}-1}{x}\right)^{2}$$

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

$$3=\frac{\left(x^{2}-1\right)^{2}}{x^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$3=\frac{\left(x^{2}\right)^{2}-2x^{2}+1}{x^{2}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 3 from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x^{2}}{x^{2}}.

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

Since \frac{x^{4}-2x^{2}+1}{x^{2}} and \frac{3x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Combine like terms in x^{4}-2x^{2}+1-3x^{2}.

$$\frac{x^{4}-5x^{2}+1}{x^{2}}=0$$

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

$$x^{4}-5x^{2}+1=0$$

Substitute t for x^{2}.

$$t^{2}-5t+1=0$$

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, -5 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{5±\sqrt{21}}{2}$$

Solve the equation t=\frac{5±\sqrt{21}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{21}+5}{2} t=\frac{5-\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}-\sqrt{7}}{2} x=\frac{\sqrt{3}-\sqrt{7}}{2}$$