Solve sqrt{1/x}=x

$$\sqrt { \frac { 1 } { x } } = x$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{\frac{1}{x}} to the power of 2 and get \frac{1}{x}.

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

Multiply both sides of the equation by x.

$$1=xx^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$1=x^{3}$$

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

$$x^{3}=1$$

Subtract 1 from both sides.

$$x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-1 by x-1 to get x^{2}+x+1. Solve the equation where the result equals to 0.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

Solve the equation x^{2}+x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}$$

Substitute 1 for x in the equation \sqrt{\frac{1}{x}}=x.

$$\sqrt{\frac{1}{1}}=1$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Substitute \frac{-\sqrt{3}i-1}{2} for x in the equation \sqrt{\frac{1}{x}}=x.

$$\sqrt{\frac{1}{\frac{-\sqrt{3}i-1}{2}}}=\frac{-\sqrt{3}i-1}{2}$$

Simplify. The value x=\frac{-\sqrt{3}i-1}{2} satisfies the equation.

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

Substitute \frac{-1+\sqrt{3}i}{2} for x in the equation \sqrt{\frac{1}{x}}=x.

$$\sqrt{\frac{1}{\frac{-1+\sqrt{3}i}{2}}}=\frac{-1+\sqrt{3}i}{2}$$

Simplify. The value x=\frac{-1+\sqrt{3}i}{2} does not satisfy the equation.

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

List all solutions of \sqrt{\frac{1}{x}}=x.

$$x=1 x=\frac{-\sqrt{3}i-1}{2}$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{\frac{1}{x}} to the power of 2 and get \frac{1}{x}.

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

Multiply both sides of the equation by x.

$$1=xx^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$1=x^{3}$$

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

$$x^{3}=1$$

Subtract 1 from both sides.

$$x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-1 by x-1 to get x^{2}+x+1. Solve the equation where the result equals to 0.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1$$

Substitute 1 for x in the equation \sqrt{\frac{1}{x}}=x.

$$\sqrt{\frac{1}{1}}=1$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Equation \sqrt{\frac{1}{x}}=x has a unique solution.

$$x=1$$