Solve sqrt{x}=1/x

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Calculate 1 to the power of 2 and get 1.

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

Subtract \frac{1}{x^{2}} from both sides.

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

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

$$\frac{xx^{2}}{x^{2}}-\frac{1}{x^{2}}=0$$

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

$$\frac{xx^{2}-1}{x^{2}}=0$$

Do the multiplications in xx^{2}-1.

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

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

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

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{x}=\frac{1}{x}.

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

$$x=1$$