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\left(\sqrt{x}\right)^{2}=\left(\frac{8}{x}\right)^{2}
Square both sides of the equation.
x=\left(\frac{8}{x}\right)^{2}
Calculate \sqrt{x} to the power of 2 and get x.
x=\frac{8^{2}}{x^{2}}
To raise \frac{8}{x} to a power, raise both numerator and denominator to the power and then divide.
x=\frac{64}{x^{2}}
Calculate 8 to the power of 2 and get 64.
x-\frac{64}{x^{2}}=0
Subtract \frac{64}{x^{2}} from both sides.
\frac{xx^{2}}{x^{2}}-\frac{64}{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}-64}{x^{2}}=0
Since \frac{xx^{2}}{x^{2}} and \frac{64}{x^{2}} have the same denominator, subtract them by subtracting their numerators.
\frac{x^{3}-64}{x^{2}}=0
Do the multiplications in xx^{2}-64.
x^{3}-64=0
Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.
±64,±32,±16,±8,±4,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=4
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^{2}+4x+16=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-64 by x-4 to get x^{2}+4x+16. Solve the equation where the result equals to 0.
x=\frac{-4±\sqrt{4^{2}-4\times 1\times 16}}{2}
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, 4 for b, and 16 for c in the quadratic formula.
x=\frac{-4±\sqrt{-48}}{2}
Do the calculations.
x\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x=4
List all found solutions.
\sqrt{4}=\frac{8}{4}
Substitute 4 for x in the equation \sqrt{x}=\frac{8}{x}.
2=2
Simplify. The value x=4 satisfies the equation.
x=4
Equation \sqrt{x}=\frac{8}{x} has a unique solution.