Solve x^2/10=100/x

$$\frac{ { x }^{ 2 } }{ 10 } = \frac{ 100 }{ x }$$

Solve for x (complex solution)

$$x=-5\sqrt{3}i-5\approx -5-8.660254038i$$

$$x=10$$

$$x=-5+5\sqrt{3}i\approx -5+8.660254038i$$

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of 10,x.

$$xx^{2}=10\times 100$$

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

$$x^{3}=10\times 100$$

Multiply 10 and 100 to get 1000.

$$x^{3}=1000$$

Subtract 1000 from both sides.

$$x^{3}-1000=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1000 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1000,±500,±250,±200,±125,±100,±50,±40,±25,±20,±10,±8,±5,±4,±2,±1$$

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.

$$x=10$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-1000 by x-10 to get x^{2}+10x+100. Solve the equation where the result equals to 0.

$$x^{2}+10x+100=0$$

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, 10 for b, and 100 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 100}}{2}$$

Do the calculations.

$$x=\frac{-10±\sqrt{-300}}{2}$$

Solve the equation x^{2}+10x+100=0 when ± is plus and when ± is minus.

$$x=-5i\sqrt{3}-5 x=-5+5i\sqrt{3}$$

List all found solutions.

$$x=10 x=-5i\sqrt{3}-5 x=-5+5i\sqrt{3}$$

Solve for x

$$x=10$$

Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of 10,x.

$$xx^{2}=10\times 100$$

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

$$x^{3}=10\times 100$$

Multiply 10 and 100 to get 1000.

$$x^{3}=1000$$

Subtract 1000 from both sides.

$$x^{3}-1000=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1000 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1000,±500,±250,±200,±125,±100,±50,±40,±25,±20,±10,±8,±5,±4,±2,±1$$

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.

$$x=10$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-1000 by x-10 to get x^{2}+10x+100. Solve the equation where the result equals to 0.

$$x^{2}+10x+100=0$$

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, 10 for b, and 100 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 100}}{2}$$

Do the calculations.

$$x=\frac{-10±\sqrt{-300}}{2}$$

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.

$$x=10$$