Solve x+sqrt{11/x}=10

$$x + \sqrt { \frac { 11 } { x } } = 10$$

Solve for x

$$x=\frac{10\left(\sqrt{3}\sin(\frac{\arccos(\frac{3\sqrt{330}}{200})}{3})+\cos(\frac{\arccos(\frac{3\sqrt{330}}{200})}{3})\right)^{2}}{3}\approx 8.887482194$$

$$x=\frac{\left(3\sqrt{10}\sin(\frac{\arccos(\frac{3\sqrt{330}}{200})}{3})-\sqrt{30}\cos(\frac{\arccos(\frac{3\sqrt{330}}{200})}{3})\right)^{2}}{9}\approx 0.112517806$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{\frac{11}{x}}=10-x$$

Square both sides of the equation.

$$\left(\sqrt{\frac{11}{x}}\right)^{2}=\left(10-x\right)^{2}$$

Calculate \sqrt{\frac{11}{x}} to the power of 2 and get \frac{11}{x}.

$$\frac{11}{x}=\left(10-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$\frac{11}{x}=100-20x+x^{2}$$

Multiply both sides of the equation by x.

$$11=x\times 100-20xx+xx^{2}$$

Multiply x and x to get x^{2}.

$$11=x\times 100-20x^{2}+xx^{2}$$

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

$$11=x\times 100-20x^{2}+x^{3}$$

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

$$x\times 100-20x^{2}+x^{3}=11$$

Subtract 11 from both sides.

$$x\times 100-20x^{2}+x^{3}-11=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}-20x^{2}+100x-11=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -11 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±11,±1$$

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.

$$x=11$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-20x^{2}+100x-11 by x-11 to get x^{2}-9x+1. Solve the equation where the result equals to 0.

$$x^{2}-9x+1=0$$

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, -9 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{9±\sqrt{77}}{2}$$

Solve the equation x^{2}-9x+1=0 when ± is plus and when ± is minus.

$$x=\frac{9-\sqrt{77}}{2} x=\frac{\sqrt{77}+9}{2}$$

List all found solutions.

$$x=11 x=\frac{9-\sqrt{77}}{2} x=\frac{\sqrt{77}+9}{2}$$

Substitute 11 for x in the equation x+\sqrt{\frac{11}{x}}=10.

$$11+\sqrt{\frac{11}{11}}=10$$

Simplify. The value x=11 does not satisfy the equation.

$$12=10$$

Substitute \frac{9-\sqrt{77}}{2} for x in the equation x+\sqrt{\frac{11}{x}}=10.

$$\frac{9-\sqrt{77}}{2}+\sqrt{\frac{11}{\frac{9-\sqrt{77}}{2}}}=10$$

Simplify. The value x=\frac{9-\sqrt{77}}{2} satisfies the equation.

$$10=10$$

Substitute \frac{\sqrt{77}+9}{2} for x in the equation x+\sqrt{\frac{11}{x}}=10.

$$\frac{\sqrt{77}+9}{2}+\sqrt{\frac{11}{\frac{\sqrt{77}+9}{2}}}=10$$

Simplify. The value x=\frac{\sqrt{77}+9}{2} satisfies the equation.

$$10=10$$

List all solutions of \sqrt{\frac{11}{x}}=10-x.

$$x=\frac{9-\sqrt{77}}{2} x=\frac{\sqrt{77}+9}{2}$$