Solve x+1/x=2/5

$$x + \frac { 1 } { x } = \frac { 2 } { 5 }$$

Solve for x (complex solution)

$$x=\frac{1+2\sqrt{6}i}{5}\approx 0.2+0.979795897i$$

$$x=\frac{-2\sqrt{6}i+1}{5}\approx 0.2-0.979795897i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5x, the least common multiple of x,5.

$$5xx+5=2x$$

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

$$5x^{2}+5=2x$$

Subtract 2x from both sides.

$$5x^{2}+5-2x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$5x^{2}-2x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-100}}{2\times 5}$$

Add 4 to -100.

$$x=\frac{-\left(-2\right)±\sqrt{-96}}{2\times 5}$$

Take the square root of -96.

$$x=\frac{-\left(-2\right)±4\sqrt{6}i}{2\times 5}$$

The opposite of -2 is 2.

$$x=\frac{2±4\sqrt{6}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{2±4\sqrt{6}i}{10}$$

Now solve the equation x=\frac{2±4\sqrt{6}i}{10} when ± is plus. Add 2 to 4i\sqrt{6}.

$$x=\frac{2+4\sqrt{6}i}{10}$$

Divide 2+4i\sqrt{6} by 10.

$$x=\frac{1+2\sqrt{6}i}{5}$$

Now solve the equation x=\frac{2±4\sqrt{6}i}{10} when ± is minus. Subtract 4i\sqrt{6} from 2.

$$x=\frac{-4\sqrt{6}i+2}{10}$$

Divide 2-4i\sqrt{6} by 10.

$$x=\frac{-2\sqrt{6}i+1}{5}$$

The equation is now solved.

$$x=\frac{1+2\sqrt{6}i}{5} x=\frac{-2\sqrt{6}i+1}{5}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5x, the least common multiple of x,5.

$$5xx+5=2x$$

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

$$5x^{2}+5=2x$$

Subtract 2x from both sides.

$$5x^{2}+5-2x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-2x=-5$$

Divide both sides by 5.

$$\frac{5x^{2}-2x}{5}=-\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{2}{5}x=-\frac{5}{5}$$

Divide -5 by 5.

$$x^{2}-\frac{2}{5}x=-1$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-1+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-1+\frac{1}{25}$$

Add -1 to \frac{1}{25}.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{24}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{24}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{24}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{2\sqrt{6}i}{5} x-\frac{1}{5}=-\frac{2\sqrt{6}i}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{1+2\sqrt{6}i}{5} x=\frac{-2\sqrt{6}i+1}{5}$$