Solve 1/x+2x=6

$$\frac{ 1 }{ x } +2x = 6$$

Solve for x

$$x = \frac{\sqrt{7} + 3}{2} \approx 2.822875656$$

$$x=\frac{3-\sqrt{7}}{2}\approx 0.177124344$$

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 x.

$$1+2xx=6x$$

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

$$1+2x^{2}=6x$$

Subtract 6x from both sides.

$$1+2x^{2}-6x=0$$

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.

$$2x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8}}{2\times 2}$$

Add 36 to -8.

$$x=\frac{-\left(-6\right)±\sqrt{28}}{2\times 2}$$

Take the square root of 28.

$$x=\frac{-\left(-6\right)±2\sqrt{7}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{7}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2\sqrt{7}}{4}$$

Now solve the equation x=\frac{6±2\sqrt{7}}{4} when ± is plus. Add 6 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+6}{4}$$

Divide 6+2\sqrt{7} by 4.

$$x=\frac{\sqrt{7}+3}{2}$$

Now solve the equation x=\frac{6±2\sqrt{7}}{4} when ± is minus. Subtract 2\sqrt{7} from 6.

$$x=\frac{6-2\sqrt{7}}{4}$$

Divide 6-2\sqrt{7} by 4.

$$x=\frac{3-\sqrt{7}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{7}+3}{2} x=\frac{3-\sqrt{7}}{2}$$

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 x.

$$1+2xx=6x$$

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

$$1+2x^{2}=6x$$

Subtract 6x from both sides.

$$1+2x^{2}-6x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-6x=-1$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=-\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=-\frac{1}{2}$$

Divide -6 by 2.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-\frac{1}{2}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-\frac{1}{2}+\frac{9}{4}$$

Add -\frac{1}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{7}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{7}}{2} x-\frac{3}{2}=-\frac{\sqrt{7}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{7}+3}{2} x=\frac{3-\sqrt{7}}{2}$$