Solve 1/x+3x=12

$$\frac{ 1 }{ x } +3x=12$$

Solve for x

$$x=\frac{\sqrt{33}}{3}+2\approx 3.914854216$$

$$x=-\frac{\sqrt{33}}{3}+2\approx 0.085145784$$

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.

$$1+3xx=12x$$

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

$$1+3x^{2}=12x$$

Subtract 12x from both sides.

$$1+3x^{2}-12x=0$$

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.

$$3x^{2}-12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3}}{2\times 3}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-12\right)±\sqrt{144-12}}{2\times 3}$$

Add 144 to -12.

$$x=\frac{-\left(-12\right)±\sqrt{132}}{2\times 3}$$

Take the square root of 132.

$$x=\frac{-\left(-12\right)±2\sqrt{33}}{2\times 3}$$

The opposite of -12 is 12.

$$x=\frac{12±2\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{12±2\sqrt{33}}{6}$$

Now solve the equation x=\frac{12±2\sqrt{33}}{6} when ± is plus. Add 12 to 2\sqrt{33}.

$$x=\frac{2\sqrt{33}+12}{6}$$

Divide 12+2\sqrt{33} by 6.

$$x=\frac{\sqrt{33}}{3}+2$$

Now solve the equation x=\frac{12±2\sqrt{33}}{6} when ± is minus. Subtract 2\sqrt{33} from 12.

$$x=\frac{12-2\sqrt{33}}{6}$$

Divide 12-2\sqrt{33} by 6.

$$x=-\frac{\sqrt{33}}{3}+2$$

The equation is now solved.

$$x=\frac{\sqrt{33}}{3}+2 x=-\frac{\sqrt{33}}{3}+2$$

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.

$$1+3xx=12x$$

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

$$1+3x^{2}=12x$$

Subtract 12x from both sides.

$$1+3x^{2}-12x=0$$

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

$$3x^{2}-12x=-1$$

Divide both sides by 3.

$$\frac{3x^{2}-12x}{3}=-\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{12}{3}\right)x=-\frac{1}{3}$$

Divide -12 by 3.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{1}{3}+\left(-2\right)^{2}$$

Square -2.

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

Add -\frac{1}{3} to 4.

$$x^{2}-4x+4=\frac{11}{3}$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=\frac{11}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{11}{3}}$$

Simplify.

$$x-2=\frac{\sqrt{33}}{3} x-2=-\frac{\sqrt{33}}{3}$$

Add 2 to both sides of the equation.

$$x=\frac{\sqrt{33}}{3}+2 x=-\frac{\sqrt{33}}{3}+2$$