Solve 2x+12/x=4

$$2x+ \frac{ 12 }{ x } = 4$$

Solve for x (complex solution)

$$x=1+\sqrt{5}i\approx 1+2.236067977i$$

$$x=-\sqrt{5}i+1\approx 1-2.236067977i$$

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.

$$2xx+12=4x$$

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

$$2x^{2}+12=4x$$

Subtract 4x from both sides.

$$2x^{2}+12-4x=0$$

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.

$$2x^{2}-4x+12=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-\left(-4\right)±\sqrt{16-96}}{2\times 2}$$

Add 16 to -96.

$$x=\frac{-\left(-4\right)±\sqrt{-80}}{2\times 2}$$

Take the square root of -80.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{5}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{5}i}{4}$$

Now solve the equation x=\frac{4±4\sqrt{5}i}{4} when ± is plus. Add 4 to 4i\sqrt{5}.

$$x=\frac{4+4\sqrt{5}i}{4}$$

Divide 4+4i\sqrt{5} by 4.

$$x=1+\sqrt{5}i$$

Now solve the equation x=\frac{4±4\sqrt{5}i}{4} when ± is minus. Subtract 4i\sqrt{5} from 4.

$$x=\frac{-4\sqrt{5}i+4}{4}$$

Divide 4-4i\sqrt{5} by 4.

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

The equation is now solved.

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

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.

$$2xx+12=4x$$

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

$$2x^{2}+12=4x$$

Subtract 4x from both sides.

$$2x^{2}+12-4x=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-4x=-12$$

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=-\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=-\frac{12}{2}$$

Divide -4 by 2.

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

Divide -12 by 2.

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

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

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

Add -6 to 1.

$$x^{2}-2x+1=-5$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=\sqrt{5}i x-1=-\sqrt{5}i$$

Add 1 to both sides of the equation.

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