Solve 0=3x+3/x+1

$$0=3x+ \frac{ 3 }{ x } +1$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{35}i}{6}\approx -0.166666667+0.986013297i$$

$$x=\frac{-\sqrt{35}i-1}{6}\approx -0.166666667-0.986013297i$$

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.

$$0=3xx+3+x$$

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

$$0=3x^{2}+3+x$$

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

$$3x^{2}+3+x=0$$

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.

$$3x^{2}+x+3=0$$

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

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

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-1±\sqrt{1-36}}{2\times 3}$$

Add 1 to -36.

$$x=\frac{-1±\sqrt{-35}}{2\times 3}$$

Take the square root of -35.

$$x=\frac{-1±\sqrt{35}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±\sqrt{35}i}{6}$$

Now solve the equation x=\frac{-1±\sqrt{35}i}{6} when ± is plus. Add -1 to i\sqrt{35}.

$$x=\frac{-1+\sqrt{35}i}{6}$$

Now solve the equation x=\frac{-1±\sqrt{35}i}{6} when ± is minus. Subtract i\sqrt{35} from -1.

$$x=\frac{-\sqrt{35}i-1}{6}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{35}i}{6} x=\frac{-\sqrt{35}i-1}{6}$$

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.

$$0=3xx+3+x$$

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

$$0=3x^{2}+3+x$$

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

$$3x^{2}+3+x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+x=-3$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

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

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=-1+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-1+\frac{1}{36}$$

Add -1 to \frac{1}{36}.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-\frac{35}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=-\frac{35}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{-\frac{35}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{35}i}{6} x+\frac{1}{6}=-\frac{\sqrt{35}i}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{-1+\sqrt{35}i}{6} x=\frac{-\sqrt{35}i-1}{6}$$