Solve 3x-1/3x^2+2x=1/x+2x-3/9x^2+12x+4

$$\frac { 3 x - 1 } { 3 x ^ { 2 } + 2 x } = \frac { 1 } { x } + \frac { 2 x - 3 } { 9 x ^ { 2 } + 12 x + 4 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-3}{2}\approx -1.5-0.866025404i$$

$$x=\frac{-3+\sqrt{3}i}{2}\approx -1.5+0.866025404i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{2}{3},0 since division by zero is not defined. Multiply both sides of the equation by x\left(3x+2\right)^{2}, the least common multiple of 3x^{2}+2x,x,9x^{2}+12x+4.

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

Use the distributive property to multiply 3x+2 by 3x-1 and combine like terms.

$$9x^{2}+3x-2=\left(3x+2\right)^{2}+x\left(2x-3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+3x-2=9x^{2}+12x+4+x\left(2x-3\right)$$

Use the distributive property to multiply x by 2x-3.

$$9x^{2}+3x-2=9x^{2}+12x+4+2x^{2}-3x$$

Combine 9x^{2} and 2x^{2} to get 11x^{2}.

$$9x^{2}+3x-2=11x^{2}+12x+4-3x$$

Combine 12x and -3x to get 9x.

$$9x^{2}+3x-2=11x^{2}+9x+4$$

Subtract 11x^{2} from both sides.

$$9x^{2}+3x-2-11x^{2}=9x+4$$

Combine 9x^{2} and -11x^{2} to get -2x^{2}.

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

Subtract 9x from both sides.

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

Combine 3x and -9x to get -6x.

$$-2x^{2}-6x-2=4$$

Subtract 4 from both sides.

$$-2x^{2}-6x-2-4=0$$

Subtract 4 from -2 to get -6.

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

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

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

Square -6.

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

Multiply -4 times -2.

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

Multiply 8 times -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-48}}{2\left(-2\right)}$$

Add 36 to -48.

$$x=\frac{-\left(-6\right)±\sqrt{-12}}{2\left(-2\right)}$$

Take the square root of -12.

$$x=\frac{-\left(-6\right)±2\sqrt{3}i}{2\left(-2\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{3}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{6±2\sqrt{3}i}{-4}$$

Now solve the equation x=\frac{6±2\sqrt{3}i}{-4} when ± is plus. Add 6 to 2i\sqrt{3}.

$$x=\frac{6+2\sqrt{3}i}{-4}$$

Divide 6+2i\sqrt{3} by -4.

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

Now solve the equation x=\frac{6±2\sqrt{3}i}{-4} when ± is minus. Subtract 2i\sqrt{3} from 6.

$$x=\frac{-2\sqrt{3}i+6}{-4}$$

Divide 6-2i\sqrt{3} by -4.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{2}{3},0 since division by zero is not defined. Multiply both sides of the equation by x\left(3x+2\right)^{2}, the least common multiple of 3x^{2}+2x,x,9x^{2}+12x+4.

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

Use the distributive property to multiply 3x+2 by 3x-1 and combine like terms.

$$9x^{2}+3x-2=\left(3x+2\right)^{2}+x\left(2x-3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$9x^{2}+3x-2=9x^{2}+12x+4+x\left(2x-3\right)$$

Use the distributive property to multiply x by 2x-3.

$$9x^{2}+3x-2=9x^{2}+12x+4+2x^{2}-3x$$

Combine 9x^{2} and 2x^{2} to get 11x^{2}.

$$9x^{2}+3x-2=11x^{2}+12x+4-3x$$

Combine 12x and -3x to get 9x.

$$9x^{2}+3x-2=11x^{2}+9x+4$$

Subtract 11x^{2} from both sides.

$$9x^{2}+3x-2-11x^{2}=9x+4$$

Combine 9x^{2} and -11x^{2} to get -2x^{2}.

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

Subtract 9x from both sides.

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

Combine 3x and -9x to get -6x.

$$-2x^{2}-6x-2=4$$

Add 2 to both sides.

$$-2x^{2}-6x=4+2$$

Add 4 and 2 to get 6.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -6 by -2.

$$x^{2}+3x=\frac{6}{-2}$$

Divide 6 by -2.

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

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.

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

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

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

Add -3 to \frac{9}{4}.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

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