Solve 3(-1/2x-1)=1/2x+1+1/2x^-1-1/2x-3

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{2}i}{3}\approx -0.333333333+0.471404521i$$

$$x=\frac{-\sqrt{2}i-1}{3}\approx -0.333333333-0.471404521i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by -\frac{1}{2}x-1.

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

Combine \frac{1}{2}x and -\frac{1}{2}x to get 0.

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

Subtract 3 from 1 to get -2.

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

Subtract -2 from both sides.

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

The opposite of -2 is 2.

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

Subtract \frac{1}{2}x^{-1} from both sides.

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

Add -3 and 2 to get -1.

$$-\frac{3}{2}x-1-\frac{1}{2}x^{-1}=0$$

Reorder the terms.

$$-\frac{3}{2}x-1-\frac{1}{2}\times \frac{1}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x.

$$-\frac{3}{2}x\times 2x+2x\left(-1\right)-\frac{1}{2}\times 2=0$$

Multiply -\frac{3}{2} and 2 to get -3.

$$-3xx+2x\left(-1\right)-\frac{1}{2}\times 2=0$$

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

$$-3x^{2}+2x\left(-1\right)-\frac{1}{2}\times 2=0$$

Multiply 2 and -1 to get -2.

$$-3x^{2}-2x-\frac{1}{2}\times 2=0$$

Multiply -\frac{1}{2} and 2 to get -1.

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

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

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

Square -2.

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

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\left(-1\right)}}{2\left(-3\right)}$$

Multiply 12 times -1.

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

Add 4 to -12.

$$x=\frac{-\left(-2\right)±\sqrt{-8}}{2\left(-3\right)}$$

Take the square root of -8.

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

The opposite of -2 is 2.

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

Multiply 2 times -3.

$$x=\frac{2±2\sqrt{2}i}{-6}$$

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

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

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

$$x=\frac{-\sqrt{2}i-1}{3}$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 3 by -\frac{1}{2}x-1.

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

Combine \frac{1}{2}x and -\frac{1}{2}x to get 0.

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

Subtract 3 from 1 to get -2.

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

Subtract \frac{1}{2}x^{-1} from both sides.

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

Add 3 to both sides.

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

Add -2 and 3 to get 1.

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

Reorder the terms.

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

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x.

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

Multiply -\frac{3}{2} and 2 to get -3.

$$-3xx-\frac{1}{2}\times 2=2x$$

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

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

Multiply -\frac{1}{2} and 2 to get -1.

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

Subtract 2x from both sides.

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

Add 1 to both sides. Anything plus zero gives itself.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -2 by -3.

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

Divide 1 by -3.

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

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

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

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{2}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{2}{9}}$$

Simplify.

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

Subtract \frac{1}{3} from both sides of the equation.

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