Solve 1-2x/x+1=x/x-1

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

Solve for x (complex solution)

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x+1,x-1.

$$\left(x-1\right)\left(1-2x\right)=\left(x+1\right)x$$

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

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

Use the distributive property to multiply x+1 by x.

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

Subtract x^{2} from both sides.

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

Combine -2x^{2} and -x^{2} to get -3x^{2}.

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

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.

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

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}.

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

Square 2.

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

Multiply -4 times -3.

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

Multiply 12 times -1.

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

Add 4 to -12.

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

Take the square root of -8.

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

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

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x+1,x-1.

$$\left(x-1\right)\left(1-2x\right)=\left(x+1\right)x$$

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

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

Use the distributive property to multiply x+1 by x.

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

Subtract x^{2} from both sides.

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

Combine -2x^{2} and -x^{2} to get -3x^{2}.

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

Subtract x from both sides.

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

Combine 3x and -x to get 2x.

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

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

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

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.

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

Add \frac{1}{3} to both sides of the equation.

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