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

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

Solve for x (complex solution)

$$x=\frac{-2\sqrt{5}i-1}{3}\approx -0.333333333-1.490711985i$$

$$x=\frac{-1+2\sqrt{5}i}{3}\approx -0.333333333+1.490711985i$$

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 4\left(x-1\right)\left(x+1\right), the least common multiple of 4x-4,x^{2}-1,x+1.

$$\left(x+1\right)\left(1-3x\right)=4\left(1+x\right)-\left(4x-4\right)$$

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

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

Use the distributive property to multiply 4 by 1+x.

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

To find the opposite of 4x-4, find the opposite of each term.

$$-2x-3x^{2}+1=4+4x-4x+4$$

Combine 4x and -4x to get 0.

$$-2x-3x^{2}+1=4+4$$

Add 4 and 4 to get 8.

$$-2x-3x^{2}+1=8$$

Subtract 8 from both sides.

$$-2x-3x^{2}+1-8=0$$

Subtract 8 from 1 to get -7.

$$-2x-3x^{2}-7=0$$

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.

$$-3x^{2}-2x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -2 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)\left(-7\right)}}{2\left(-3\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)\left(-7\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\left(-7\right)}}{2\left(-3\right)}$$

Multiply 12 times -7.

$$x=\frac{-\left(-2\right)±\sqrt{4-84}}{2\left(-3\right)}$$

Add 4 to -84.

$$x=\frac{-\left(-2\right)±\sqrt{-80}}{2\left(-3\right)}$$

Take the square root of -80.

$$x=\frac{-\left(-2\right)±4\sqrt{5}i}{2\left(-3\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±4\sqrt{5}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{2±4\sqrt{5}i}{-6}$$

Now solve the equation x=\frac{2±4\sqrt{5}i}{-6} when ± is plus. Add 2 to 4i\sqrt{5}.

$$x=\frac{2+4\sqrt{5}i}{-6}$$

Divide 2+4i\sqrt{5} by -6.

$$x=\frac{-2\sqrt{5}i-1}{3}$$

Now solve the equation x=\frac{2±4\sqrt{5}i}{-6} when ± is minus. Subtract 4i\sqrt{5} from 2.

$$x=\frac{-4\sqrt{5}i+2}{-6}$$

Divide 2-4i\sqrt{5} by -6.

$$x=\frac{-1+2\sqrt{5}i}{3}$$

The equation is now solved.

$$x=\frac{-2\sqrt{5}i-1}{3} x=\frac{-1+2\sqrt{5}i}{3}$$

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 4\left(x-1\right)\left(x+1\right), the least common multiple of 4x-4,x^{2}-1,x+1.

$$\left(x+1\right)\left(1-3x\right)=4\left(1+x\right)-\left(4x-4\right)$$

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

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

Use the distributive property to multiply 4 by 1+x.

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

To find the opposite of 4x-4, find the opposite of each term.

$$-2x-3x^{2}+1=4+4x-4x+4$$

Combine 4x and -4x to get 0.

$$-2x-3x^{2}+1=4+4$$

Add 4 and 4 to get 8.

$$-2x-3x^{2}+1=8$$

Subtract 1 from both sides.

$$-2x-3x^{2}=8-1$$

Subtract 1 from 8 to get 7.

$$-2x-3x^{2}=7$$

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.

$$-3x^{2}-2x=7$$

Divide both sides by -3.

$$\frac{-3x^{2}-2x}{-3}=\frac{7}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{2}{-3}\right)x=\frac{7}{-3}$$

Divide -2 by -3.

$$x^{2}+\frac{2}{3}x=\frac{7}{-3}$$

Divide 7 by -3.

$$x^{2}+\frac{2}{3}x=-\frac{7}{3}$$

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{7}{3}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{7}{3}+\frac{1}{9}$$

Add -\frac{7}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{20}{9}$$

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}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{20}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{20}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{2\sqrt{5}i}{3} x+\frac{1}{3}=-\frac{2\sqrt{5}i}{3}$$

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

$$x=\frac{-1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i-1}{3}$$