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

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

Solve for x

$$x=-2$$

$$x=\frac{1}{2}=0.5$$

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.

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

Use the distributive property to multiply x+1 by 3.

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

Use the distributive property to multiply x-1 by 3.

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

Combine 3x and 3x to get 6x.

$$6x+3-3=-4\left(x-1\right)\left(x+1\right)$$

Subtract 3 from 3 to get 0.

$$6x=-4\left(x-1\right)\left(x+1\right)$$

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

$$6x=\left(-4x+4\right)\left(x+1\right)$$

Use the distributive property to multiply -4x+4 by x+1 and combine like terms.

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

Add 4x^{2} to both sides.

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

Subtract 4 from both sides.

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

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.

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

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

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

Square 6.

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

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\left(-4\right)}}{2\times 4}$$

Multiply -16 times -4.

$$x=\frac{-6±\sqrt{36+64}}{2\times 4}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\times 4}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±10}{8}$$

Now solve the equation x=\frac{-6±10}{8} when ± is plus. Add -6 to 10.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-6±10}{8} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

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

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.

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

Use the distributive property to multiply x+1 by 3.

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

Use the distributive property to multiply x-1 by 3.

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

Combine 3x and 3x to get 6x.

$$6x+3-3=-4\left(x-1\right)\left(x+1\right)$$

Subtract 3 from 3 to get 0.

$$6x=-4\left(x-1\right)\left(x+1\right)$$

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

$$6x=\left(-4x+4\right)\left(x+1\right)$$

Use the distributive property to multiply -4x+4 by x+1 and combine like terms.

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

Add 4x^{2} to both sides.

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

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.

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

Divide both sides by 4.

$$\frac{4x^{2}+6x}{4}=\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{6}{4}x=\frac{4}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Divide 4 by 4.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{5}{4} x+\frac{3}{4}=-\frac{5}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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