Solve 4/x^2-1+3/2x+2=2x-5/2x^2-4x+2

$$\frac { 4 } { x ^ { 2 } - 1 } + \frac { 3 } { 2 x + 2 } = \frac { 2 x - 5 } { 2 x ^ { 2 } - 4 x + 2 }$$

Solve for x

$$x=-5$$

$$x=0$$

Steps Using Factoring

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

$$\left(2x-2\right)\times 4+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply 2x-2 by 4.

$$8x-8+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$8x-8+\left(x^{2}-2x+1\right)\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x^{2}-2x+1 by 3.

$$8x-8+3x^{2}-6x+3=\left(x+1\right)\left(2x-5\right)$$

Combine 8x and -6x to get 2x.

$$2x-8+3x^{2}+3=\left(x+1\right)\left(2x-5\right)$$

Add -8 and 3 to get -5.

$$2x-5+3x^{2}=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x+1 by 2x-5 and combine like terms.

$$2x-5+3x^{2}=2x^{2}-3x-5$$

Subtract 2x^{2} from both sides.

$$2x-5+3x^{2}-2x^{2}=-3x-5$$

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

$$2x-5+x^{2}=-3x-5$$

Add 3x to both sides.

$$2x-5+x^{2}+3x=-5$$

Combine 2x and 3x to get 5x.

$$5x-5+x^{2}=-5$$

Add 5 to both sides.

$$5x-5+x^{2}+5=0$$

Add -5 and 5 to get 0.

$$5x+x^{2}=0$$

Factor out x.

$$x\left(5+x\right)=0$$

To find equation solutions, solve x=0 and 5+x=0.

$$x=0 x=-5$$

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

$$\left(2x-2\right)\times 4+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply 2x-2 by 4.

$$8x-8+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$8x-8+\left(x^{2}-2x+1\right)\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x^{2}-2x+1 by 3.

$$8x-8+3x^{2}-6x+3=\left(x+1\right)\left(2x-5\right)$$

Combine 8x and -6x to get 2x.

$$2x-8+3x^{2}+3=\left(x+1\right)\left(2x-5\right)$$

Add -8 and 3 to get -5.

$$2x-5+3x^{2}=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x+1 by 2x-5 and combine like terms.

$$2x-5+3x^{2}=2x^{2}-3x-5$$

Subtract 2x^{2} from both sides.

$$2x-5+3x^{2}-2x^{2}=-3x-5$$

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

$$2x-5+x^{2}=-3x-5$$

Add 3x to both sides.

$$2x-5+x^{2}+3x=-5$$

Combine 2x and 3x to get 5x.

$$5x-5+x^{2}=-5$$

Add 5 to both sides.

$$5x-5+x^{2}+5=0$$

Add -5 and 5 to get 0.

$$5x+x^{2}=0$$

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.

$$x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2}$$

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

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

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

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=0 x=-5$$

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

$$\left(2x-2\right)\times 4+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply 2x-2 by 4.

$$8x-8+\left(x-1\right)^{2}\times 3=\left(x+1\right)\left(2x-5\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$8x-8+\left(x^{2}-2x+1\right)\times 3=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x^{2}-2x+1 by 3.

$$8x-8+3x^{2}-6x+3=\left(x+1\right)\left(2x-5\right)$$

Combine 8x and -6x to get 2x.

$$2x-8+3x^{2}+3=\left(x+1\right)\left(2x-5\right)$$

Add -8 and 3 to get -5.

$$2x-5+3x^{2}=\left(x+1\right)\left(2x-5\right)$$

Use the distributive property to multiply x+1 by 2x-5 and combine like terms.

$$2x-5+3x^{2}=2x^{2}-3x-5$$

Subtract 2x^{2} from both sides.

$$2x-5+3x^{2}-2x^{2}=-3x-5$$

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

$$2x-5+x^{2}=-3x-5$$

Add 3x to both sides.

$$2x-5+x^{2}+3x=-5$$

Combine 2x and 3x to get 5x.

$$5x-5+x^{2}=-5$$

Add 5 to both sides.

$$5x-5+x^{2}+5=0$$

Add -5 and 5 to get 0.

$$5x+x^{2}=0$$

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.

$$x^{2}+5x=0$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{5}{2} x+\frac{5}{2}=-\frac{5}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=0 x=-5$$