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

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

Solve for x

$$x=-4$$

$$x=1$$

Steps Using Factoring

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,x-2,x^{2}-4.

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

Use the distributive property to multiply x+2 by x.

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

Combine x and 2x to get 3x.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

$$3x-4+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+3x-4=0$$

To solve the equation, factor x^{2}+3x-4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

To find equation solutions, solve x-1=0 and x+4=0.

$$x=1 x=-4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,x-2,x^{2}-4.

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

Use the distributive property to multiply x+2 by x.

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

Combine x and 2x to get 3x.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

$$3x-4+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+3x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-4\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite x^{2}+3x-4 as \left(x^{2}-x\right)+\left(4x-4\right).

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

Factor out x in the first and 4 in the second group.

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

Factor out common term x-1 by using distributive property.

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

To find equation solutions, solve x-1=0 and x+4=0.

$$x=1 x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,x-2,x^{2}-4.

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

Use the distributive property to multiply x+2 by x.

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

Combine x and 2x to get 3x.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

$$3x-4+x^{2}=0$$

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.

$$x^{2}+3x-4=0$$

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-3±\sqrt{9+16}}{2}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-3±5}{2}$$

Now solve the equation x=\frac{-3±5}{2} when ± is plus. Add -3 to 5.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-3±5}{2} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=1 x=-4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right), the least common multiple of x+2,x-2,x^{2}-4.

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

Use the distributive property to multiply x+2 by x.

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

Combine x and 2x to get 3x.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

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

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.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=1 x=-4$$