Solve x+2/x-2-16/x^2-4=3/x+2

$$\frac { x + 2 } { x - 2 } - \frac { 16 } { x ^ { 2 } - 4 } = \frac { 3 } { x + 2 }$$

Solve for x

$$x=-3$$

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}-4,x+2.

$$\left(x+2\right)\left(x+2\right)-16=\left(x-2\right)\times 3$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

$$\left(x+2\right)^{2}-16=\left(x-2\right)\times 3$$

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

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

Subtract 16 from 4 to get -12.

$$x^{2}+4x-12=\left(x-2\right)\times 3$$

Use the distributive property to multiply x-2 by 3.

$$x^{2}+4x-12=3x-6$$

Subtract 3x from both sides.

$$x^{2}+4x-12-3x=-6$$

Combine 4x and -3x to get x.

$$x^{2}+x-12=-6$$

Add 6 to both sides.

$$x^{2}+x-12+6=0$$

Add -12 and 6 to get -6.

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

To solve the equation, factor x^{2}+x-6 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.

$$a+b=1 ab=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

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

$$\left(x-2\right)\left(x+3\right)$$

To find equation solutions, solve x-2=0 and x+3=0.

$$x=2 x=-3$$

Variable x cannot be equal to 2.

$$x=-3$$

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}-4,x+2.

$$\left(x+2\right)\left(x+2\right)-16=\left(x-2\right)\times 3$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

$$\left(x+2\right)^{2}-16=\left(x-2\right)\times 3$$

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

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

Subtract 16 from 4 to get -12.

$$x^{2}+4x-12=\left(x-2\right)\times 3$$

Use the distributive property to multiply x-2 by 3.

$$x^{2}+4x-12=3x-6$$

Subtract 3x from both sides.

$$x^{2}+4x-12-3x=-6$$

Combine 4x and -3x to get x.

$$x^{2}+x-12=-6$$

Add 6 to both sides.

$$x^{2}+x-12+6=0$$

Add -12 and 6 to get -6.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite x^{2}+x-6 as \left(x^{2}-2x\right)+\left(3x-6\right).

$$\left(x^{2}-2x\right)+\left(3x-6\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-2\right)+3\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+3\right)$$

To find equation solutions, solve x-2=0 and x+3=0.

$$x=2 x=-3$$

Variable x cannot be equal to 2.

$$x=-3$$

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}-4,x+2.

$$\left(x+2\right)\left(x+2\right)-16=\left(x-2\right)\times 3$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

$$\left(x+2\right)^{2}-16=\left(x-2\right)\times 3$$

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

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

Subtract 16 from 4 to get -12.

$$x^{2}+4x-12=\left(x-2\right)\times 3$$

Use the distributive property to multiply x-2 by 3.

$$x^{2}+4x-12=3x-6$$

Subtract 3x from both sides.

$$x^{2}+4x-12-3x=-6$$

Combine 4x and -3x to get x.

$$x^{2}+x-12=-6$$

Add 6 to both sides.

$$x^{2}+x-12+6=0$$

Add -12 and 6 to get -6.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-1±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-1±5}{2}$$

Now solve the equation x=\frac{-1±5}{2} when ± is plus. Add -1 to 5.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-1±5}{2} when ± is minus. Subtract 5 from -1.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=2 x=-3$$

Variable x cannot be equal to 2.

$$x=-3$$

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}-4,x+2.

$$\left(x+2\right)\left(x+2\right)-16=\left(x-2\right)\times 3$$

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

$$\left(x+2\right)^{2}-16=\left(x-2\right)\times 3$$

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

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

Subtract 16 from 4 to get -12.

$$x^{2}+4x-12=\left(x-2\right)\times 3$$

Use the distributive property to multiply x-2 by 3.

$$x^{2}+4x-12=3x-6$$

Subtract 3x from both sides.

$$x^{2}+4x-12-3x=-6$$

Combine 4x and -3x to get x.

$$x^{2}+x-12=-6$$

Add 12 to both sides.

$$x^{2}+x=-6+12$$

Add -6 and 12 to get 6.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=2 x=-3$$

Variable x cannot be equal to 2.

$$x=-3$$