Solve {l}{x-2/x+2=16/x^2-4}{=}

$$\left. \begin{array} { l } { \frac { x - 2 } { x + 2 } = \frac { 16 } { x ^ { 2 } - 4 } } \\ { = } \end{array} \right.$$

Solve for x

$$x=6$$

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.

$$\left(x-2\right)\left(x-2\right)=16$$

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

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

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

$$x^{2}-4x+4=16$$

Subtract 16 from both sides.

$$x^{2}-4x+4-16=0$$

Subtract 16 from 4 to get -12.

$$x^{2}-4x-12=0$$

To solve the equation, factor x^{2}-4x-12 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.

$$a+b=-4 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

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

$$\left(x-6\right)\left(x+2\right)$$

To find equation solutions, solve x-6=0 and x+2=0.

$$x=6 x=-2$$

Variable x cannot be equal to -2.

$$x=6$$

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.

$$\left(x-2\right)\left(x-2\right)=16$$

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

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

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

$$x^{2}-4x+4=16$$

Subtract 16 from both sides.

$$x^{2}-4x+4-16=0$$

Subtract 16 from 4 to get -12.

$$x^{2}-4x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite x^{2}-4x-12 as \left(x^{2}-6x\right)+\left(2x-12\right).

$$\left(x^{2}-6x\right)+\left(2x-12\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-6\right)+2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+2\right)$$

To find equation solutions, solve x-6=0 and x+2=0.

$$x=6 x=-2$$

Variable x cannot be equal to -2.

$$x=6$$

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.

$$\left(x-2\right)\left(x-2\right)=16$$

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

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

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

$$x^{2}-4x+4=16$$

Subtract 16 from both sides.

$$x^{2}-4x+4-16=0$$

Subtract 16 from 4 to get -12.

$$x^{2}-4x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-\left(-4\right)±8}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±8}{2}$$

Now solve the equation x=\frac{4±8}{2} when ± is plus. Add 4 to 8.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{4±8}{2} when ± is minus. Subtract 8 from 4.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=6 x=-2$$

Variable x cannot be equal to -2.

$$x=6$$

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.

$$\left(x-2\right)\left(x-2\right)=16$$

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

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-2=4 x-2=-4$$

Add 2 to both sides of the equation.

$$x=6 x=-2$$

Variable x cannot be equal to -2.

$$x=6$$