Solve 2x/x-4-1/x+4=8/x^2-16

$$\frac { 2 x } { x - 4 } - \frac { 1 } { x + 4 } = \frac { 8 } { x ^ { 2 } - 16 }$$

Solve for x

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

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x-4,x+4,x^{2}-16.

$$\left(x+4\right)\times 2x-\left(x-4\right)=8$$

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

$$\left(2x+8\right)x-\left(x-4\right)=8$$

Use the distributive property to multiply 2x+8 by x.

$$2x^{2}+8x-\left(x-4\right)=8$$

To find the opposite of x-4, find the opposite of each term.

$$2x^{2}+8x-x+4=8$$

Combine 8x and -x to get 7x.

$$2x^{2}+7x+4=8$$

Subtract 8 from both sides.

$$2x^{2}+7x+4-8=0$$

Subtract 8 from 4 to get -4.

$$2x^{2}+7x-4=0$$

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

$$a+b=7 ab=2\left(-4\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite 2x^{2}+7x-4 as \left(2x^{2}-x\right)+\left(8x-4\right).

$$\left(2x^{2}-x\right)+\left(8x-4\right)$$

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

$$x\left(2x-1\right)+4\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+4\right)$$

To find equation solutions, solve 2x-1=0 and x+4=0.

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

Variable x cannot be equal to -4.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x-4,x+4,x^{2}-16.

$$\left(x+4\right)\times 2x-\left(x-4\right)=8$$

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

$$\left(2x+8\right)x-\left(x-4\right)=8$$

Use the distributive property to multiply 2x+8 by x.

$$2x^{2}+8x-\left(x-4\right)=8$$

To find the opposite of x-4, find the opposite of each term.

$$2x^{2}+8x-x+4=8$$

Combine 8x and -x to get 7x.

$$2x^{2}+7x+4=8$$

Subtract 8 from both sides.

$$2x^{2}+7x+4-8=0$$

Subtract 8 from 4 to get -4.

$$2x^{2}+7x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-4\right)}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-4\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-7±\sqrt{49+32}}{2\times 2}$$

Add 49 to 32.

$$x=\frac{-7±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-7±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±9}{4}$$

Now solve the equation x=\frac{-7±9}{4} when ± is plus. Add -7 to 9.

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

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

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

Now solve the equation x=\frac{-7±9}{4} when ± is minus. Subtract 9 from -7.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

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

Variable x cannot be equal to -4.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x-4,x+4,x^{2}-16.

$$\left(x+4\right)\times 2x-\left(x-4\right)=8$$

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

$$\left(2x+8\right)x-\left(x-4\right)=8$$

Use the distributive property to multiply 2x+8 by x.

$$2x^{2}+8x-\left(x-4\right)=8$$

To find the opposite of x-4, find the opposite of each term.

$$2x^{2}+8x-x+4=8$$

Combine 8x and -x to get 7x.

$$2x^{2}+7x+4=8$$

Subtract 4 from both sides.

$$2x^{2}+7x=8-4$$

Subtract 4 from 8 to get 4.

$$2x^{2}+7x=4$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=\frac{4}{2}$$

Divide 4 by 2.

$$x^{2}+\frac{7}{2}x=2$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=2+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=2+\frac{49}{16}$$

Add 2 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{9}{4} x+\frac{7}{4}=-\frac{9}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

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

Variable x cannot be equal to -4.

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