Solve x^2-8/x^2-4+2/x+2=5/x-2

$$\frac { x ^ { 2 } - 8 } { x ^ { 2 } - 4 } + \frac { 2 } { x + 2 } = \frac { 5 } { x - 2 }$$

Solve for x

$$x = \frac{\sqrt{97} + 3}{2} \approx 6.424428901$$

$$x=\frac{3-\sqrt{97}}{2}\approx -3.424428901$$

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}-4,x+2,x-2.

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

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

$$x^{2}-8+2x-4=\left(x+2\right)\times 5$$

Subtract 4 from -8 to get -12.

$$x^{2}-12+2x=\left(x+2\right)\times 5$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-12+2x=5x+10$$

Subtract 5x from both sides.

$$x^{2}-12+2x-5x=10$$

Combine 2x and -5x to get -3x.

$$x^{2}-12-3x=10$$

Subtract 10 from both sides.

$$x^{2}-12-3x-10=0$$

Subtract 10 from -12 to get -22.

$$x^{2}-22-3x=0$$

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.

$$x^{2}-3x-22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-22\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-22\right)}}{2}$$

Multiply -4 times -22.

$$x=\frac{-\left(-3\right)±\sqrt{9+88}}{2}$$

Add 9 to 88.

$$x=\frac{-\left(-3\right)±\sqrt{97}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{97}}{2}$$

Now solve the equation x=\frac{3±\sqrt{97}}{2} when ± is plus. Add 3 to \sqrt{97}.

$$x=\frac{\sqrt{97}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{97}}{2} when ± is minus. Subtract \sqrt{97} from 3.

$$x=\frac{3-\sqrt{97}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{97}+3}{2} x=\frac{3-\sqrt{97}}{2}$$

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}-4,x+2,x-2.

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

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

$$x^{2}-8+2x-4=\left(x+2\right)\times 5$$

Subtract 4 from -8 to get -12.

$$x^{2}-12+2x=\left(x+2\right)\times 5$$

Use the distributive property to multiply x+2 by 5.

$$x^{2}-12+2x=5x+10$$

Subtract 5x from both sides.

$$x^{2}-12+2x-5x=10$$

Combine 2x and -5x to get -3x.

$$x^{2}-12-3x=10$$

Add 12 to both sides.

$$x^{2}-3x=10+12$$

Add 10 and 12 to get 22.

$$x^{2}-3x=22$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=22+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=22+\frac{9}{4}$$

Add 22 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{97}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{97}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{97}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{97}}{2} x-\frac{3}{2}=-\frac{\sqrt{97}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{97}+3}{2} x=\frac{3-\sqrt{97}}{2}$$