Solve 9/x-2+6x/x+2=9x^2/x^2-4

$$\frac{ 9 }{ x-2 } + \frac{ 6x }{ x+2 } = \frac{ 9 { x }^{ 2 } }{ { x }^{ 2 } -4 }$$

Solve for x

$$x=-3$$

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,x^{2}-4.

$$\left(x+2\right)\times 9+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x+2 by 9.

$$9x+18+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x-2 by 6.

$$9x+18+\left(6x-12\right)x=9x^{2}$$

Use the distributive property to multiply 6x-12 by x.

$$9x+18+6x^{2}-12x=9x^{2}$$

Combine 9x and -12x to get -3x.

$$-3x+18+6x^{2}=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-3x+18+6x^{2}-9x^{2}=0$$

Combine 6x^{2} and -9x^{2} to get -3x^{2}.

$$-3x+18-3x^{2}=0$$

Divide both sides by 3.

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

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

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

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.

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

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 -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,x^{2}-4.

$$\left(x+2\right)\times 9+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x+2 by 9.

$$9x+18+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x-2 by 6.

$$9x+18+\left(6x-12\right)x=9x^{2}$$

Use the distributive property to multiply 6x-12 by x.

$$9x+18+6x^{2}-12x=9x^{2}$$

Combine 9x and -12x to get -3x.

$$-3x+18+6x^{2}=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-3x+18+6x^{2}-9x^{2}=0$$

Combine 6x^{2} and -9x^{2} to get -3x^{2}.

$$-3x+18-3x^{2}=0$$

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.

$$-3x^{2}-3x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -3 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-3\right)\times 18}}{2\left(-3\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-3\right)\times 18}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+12\times 18}}{2\left(-3\right)}$$

Multiply 12 times 18.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\left(-3\right)}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\left(-3\right)}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\left(-3\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{3±15}{-6}$$

Now solve the equation x=\frac{3±15}{-6} when ± is plus. Add 3 to 15.

$$x=\frac{18}{-6}$$

Divide 18 by -6.

$$x=-3$$

Now solve the equation x=\frac{3±15}{-6} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=-3 x=2$$

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,x^{2}-4.

$$\left(x+2\right)\times 9+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x+2 by 9.

$$9x+18+\left(x-2\right)\times 6x=9x^{2}$$

Use the distributive property to multiply x-2 by 6.

$$9x+18+\left(6x-12\right)x=9x^{2}$$

Use the distributive property to multiply 6x-12 by x.

$$9x+18+6x^{2}-12x=9x^{2}$$

Combine 9x and -12x to get -3x.

$$-3x+18+6x^{2}=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-3x+18+6x^{2}-9x^{2}=0$$

Combine 6x^{2} and -9x^{2} to get -3x^{2}.

$$-3x+18-3x^{2}=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$-3x-3x^{2}=-18$$

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.

$$-3x^{2}-3x=-18$$

Divide both sides by -3.

$$\frac{-3x^{2}-3x}{-3}=-\frac{18}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{3}{-3}\right)x=-\frac{18}{-3}$$

Divide -3 by -3.

$$x^{2}+x=-\frac{18}{-3}$$

Divide -18 by -3.

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

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.

$$x=-3$$