Solve 0/x^2-36+6/(x+6)^2+1/2x+12=0

$$\frac { 0 } { x ^ { 2 } - 36 } + \frac { 6 } { ( x + 6 ) ^ { 2 } } + \frac { 1 } { 2 x + 12 } = 0$$

Solve for x

$$x=-18$$

Steps Using Factoring

Variable x cannot be equal to any of the values -6,6 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-6\right)\left(x+6\right)^{2}, the least common multiple of x^{2}-36,\left(x+6\right)^{2},2x+12.

$$\left(2x+12\right)\times 0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Anything times zero gives zero.

$$0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Use the distributive property to multiply 2x-12 by 6.

$$0+12x-72+x^{2}-36=0$$

Subtract 72 from 0 to get -72.

$$-72+12x+x^{2}-36=0$$

Subtract 36 from -72 to get -108.

$$-108+12x+x^{2}=0$$

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

$$x^{2}+12x-108=0$$

To solve the equation, factor x^{2}+12x-108 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.

$$a+b=12 ab=-108$$

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 -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 12.

$$a=-6 b=18$$

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

$$\left(x-6\right)\left(x+18\right)$$

To find equation solutions, solve x-6=0 and x+18=0.

$$x=6 x=-18$$

Variable x cannot be equal to 6.

$$x=-18$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -6,6 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-6\right)\left(x+6\right)^{2}, the least common multiple of x^{2}-36,\left(x+6\right)^{2},2x+12.

$$\left(2x+12\right)\times 0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Anything times zero gives zero.

$$0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Use the distributive property to multiply 2x-12 by 6.

$$0+12x-72+x^{2}-36=0$$

Subtract 72 from 0 to get -72.

$$-72+12x+x^{2}-36=0$$

Subtract 36 from -72 to get -108.

$$-108+12x+x^{2}=0$$

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

$$x^{2}+12x-108=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-108. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\left(-108\right)=-108$$

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 -108.

$$-1,108 -2,54 -3,36 -4,27 -6,18 -9,12$$

Calculate the sum for each pair.

$$-1+108=107 -2+54=52 -3+36=33 -4+27=23 -6+18=12 -9+12=3$$

The solution is the pair that gives sum 12.

$$a=-6 b=18$$

Rewrite x^{2}+12x-108 as \left(x^{2}-6x\right)+\left(18x-108\right).

$$\left(x^{2}-6x\right)+\left(18x-108\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-6\right)+18\left(x-6\right)$$

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

$$\left(x-6\right)\left(x+18\right)$$

To find equation solutions, solve x-6=0 and x+18=0.

$$x=6 x=-18$$

Variable x cannot be equal to 6.

$$x=-18$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -6,6 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-6\right)\left(x+6\right)^{2}, the least common multiple of x^{2}-36,\left(x+6\right)^{2},2x+12.

$$\left(2x+12\right)\times 0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Anything times zero gives zero.

$$0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Use the distributive property to multiply 2x-12 by 6.

$$0+12x-72+x^{2}-36=0$$

Subtract 72 from 0 to get -72.

$$-72+12x+x^{2}-36=0$$

Subtract 36 from -72 to get -108.

$$-108+12x+x^{2}=0$$

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.

$$x^{2}+12x-108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and -108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-108\right)}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-108\right)}}{2}$$

Multiply -4 times -108.

$$x=\frac{-12±\sqrt{144+432}}{2}$$

Add 144 to 432.

$$x=\frac{-12±\sqrt{576}}{2}$$

Take the square root of 576.

$$x=\frac{-12±24}{2}$$

Now solve the equation x=\frac{-12±24}{2} when ± is plus. Add -12 to 24.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{-12±24}{2} when ± is minus. Subtract 24 from -12.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

The equation is now solved.

$$x=6 x=-18$$

Variable x cannot be equal to 6.

$$x=-18$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -6,6 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-6\right)\left(x+6\right)^{2}, the least common multiple of x^{2}-36,\left(x+6\right)^{2},2x+12.

$$\left(2x+12\right)\times 0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Anything times zero gives zero.

$$0+\left(2x-12\right)\times 6+x^{2}-36=0$$

Use the distributive property to multiply 2x-12 by 6.

$$0+12x-72+x^{2}-36=0$$

Subtract 72 from 0 to get -72.

$$-72+12x+x^{2}-36=0$$

Subtract 36 from -72 to get -108.

$$-108+12x+x^{2}=0$$

Add 108 to both sides. Anything plus zero gives itself.

$$12x+x^{2}=108$$

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.

$$x^{2}+12x=108$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=108+6^{2}$$

Square 6.

$$x^{2}+12x+36=108+36$$

Add 108 to 36.

$$x^{2}+12x+36=144$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{144}$$

Simplify.

$$x+6=12 x+6=-12$$

Subtract 6 from both sides of the equation.

$$x=6 x=-18$$

Variable x cannot be equal to 6.

$$x=-18$$