Solve 0=-12x^2-6x+126/left(x^2-6x+9right)^2

$$0 = \frac{ -12 { x }^{ 2 } -6x+126 }{ { \left( { x }^{ 2 } -6x+9 \right) }^{ 2 } }$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{4}.

$$0=-12x^{2}-6x+126$$

Swap sides so that all variable terms are on the left hand side.

$$-12x^{2}-6x+126=0$$

Divide both sides by 6.

$$-2x^{2}-x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2\times 21=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -1.

$$a=6 b=-7$$

Rewrite -2x^{2}-x+21 as \left(-2x^{2}+6x\right)+\left(-7x+21\right).

$$\left(-2x^{2}+6x\right)+\left(-7x+21\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(-x+3\right)+7\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x+7\right)$$

To find equation solutions, solve -x+3=0 and 2x+7=0.

$$x=3 x=-\frac{7}{2}$$

Variable x cannot be equal to 3.

$$x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{4}.

$$0=-12x^{2}-6x+126$$

Swap sides so that all variable terms are on the left hand side.

$$-12x^{2}-6x+126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -6 for b, and 126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-12\right)\times 126}}{2\left(-12\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-12\right)\times 126}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-6\right)±\sqrt{36+48\times 126}}{2\left(-12\right)}$$

Multiply 48 times 126.

$$x=\frac{-\left(-6\right)±\sqrt{36+6048}}{2\left(-12\right)}$$

Add 36 to 6048.

$$x=\frac{-\left(-6\right)±\sqrt{6084}}{2\left(-12\right)}$$

Take the square root of 6084.

$$x=\frac{-\left(-6\right)±78}{2\left(-12\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±78}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{6±78}{-24}$$

Now solve the equation x=\frac{6±78}{-24} when ± is plus. Add 6 to 78.

$$x=\frac{84}{-24}$$

Reduce the fraction \frac{84}{-24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{7}{2}$$

Now solve the equation x=\frac{6±78}{-24} when ± is minus. Subtract 78 from 6.

$$x=-\frac{72}{-24}$$

Divide -72 by -24.

$$x=3$$

The equation is now solved.

$$x=-\frac{7}{2} x=3$$

Variable x cannot be equal to 3.

$$x=-\frac{7}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{4}.

$$0=-12x^{2}-6x+126$$

Swap sides so that all variable terms are on the left hand side.

$$-12x^{2}-6x+126=0$$

Subtract 126 from both sides. Anything subtracted from zero gives its negation.

$$-12x^{2}-6x=-126$$

Divide both sides by -12.

$$\frac{-12x^{2}-6x}{-12}=-\frac{126}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{6}{-12}\right)x=-\frac{126}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{1}{2}x=-\frac{126}{-12}$$

Reduce the fraction \frac{-126}{-12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{1}{2}x=\frac{21}{2}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{21}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{21}{2}+\frac{1}{16}$$

Add \frac{21}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{169}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{13}{4} x+\frac{1}{4}=-\frac{13}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=3 x=-\frac{7}{2}$$

Variable x cannot be equal to 3.

$$x=-\frac{7}{2}$$