Solve x^2-9=2(x+3)^2

$$x ^ { 2 } - 9 = 2 ( x + 3 ) ^ { 2 }$$

Solve for x

$$x=-9$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Subtract 12x from both sides.

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

Subtract 18 from both sides.

$$-x^{2}-9-12x-18=0$$

Subtract 18 from -9 to get -27.

$$-x^{2}-27-12x=0$$

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

$$-x^{2}-12x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-\left(-27\right)=27$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 27.

$$-1,-27 -3,-9$$

Calculate the sum for each pair.

$$-1-27=-28 -3-9=-12$$

The solution is the pair that gives sum -12.

$$a=-3 b=-9$$

Rewrite -x^{2}-12x-27 as \left(-x^{2}-3x\right)+\left(-9x-27\right).

$$\left(-x^{2}-3x\right)+\left(-9x-27\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(-x-3\right)+9\left(-x-3\right)$$

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

$$\left(-x-3\right)\left(x+9\right)$$

To find equation solutions, solve -x-3=0 and x+9=0.

$$x=-3 x=-9$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Subtract 12x from both sides.

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

Subtract 18 from both sides.

$$-x^{2}-9-12x-18=0$$

Subtract 18 from -9 to get -27.

$$-x^{2}-27-12x=0$$

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.

$$-x^{2}-12x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\left(-27\right)}}{2\left(-1\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\left(-27\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-12\right)±\sqrt{144+4\left(-27\right)}}{2\left(-1\right)}$$

Multiply 4 times -27.

$$x=\frac{-\left(-12\right)±\sqrt{144-108}}{2\left(-1\right)}$$

Add 144 to -108.

$$x=\frac{-\left(-12\right)±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{-\left(-12\right)±6}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{12±6}{-2}$$

Now solve the equation x=\frac{12±6}{-2} when ± is plus. Add 12 to 6.

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

Divide 18 by -2.

$$x=-9$$

Now solve the equation x=\frac{12±6}{-2} when ± is minus. Subtract 6 from 12.

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

The equation is now solved.

$$x=-9 x=-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Subtract 12x from both sides.

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

Add 9 to both sides.

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

Add 18 and 9 to get 27.

$$-x^{2}-12x=27$$

Divide both sides by -1.

$$\frac{-x^{2}-12x}{-1}=\frac{27}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{12}{-1}\right)x=\frac{27}{-1}$$

Divide -12 by -1.

$$x^{2}+12x=\frac{27}{-1}$$

Divide 27 by -1.

$$x^{2}+12x=-27$$

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.

$$x^{2}+12x+6^{2}=-27+6^{2}$$

Square 6.

$$x^{2}+12x+36=-27+36$$

Add -27 to 36.

$$x^{2}+12x+36=9$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+6=3 x+6=-3$$

Subtract 6 from both sides of the equation.

$$x=-3 x=-9$$