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

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

Solve for x

$$x=3$$

$$x=5$$

Steps Using Factoring By Grouping

Multiply 2 and 2 to get 4.

$$4\left(x-3\right)^{2}=x^{2}-9$$

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

$$4\left(x^{2}-6x+9\right)=x^{2}-9$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x^{2}-9$$

Subtract x^{2} from both sides.

$$4x^{2}-24x+36-x^{2}=-9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-24x+36=-9$$

Add 9 to both sides.

$$3x^{2}-24x+36+9=0$$

Add 36 and 9 to get 45.

$$3x^{2}-24x+45=0$$

Divide both sides by 3.

$$x^{2}-8x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-3\right)$$

To find equation solutions, solve x-5=0 and x-3=0.

$$x=5 x=3$$

Steps Using the Quadratic Formula

Multiply 2 and 2 to get 4.

$$4\left(x-3\right)^{2}=x^{2}-9$$

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

$$4\left(x^{2}-6x+9\right)=x^{2}-9$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x^{2}-9$$

Subtract x^{2} from both sides.

$$4x^{2}-24x+36-x^{2}=-9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-24x+36=-9$$

Add 9 to both sides.

$$3x^{2}-24x+36+9=0$$

Add 36 and 9 to get 45.

$$3x^{2}-24x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -24 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 3\times 45}}{2\times 3}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 3\times 45}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-24\right)±\sqrt{576-12\times 45}}{2\times 3}$$

Multiply -12 times 45.

$$x=\frac{-\left(-24\right)±\sqrt{576-540}}{2\times 3}$$

Add 576 to -540.

$$x=\frac{-\left(-24\right)±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$x=\frac{-\left(-24\right)±6}{2\times 3}$$

The opposite of -24 is 24.

$$x=\frac{24±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{24±6}{6}$$

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

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

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

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

Divide 18 by 6.

$$x=3$$

The equation is now solved.

$$x=5 x=3$$

Steps for Completing the Square

Multiply 2 and 2 to get 4.

$$4\left(x-3\right)^{2}=x^{2}-9$$

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

$$4\left(x^{2}-6x+9\right)=x^{2}-9$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x^{2}-9$$

Subtract x^{2} from both sides.

$$4x^{2}-24x+36-x^{2}=-9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-24x+36=-9$$

Subtract 36 from both sides.

$$3x^{2}-24x=-9-36$$

Subtract 36 from -9 to get -45.

$$3x^{2}-24x=-45$$

Divide both sides by 3.

$$\frac{3x^{2}-24x}{3}=-\frac{45}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{24}{3}\right)x=-\frac{45}{3}$$

Divide -24 by 3.

$$x^{2}-8x=-\frac{45}{3}$$

Divide -45 by 3.

$$x^{2}-8x=-15$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

$$x^{2}-8x+16=1$$

Factor x^{2}-8x+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}.

$$\left(x-4\right)^{2}=1$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$