Solve (x-2)^2=3(x^2-4)

$$( x - 2 ) ^ { 2 } = 3 ( x ^ { 2 } - 4 )$$

Solve for x

$$x=2$$

$$x=-4$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 3 by x^{2}-4.

$$x^{2}-4x+4=3x^{2}-12$$

Subtract 3x^{2} from both sides.

$$x^{2}-4x+4-3x^{2}=-12$$

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

$$-2x^{2}-4x+4=-12$$

Add 12 to both sides.

$$-2x^{2}-4x+4+12=0$$

Add 4 and 12 to get 16.

$$-2x^{2}-4x+16=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-8=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$a=2 b=-4$$

Rewrite -x^{2}-2x+8 as \left(-x^{2}+2x\right)+\left(-4x+8\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(-x+2\right)+4\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+4\right)$$

To find equation solutions, solve -x+2=0 and x+4=0.

$$x=2 x=-4$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 3 by x^{2}-4.

$$x^{2}-4x+4=3x^{2}-12$$

Subtract 3x^{2} from both sides.

$$x^{2}-4x+4-3x^{2}=-12$$

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

$$-2x^{2}-4x+4=-12$$

Add 12 to both sides.

$$-2x^{2}-4x+4+12=0$$

Add 4 and 12 to get 16.

$$-2x^{2}-4x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -4 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 16}}{2\left(-2\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 16}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 16}}{2\left(-2\right)}$$

Multiply 8 times 16.

$$x=\frac{-\left(-4\right)±\sqrt{16+128}}{2\left(-2\right)}$$

Add 16 to 128.

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

Take the square root of 144.

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

The opposite of -4 is 4.

$$x=\frac{4±12}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±12}{-4}$$

Now solve the equation x=\frac{4±12}{-4} when ± is plus. Add 4 to 12.

$$x=\frac{16}{-4}$$

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{4±12}{-4} when ± is minus. Subtract 12 from 4.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-4 x=2$$

Steps for Completing the Square

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

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

Use the distributive property to multiply 3 by x^{2}-4.

$$x^{2}-4x+4=3x^{2}-12$$

Subtract 3x^{2} from both sides.

$$x^{2}-4x+4-3x^{2}=-12$$

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

$$-2x^{2}-4x+4=-12$$

Subtract 4 from both sides.

$$-2x^{2}-4x=-12-4$$

Subtract 4 from -12 to get -16.

$$-2x^{2}-4x=-16$$

Divide both sides by -2.

$$\frac{-2x^{2}-4x}{-2}=-\frac{16}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

$$x^{2}+2x=-\frac{16}{-2}$$

Divide -16 by -2.

$$x^{2}+2x=8$$

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

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

Square 1.

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

Add 8 to 1.

$$x^{2}+2x+1=9$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+1=3 x+1=-3$$

Subtract 1 from both sides of the equation.

$$x=2 x=-4$$