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

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

Solve for x

$$x=9$$

$$x=1$$

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}.

$$x^{2}+6x+9=4\left(x-3\right)^{2}$$

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

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

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

$$x^{2}+6x+9=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+6x+9-4x^{2}=-24x+36$$

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

$$-3x^{2}+6x+9=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+6x+9+24x=36$$

Combine 6x and 24x to get 30x.

$$-3x^{2}+30x+9=36$$

Subtract 36 from both sides.

$$-3x^{2}+30x+9-36=0$$

Subtract 36 from 9 to get -27.

$$-3x^{2}+30x-27=0$$

Divide both sides by 3.

$$-x^{2}+10x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -x^{2}+10x-9 as \left(-x^{2}+9x\right)+\left(x-9\right).

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

Factor out -x in -x^{2}+9x.

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

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x+1\right)$$

To find equation solutions, solve x-9=0 and -x+1=0.

$$x=9 x=1$$

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}.

$$x^{2}+6x+9=4\left(x-3\right)^{2}$$

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

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

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

$$x^{2}+6x+9=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+6x+9-4x^{2}=-24x+36$$

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

$$-3x^{2}+6x+9=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+6x+9+24x=36$$

Combine 6x and 24x to get 30x.

$$-3x^{2}+30x+9=36$$

Subtract 36 from both sides.

$$-3x^{2}+30x+9-36=0$$

Subtract 36 from 9 to get -27.

$$-3x^{2}+30x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 30 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-3\right)\left(-27\right)}}{2\left(-3\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-3\right)\left(-27\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-30±\sqrt{900+12\left(-27\right)}}{2\left(-3\right)}$$

Multiply 12 times -27.

$$x=\frac{-30±\sqrt{900-324}}{2\left(-3\right)}$$

Add 900 to -324.

$$x=\frac{-30±\sqrt{576}}{2\left(-3\right)}$$

Take the square root of 576.

$$x=\frac{-30±24}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-30±24}{-6}$$

Now solve the equation x=\frac{-30±24}{-6} when ± is plus. Add -30 to 24.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-30±24}{-6} when ± is minus. Subtract 24 from -30.

$$x=-\frac{54}{-6}$$

Divide -54 by -6.

$$x=9$$

The equation is now solved.

$$x=1 x=9$$

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}.

$$x^{2}+6x+9=4\left(x-3\right)^{2}$$

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

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

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

$$x^{2}+6x+9=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+6x+9-4x^{2}=-24x+36$$

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

$$-3x^{2}+6x+9=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+6x+9+24x=36$$

Combine 6x and 24x to get 30x.

$$-3x^{2}+30x+9=36$$

Subtract 9 from both sides.

$$-3x^{2}+30x=36-9$$

Subtract 9 from 36 to get 27.

$$-3x^{2}+30x=27$$

Divide both sides by -3.

$$\frac{-3x^{2}+30x}{-3}=\frac{27}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{30}{-3}x=\frac{27}{-3}$$

Divide 30 by -3.

$$x^{2}-10x=\frac{27}{-3}$$

Divide 27 by -3.

$$x^{2}-10x=-9$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$