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

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

Solve for x

$$x=-6$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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}.

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

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

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

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

$$x^{2}-6x+9=81+54x+9x^{2}$$

Subtract 81 from both sides.

$$x^{2}-6x+9-81=54x+9x^{2}$$

Subtract 81 from 9 to get -72.

$$x^{2}-6x-72=54x+9x^{2}$$

Subtract 54x from both sides.

$$x^{2}-6x-72-54x=9x^{2}$$

Combine -6x and -54x to get -60x.

$$x^{2}-60x-72=9x^{2}$$

Subtract 9x^{2} from both sides.

$$x^{2}-60x-72-9x^{2}=0$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}-60x-72=0$$

Divide both sides by 4.

$$-2x^{2}-15x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=-15 ab=-2\left(-18\right)=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -15.

$$a=-3 b=-12$$

Rewrite -2x^{2}-15x-18 as \left(-2x^{2}-3x\right)+\left(-12x-18\right).

$$\left(-2x^{2}-3x\right)+\left(-12x-18\right)$$

Factor out -x in the first and -6 in the second group.

$$-x\left(2x+3\right)-6\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(-x-6\right)$$

To find equation solutions, solve 2x+3=0 and -x-6=0.

$$x=-\frac{3}{2} x=-6$$

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}.

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

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

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

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

$$x^{2}-6x+9=81+54x+9x^{2}$$

Subtract 81 from both sides.

$$x^{2}-6x+9-81=54x+9x^{2}$$

Subtract 81 from 9 to get -72.

$$x^{2}-6x-72=54x+9x^{2}$$

Subtract 54x from both sides.

$$x^{2}-6x-72-54x=9x^{2}$$

Combine -6x and -54x to get -60x.

$$x^{2}-60x-72=9x^{2}$$

Subtract 9x^{2} from both sides.

$$x^{2}-60x-72-9x^{2}=0$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}-60x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, -60 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\left(-8\right)\left(-72\right)}}{2\left(-8\right)}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\left(-8\right)\left(-72\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-\left(-60\right)±\sqrt{3600+32\left(-72\right)}}{2\left(-8\right)}$$

Multiply 32 times -72.

$$x=\frac{-\left(-60\right)±\sqrt{3600-2304}}{2\left(-8\right)}$$

Add 3600 to -2304.

$$x=\frac{-\left(-60\right)±\sqrt{1296}}{2\left(-8\right)}$$

Take the square root of 1296.

$$x=\frac{-\left(-60\right)±36}{2\left(-8\right)}$$

The opposite of -60 is 60.

$$x=\frac{60±36}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{60±36}{-16}$$

Now solve the equation x=\frac{60±36}{-16} when ± is plus. Add 60 to 36.

$$x=\frac{96}{-16}$$

Divide 96 by -16.

$$x=-6$$

Now solve the equation x=\frac{60±36}{-16} when ± is minus. Subtract 36 from 60.

$$x=\frac{24}{-16}$$

Reduce the fraction \frac{24}{-16} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=-6 x=-\frac{3}{2}$$

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}.

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

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

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

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

$$x^{2}-6x+9=81+54x+9x^{2}$$

Subtract 54x from both sides.

$$x^{2}-6x+9-54x=81+9x^{2}$$

Combine -6x and -54x to get -60x.

$$x^{2}-60x+9=81+9x^{2}$$

Subtract 9x^{2} from both sides.

$$x^{2}-60x+9-9x^{2}=81$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}-60x+9=81$$

Subtract 9 from both sides.

$$-8x^{2}-60x=81-9$$

Subtract 9 from 81 to get 72.

$$-8x^{2}-60x=72$$

Divide both sides by -8.

$$\frac{-8x^{2}-60x}{-8}=\frac{72}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\left(-\frac{60}{-8}\right)x=\frac{72}{-8}$$

Reduce the fraction \frac{-60}{-8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{15}{2}x=\frac{72}{-8}$$

Divide 72 by -8.

$$x^{2}+\frac{15}{2}x=-9$$

Divide \frac{15}{2}, the coefficient of the x term, by 2 to get \frac{15}{4}. Then add the square of \frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{15}{2}x+\left(\frac{15}{4}\right)^{2}=-9+\left(\frac{15}{4}\right)^{2}$$

Square \frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=-9+\frac{225}{16}$$

Add -9 to \frac{225}{16}.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{15}{2}x+\frac{225}{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}.

$$\left(x+\frac{15}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{15}{4}=\frac{9}{4} x+\frac{15}{4}=-\frac{9}{4}$$

Subtract \frac{15}{4} from both sides of the equation.

$$x=-\frac{3}{2} x=-6$$