Solve x^2+(x+3)^2=117

$$x ^ { 2 } + ( x + 3 ) ^ { 2 } = 117$$

Solve for x

$$x=-9$$

$$x=6$$

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}.

$$x^{2}+x^{2}+6x+9=117$$

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

$$2x^{2}+6x+9=117$$

Subtract 117 from both sides.

$$2x^{2}+6x+9-117=0$$

Subtract 117 from 9 to get -108.

$$2x^{2}+6x-108=0$$

Divide both sides by 2.

$$x^{2}+3x-54=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-54. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-54\right)=-54$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=-6 b=9$$

Rewrite x^{2}+3x-54 as \left(x^{2}-6x\right)+\left(9x-54\right).

$$\left(x^{2}-6x\right)+\left(9x-54\right)$$

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

$$x\left(x-6\right)+9\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+9\right)$$

To find equation solutions, solve x-6=0 and x+9=0.

$$x=6 x=-9$$

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}.

$$x^{2}+x^{2}+6x+9=117$$

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

$$2x^{2}+6x+9=117$$

Subtract 117 from both sides.

$$2x^{2}+6x+9-117=0$$

Subtract 117 from 9 to get -108.

$$2x^{2}+6x-108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-108\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-108\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-108\right)}}{2\times 2}$$

Multiply -8 times -108.

$$x=\frac{-6±\sqrt{36+864}}{2\times 2}$$

Add 36 to 864.

$$x=\frac{-6±\sqrt{900}}{2\times 2}$$

Take the square root of 900.

$$x=\frac{-6±30}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±30}{4}$$

Now solve the equation x=\frac{-6±30}{4} when ± is plus. Add -6 to 30.

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

Now solve the equation x=\frac{-6±30}{4} when ± is minus. Subtract 30 from -6.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=6 x=-9$$

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}.

$$x^{2}+x^{2}+6x+9=117$$

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

$$2x^{2}+6x+9=117$$

Subtract 9 from both sides.

$$2x^{2}+6x=117-9$$

Subtract 9 from 117 to get 108.

$$2x^{2}+6x=108$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{108}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{108}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{108}{2}$$

Divide 108 by 2.

$$x^{2}+3x=54$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=54+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=54+\frac{9}{4}$$

Add 54 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{225}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{15}{2} x+\frac{3}{2}=-\frac{15}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=6 x=-9$$