Solve 113=x^2+(x+1)^2

$$113 = x ^ { 2 } + ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=-8$$

$$x=7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$113=x^{2}+x^{2}+2x+1$$

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

$$113=2x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+2x+1=113$$

Subtract 113 from both sides.

$$2x^{2}+2x+1-113=0$$

Subtract 113 from 1 to get -112.

$$2x^{2}+2x-112=0$$

Divide both sides by 2.

$$x^{2}+x-56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-56\right)=-56$$

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 -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 1.

$$a=-7 b=8$$

Rewrite x^{2}+x-56 as \left(x^{2}-7x\right)+\left(8x-56\right).

$$\left(x^{2}-7x\right)+\left(8x-56\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-7\right)+8\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+8\right)$$

To find equation solutions, solve x-7=0 and x+8=0.

$$x=7 x=-8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$113=x^{2}+x^{2}+2x+1$$

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

$$113=2x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+2x+1=113$$

Subtract 113 from both sides.

$$2x^{2}+2x+1-113=0$$

Subtract 113 from 1 to get -112.

$$2x^{2}+2x-112=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -112 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-112\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-112\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-112\right)}}{2\times 2}$$

Multiply -8 times -112.

$$x=\frac{-2±\sqrt{4+896}}{2\times 2}$$

Add 4 to 896.

$$x=\frac{-2±\sqrt{900}}{2\times 2}$$

Take the square root of 900.

$$x=\frac{-2±30}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±30}{4}$$

Now solve the equation x=\frac{-2±30}{4} when ± is plus. Add -2 to 30.

$$x=\frac{28}{4}$$

Divide 28 by 4.

$$x=7$$

Now solve the equation x=\frac{-2±30}{4} when ± is minus. Subtract 30 from -2.

$$x=-\frac{32}{4}$$

Divide -32 by 4.

$$x=-8$$

The equation is now solved.

$$x=7 x=-8$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$113=x^{2}+x^{2}+2x+1$$

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

$$113=2x^{2}+2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+2x+1=113$$

Subtract 1 from both sides.

$$2x^{2}+2x=113-1$$

Subtract 1 from 113 to get 112.

$$2x^{2}+2x=112$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{112}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{112}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{112}{2}$$

Divide 112 by 2.

$$x^{2}+x=56$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=56+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=56+\frac{1}{4}$$

Add 56 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{225}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{15}{2} x+\frac{1}{2}=-\frac{15}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=7 x=-8$$