Solve x^2+(x+1)^2-61=0

$$x ^ { 2 } + ( x + 1 ) ^ { 2 } - 61 = 0$$

Solve for x

$$x=-6$$

$$x=5$$

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}.

$$x^{2}+x^{2}+2x+1-61=0$$

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

$$2x^{2}+2x+1-61=0$$

Subtract 61 from 1 to get -60.

$$2x^{2}+2x-60=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite x^{2}+x-30 as \left(x^{2}-5x\right)+\left(6x-30\right).

$$\left(x^{2}-5x\right)+\left(6x-30\right)$$

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

$$x\left(x-5\right)+6\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

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}.

$$x^{2}+x^{2}+2x+1-61=0$$

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

$$2x^{2}+2x+1-61=0$$

Subtract 61 from 1 to get -60.

$$2x^{2}+2x-60=0$$

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

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

Square 2.

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

Multiply -4 times 2.

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

Multiply -8 times -60.

$$x=\frac{-2±\sqrt{4+480}}{2\times 2}$$

Add 4 to 480.

$$x=\frac{-2±\sqrt{484}}{2\times 2}$$

Take the square root of 484.

$$x=\frac{-2±22}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±22}{4}$$

Now solve the equation x=\frac{-2±22}{4} when ± is plus. Add -2 to 22.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{-2±22}{4} when ± is minus. Subtract 22 from -2.

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

Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=5 x=-6$$

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}.

$$x^{2}+x^{2}+2x+1-61=0$$

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

$$2x^{2}+2x+1-61=0$$

Subtract 61 from 1 to get -60.

$$2x^{2}+2x-60=0$$

Add 60 to both sides. Anything plus zero gives itself.

$$2x^{2}+2x=60$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{60}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{60}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{60}{2}$$

Divide 60 by 2.

$$x^{2}+x=30$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{121}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{11}{2} x+\frac{1}{2}=-\frac{11}{2}$$

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

$$x=5 x=-6$$