Solve x^2+left(x+1right)^2=25

$${ x }^{ 2 } + { \left(x+1 \right) }^{ 2 } =25$$

Solve for x

$$x=-4$$

$$x=3$$

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}.

$$x^{2}+x^{2}+2x+1=25$$

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

$$2x^{2}+2x+1=25$$

Subtract 25 from both sides.

$$2x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$2x^{2}+2x-24=0$$

Divide both sides by 2.

$$x^{2}+x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-12\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite x^{2}+x-12 as \left(x^{2}-3x\right)+\left(4x-12\right).

$$\left(x^{2}-3x\right)+\left(4x-12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-3\right)+4\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+4\right)$$

To find equation solutions, solve x-3=0 and x+4=0.

$$x=3 x=-4$$

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}.

$$x^{2}+x^{2}+2x+1=25$$

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

$$2x^{2}+2x+1=25$$

Subtract 25 from both sides.

$$2x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$2x^{2}+2x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$x=\frac{-2±\sqrt{4+192}}{2\times 2}$$

Add 4 to 192.

$$x=\frac{-2±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$x=\frac{-2±14}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±14}{4}$$

Now solve the equation x=\frac{-2±14}{4} when ± is plus. Add -2 to 14.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{-2±14}{4} when ± is minus. Subtract 14 from -2.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=3 x=-4$$

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}.

$$x^{2}+x^{2}+2x+1=25$$

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

$$2x^{2}+2x+1=25$$

Subtract 1 from both sides.

$$2x^{2}+2x=25-1$$

Subtract 1 from 25 to get 24.

$$2x^{2}+2x=24$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{24}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{24}{2}$$

Divide 24 by 2.

$$x^{2}+x=12$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=12+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{49}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{7}{2} x+\frac{1}{2}=-\frac{7}{2}$$

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

$$x=3 x=-4$$