Solve x^2+(x+1)^2=5

$$x ^ { 2 } + ( x + 1 ) ^ { 2 } = 5$$

Solve for x

$$x=-2$$

$$x=1$$

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}.

$$x^{2}+x^{2}+2x+1=5$$

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

$$2x^{2}+2x+1=5$$

Subtract 5 from both sides.

$$2x^{2}+2x+1-5=0$$

Subtract 5 from 1 to get -4.

$$2x^{2}+2x-4=0$$

Divide both sides by 2.

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

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

$$a+b=1 ab=1\left(-2\right)=-2$$

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. The only such pair is the system solution.

$$a=-1 b=2$$

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

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

Factor out x in the first and 2 in the second group.

$$x\left(x-1\right)+2\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+2\right)$$

To find equation solutions, solve x-1=0 and x+2=0.

$$x=1 x=-2$$

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}.

$$x^{2}+x^{2}+2x+1=5$$

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

$$2x^{2}+2x+1=5$$

Subtract 5 from both sides.

$$2x^{2}+2x+1-5=0$$

Subtract 5 from 1 to get -4.

$$2x^{2}+2x-4=0$$

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

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

Square 2.

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

Multiply -4 times 2.

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

Multiply -8 times -4.

$$x=\frac{-2±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$x=\frac{-2±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±6}{4}$$

Now solve the equation x=\frac{-2±6}{4} when ± is plus. Add -2 to 6.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-2±6}{4} when ± is minus. Subtract 6 from -2.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=1 x=-2$$

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}.

$$x^{2}+x^{2}+2x+1=5$$

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

$$2x^{2}+2x+1=5$$

Subtract 1 from both sides.

$$2x^{2}+2x=5-1$$

Subtract 1 from 5 to get 4.

$$2x^{2}+2x=4$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

Divide 4 by 2.

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

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.

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

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

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

Add 2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$x=1 x=-2$$