Solve (x-1)^2+(x+1-1)^2=13

$$( x - 1 ) ^ { 2 } + ( x + 1 - 1 ) ^ { 2 } = 13$$

Solve for x

$$x=-2$$

$$x=3$$

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}.

$$x^{2}-2x+1+\left(x+1-1\right)^{2}=13$$

Subtract 1 from 1 to get 0.

$$x^{2}-2x+1+x^{2}=13$$

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

$$2x^{2}-2x+1=13$$

Subtract 13 from both sides.

$$2x^{2}-2x+1-13=0$$

Subtract 13 from 1 to get -12.

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

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(x^{2}-3x\right)+\left(2x-6\right)$$

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

$$x\left(x-3\right)+2\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

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}.

$$x^{2}-2x+1+\left(x+1-1\right)^{2}=13$$

Subtract 1 from 1 to get 0.

$$x^{2}-2x+1+x^{2}=13$$

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

$$2x^{2}-2x+1=13$$

Subtract 13 from both sides.

$$2x^{2}-2x+1-13=0$$

Subtract 13 from 1 to get -12.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 2}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2±10}{4}$$

Now solve the equation x=\frac{2±10}{4} when ± is plus. Add 2 to 10.

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

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{2±10}{4} when ± is minus. Subtract 10 from 2.

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=3 x=-2$$

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}.

$$x^{2}-2x+1+\left(x+1-1\right)^{2}=13$$

Subtract 1 from 1 to get 0.

$$x^{2}-2x+1+x^{2}=13$$

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

$$2x^{2}-2x+1=13$$

Subtract 1 from both sides.

$$2x^{2}-2x=13-1$$

Subtract 1 from 13 to get 12.

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

Divide both sides by 2.

$$\frac{2x^{2}-2x}{2}=\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

$$x^{2}-x=\frac{12}{2}$$

Divide 12 by 2.

$$x^{2}-x=6$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$