Solve 2(x+1)^2-1=71

$$2 ( x + 1 ) ^ { 2 } - 1 = 71$$

Solve for x

$$x=5$$

$$x=-7$$

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}.

$$2\left(x^{2}+2x+1\right)-1=71$$

Use the distributive property to multiply 2 by x^{2}+2x+1.

$$2x^{2}+4x+2-1=71$$

Subtract 1 from 2 to get 1.

$$2x^{2}+4x+1=71$$

Subtract 71 from both sides.

$$2x^{2}+4x+1-71=0$$

Subtract 71 from 1 to get -70.

$$2x^{2}+4x-70=0$$

Divide both sides by 2.

$$x^{2}+2x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-35\right)=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

Rewrite x^{2}+2x-35 as \left(x^{2}-5x\right)+\left(7x-35\right).

$$\left(x^{2}-5x\right)+\left(7x-35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-5\right)+7\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+7\right)$$

To find equation solutions, solve x-5=0 and x+7=0.

$$x=5 x=-7$$

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}.

$$2\left(x^{2}+2x+1\right)-1=71$$

Use the distributive property to multiply 2 by x^{2}+2x+1.

$$2x^{2}+4x+2-1=71$$

Subtract 1 from 2 to get 1.

$$2x^{2}+4x+1=71$$

Subtract 71 from both sides.

$$2x^{2}+4x+1-71=0$$

Subtract 71 from 1 to get -70.

$$2x^{2}+4x-70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-70\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-70\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-70\right)}}{2\times 2}$$

Multiply -8 times -70.

$$x=\frac{-4±\sqrt{16+560}}{2\times 2}$$

Add 16 to 560.

$$x=\frac{-4±\sqrt{576}}{2\times 2}$$

Take the square root of 576.

$$x=\frac{-4±24}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±24}{4}$$

Now solve the equation x=\frac{-4±24}{4} when ± is plus. Add -4 to 24.

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

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{-4±24}{4} when ± is minus. Subtract 24 from -4.

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

Divide -28 by 4.

$$x=-7$$

The equation is now solved.

$$x=5 x=-7$$

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}.

$$2\left(x^{2}+2x+1\right)-1=71$$

Use the distributive property to multiply 2 by x^{2}+2x+1.

$$2x^{2}+4x+2-1=71$$

Subtract 1 from 2 to get 1.

$$2x^{2}+4x+1=71$$

Subtract 1 from both sides.

$$2x^{2}+4x=71-1$$

Subtract 1 from 71 to get 70.

$$2x^{2}+4x=70$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{70}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{70}{2}$$

Divide 4 by 2.

$$x^{2}+2x=\frac{70}{2}$$

Divide 70 by 2.

$$x^{2}+2x=35$$

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

$$x^{2}+2x+1^{2}=35+1^{2}$$

Square 1.

$$x^{2}+2x+1=35+1$$

Add 35 to 1.

$$x^{2}+2x+1=36$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+1=6 x+1=-6$$

Subtract 1 from both sides of the equation.

$$x=5 x=-7$$