Solve left(x+1right)^2=81

$${ \left(x+1 \right) }^{ 2 } = 81$$

Solve for x

$$x=8$$

$$x=-10$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1=81$$

Subtract 81 from both sides.

$$x^{2}+2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}+2x-80=0$$

To solve the equation, factor x^{2}+2x-80 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 2.

$$a=-8 b=10$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x+10\right)$$

To find equation solutions, solve x-8=0 and x+10=0.

$$x=8 x=-10$$

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}.

$$x^{2}+2x+1=81$$

Subtract 81 from both sides.

$$x^{2}+2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}+2x-80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-80. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-80\right)=-80$$

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 -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 2.

$$a=-8 b=10$$

Rewrite x^{2}+2x-80 as \left(x^{2}-8x\right)+\left(10x-80\right).

$$\left(x^{2}-8x\right)+\left(10x-80\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-8\right)+10\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+10\right)$$

To find equation solutions, solve x-8=0 and x+10=0.

$$x=8 x=-10$$

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}.

$$x^{2}+2x+1=81$$

Subtract 81 from both sides.

$$x^{2}+2x+1-81=0$$

Subtract 81 from 1 to get -80.

$$x^{2}+2x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-80\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-80\right)}}{2}$$

Multiply -4 times -80.

$$x=\frac{-2±\sqrt{4+320}}{2}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-2±18}{2}$$

Now solve the equation x=\frac{-2±18}{2} when ± is plus. Add -2 to 18.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{-2±18}{2} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=8 x=-10$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{81}$$

Simplify.

$$x+1=9 x+1=-9$$

Subtract 1 from both sides of the equation.

$$x=8 x=-10$$