Solve 100=900/(1+r)^2

$$100 = \frac { 900 } { ( 1 + r ) ^ { 2 } }$$

Solve for r

$$r=-4$$

$$r=2$$

Steps Using Factoring By Grouping

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}.

$$100\left(r+1\right)^{2}=900$$

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

$$100\left(r^{2}+2r+1\right)=900$$

Use the distributive property to multiply 100 by r^{2}+2r+1.

$$100r^{2}+200r+100=900$$

Subtract 900 from both sides.

$$100r^{2}+200r+100-900=0$$

Subtract 900 from 100 to get -800.

$$100r^{2}+200r-800=0$$

Divide both sides by 100.

$$r^{2}+2r-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite r^{2}+2r-8 as \left(r^{2}-2r\right)+\left(4r-8\right).

$$\left(r^{2}-2r\right)+\left(4r-8\right)$$

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

$$r\left(r-2\right)+4\left(r-2\right)$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(r+4\right)$$

To find equation solutions, solve r-2=0 and r+4=0.

$$r=2 r=-4$$

Steps Using the Quadratic Formula

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}.

$$100\left(r+1\right)^{2}=900$$

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

$$100\left(r^{2}+2r+1\right)=900$$

Use the distributive property to multiply 100 by r^{2}+2r+1.

$$100r^{2}+200r+100=900$$

Subtract 900 from both sides.

$$100r^{2}+200r+100-900=0$$

Subtract 900 from 100 to get -800.

$$100r^{2}+200r-800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, 200 for b, and -800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-200±\sqrt{200^{2}-4\times 100\left(-800\right)}}{2\times 100}$$

Square 200.

$$r=\frac{-200±\sqrt{40000-4\times 100\left(-800\right)}}{2\times 100}$$

Multiply -4 times 100.

$$r=\frac{-200±\sqrt{40000-400\left(-800\right)}}{2\times 100}$$

Multiply -400 times -800.

$$r=\frac{-200±\sqrt{40000+320000}}{2\times 100}$$

Add 40000 to 320000.

$$r=\frac{-200±\sqrt{360000}}{2\times 100}$$

Take the square root of 360000.

$$r=\frac{-200±600}{2\times 100}$$

Multiply 2 times 100.

$$r=\frac{-200±600}{200}$$

Now solve the equation r=\frac{-200±600}{200} when ± is plus. Add -200 to 600.

$$r=\frac{400}{200}$$

Divide 400 by 200.

$$r=2$$

Now solve the equation r=\frac{-200±600}{200} when ± is minus. Subtract 600 from -200.

$$r=-\frac{800}{200}$$

Divide -800 by 200.

$$r=-4$$

The equation is now solved.

$$r=2 r=-4$$

Steps for Completing the Square

Variable r cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(r+1\right)^{2}.

$$100\left(r+1\right)^{2}=900$$

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

$$100\left(r^{2}+2r+1\right)=900$$

Use the distributive property to multiply 100 by r^{2}+2r+1.

$$100r^{2}+200r+100=900$$

Subtract 100 from both sides.

$$100r^{2}+200r=900-100$$

Subtract 100 from 900 to get 800.

$$100r^{2}+200r=800$$

Divide both sides by 100.

$$\frac{100r^{2}+200r}{100}=\frac{800}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$r^{2}+\frac{200}{100}r=\frac{800}{100}$$

Divide 200 by 100.

$$r^{2}+2r=\frac{800}{100}$$

Divide 800 by 100.

$$r^{2}+2r=8$$

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.

$$r^{2}+2r+1^{2}=8+1^{2}$$

Square 1.

$$r^{2}+2r+1=8+1$$

Add 8 to 1.

$$r^{2}+2r+1=9$$

Factor r^{2}+2r+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}.

$$\left(r+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$r+1=3 r+1=-3$$

Subtract 1 from both sides of the equation.

$$r=2 r=-4$$