Solve 1+r+r^2/1-2r+r^2=19

$$\frac{ 1+r+ { r }^{ 2 } }{ 1-2r+ { r }^{ 2 } } =19$$

Solve for r

$$r=\frac{2}{3}\approx 0.666666667$$

$$r = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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}.

$$1+r+r^{2}=19\left(r-1\right)^{2}$$

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

$$1+r+r^{2}=19\left(r^{2}-2r+1\right)$$

Use the distributive property to multiply 19 by r^{2}-2r+1.

$$1+r+r^{2}=19r^{2}-38r+19$$

Subtract 19r^{2} from both sides.

$$1+r+r^{2}-19r^{2}=-38r+19$$

Combine r^{2} and -19r^{2} to get -18r^{2}.

$$1+r-18r^{2}=-38r+19$$

Add 38r to both sides.

$$1+r-18r^{2}+38r=19$$

Combine r and 38r to get 39r.

$$1+39r-18r^{2}=19$$

Subtract 19 from both sides.

$$1+39r-18r^{2}-19=0$$

Subtract 19 from 1 to get -18.

$$-18+39r-18r^{2}=0$$

Divide both sides by 3.

$$-6+13r-6r^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-6r^{2}+13r-6=0$$

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

$$a+b=13 ab=-6\left(-6\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=9 b=4$$

Rewrite -6r^{2}+13r-6 as \left(-6r^{2}+9r\right)+\left(4r-6\right).

$$\left(-6r^{2}+9r\right)+\left(4r-6\right)$$

Factor out -3r in the first and 2 in the second group.

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

Factor out common term 2r-3 by using distributive property.

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

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

$$r=\frac{3}{2} r=\frac{2}{3}$$

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}.

$$1+r+r^{2}=19\left(r-1\right)^{2}$$

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

$$1+r+r^{2}=19\left(r^{2}-2r+1\right)$$

Use the distributive property to multiply 19 by r^{2}-2r+1.

$$1+r+r^{2}=19r^{2}-38r+19$$

Subtract 19r^{2} from both sides.

$$1+r+r^{2}-19r^{2}=-38r+19$$

Combine r^{2} and -19r^{2} to get -18r^{2}.

$$1+r-18r^{2}=-38r+19$$

Add 38r to both sides.

$$1+r-18r^{2}+38r=19$$

Combine r and 38r to get 39r.

$$1+39r-18r^{2}=19$$

Subtract 19 from both sides.

$$1+39r-18r^{2}-19=0$$

Subtract 19 from 1 to get -18.

$$-18+39r-18r^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-18r^{2}+39r-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 39 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-39±\sqrt{39^{2}-4\left(-18\right)\left(-18\right)}}{2\left(-18\right)}$$

Square 39.

$$r=\frac{-39±\sqrt{1521-4\left(-18\right)\left(-18\right)}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$r=\frac{-39±\sqrt{1521+72\left(-18\right)}}{2\left(-18\right)}$$

Multiply 72 times -18.

$$r=\frac{-39±\sqrt{1521-1296}}{2\left(-18\right)}$$

Add 1521 to -1296.

$$r=\frac{-39±\sqrt{225}}{2\left(-18\right)}$$

Take the square root of 225.

$$r=\frac{-39±15}{2\left(-18\right)}$$

Multiply 2 times -18.

$$r=\frac{-39±15}{-36}$$

Now solve the equation r=\frac{-39±15}{-36} when ± is plus. Add -39 to 15.

$$r=-\frac{24}{-36}$$

Reduce the fraction \frac{-24}{-36} to lowest terms by extracting and canceling out 12.

$$r=\frac{2}{3}$$

Now solve the equation r=\frac{-39±15}{-36} when ± is minus. Subtract 15 from -39.

$$r=-\frac{54}{-36}$$

Reduce the fraction \frac{-54}{-36} to lowest terms by extracting and canceling out 18.

$$r=\frac{3}{2}$$

The equation is now solved.

$$r=\frac{2}{3} r=\frac{3}{2}$$

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}.

$$1+r+r^{2}=19\left(r-1\right)^{2}$$

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

$$1+r+r^{2}=19\left(r^{2}-2r+1\right)$$

Use the distributive property to multiply 19 by r^{2}-2r+1.

$$1+r+r^{2}=19r^{2}-38r+19$$

Subtract 19r^{2} from both sides.

$$1+r+r^{2}-19r^{2}=-38r+19$$

Combine r^{2} and -19r^{2} to get -18r^{2}.

$$1+r-18r^{2}=-38r+19$$

Add 38r to both sides.

$$1+r-18r^{2}+38r=19$$

Combine r and 38r to get 39r.

$$1+39r-18r^{2}=19$$

Subtract 1 from both sides.

$$39r-18r^{2}=19-1$$

Subtract 1 from 19 to get 18.

$$39r-18r^{2}=18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-18r^{2}+39r=18$$

Divide both sides by -18.

$$\frac{-18r^{2}+39r}{-18}=\frac{18}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$r^{2}+\frac{39}{-18}r=\frac{18}{-18}$$

Reduce the fraction \frac{39}{-18} to lowest terms by extracting and canceling out 3.

$$r^{2}-\frac{13}{6}r=\frac{18}{-18}$$

Divide 18 by -18.

$$r^{2}-\frac{13}{6}r=-1$$

Divide -\frac{13}{6}, the coefficient of the x term, by 2 to get -\frac{13}{12}. Then add the square of -\frac{13}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{13}{6}r+\left(-\frac{13}{12}\right)^{2}=-1+\left(-\frac{13}{12}\right)^{2}$$

Square -\frac{13}{12} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{13}{6}r+\frac{169}{144}=-1+\frac{169}{144}$$

Add -1 to \frac{169}{144}.

$$r^{2}-\frac{13}{6}r+\frac{169}{144}=\frac{25}{144}$$

Factor r^{2}-\frac{13}{6}r+\frac{169}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{13}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{13}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$r-\frac{13}{12}=\frac{5}{12} r-\frac{13}{12}=-\frac{5}{12}$$

Add \frac{13}{12} to both sides of the equation.

$$r=\frac{3}{2} r=\frac{2}{3}$$