Solve 1/2+θ-1/(2+θ)^2=5/12-5/18

$$\frac { 1 } { 2 + \theta } - \frac { 1 } { ( 2 + \theta ) ^ { 2 } } = \frac { 5 } { 12 } - \frac { 5 } { 18 }$$

Solve for θ

$$\theta =4$$

$$\theta =-\frac{4}{5}=-0.8$$

Steps Using Factoring By Grouping

Variable \theta cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 36\left(\theta +2\right)^{2}, the least common multiple of 2+\theta ,\left(2+\theta \right)^{2},12,18.

$$36\theta +72-36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Subtract 36 from 72 to get 36.

$$36\theta +36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=36\left(\theta ^{2}+4\theta +4\right)\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Multiply 36 and \frac{5}{12} to get 15.

$$36\theta +36=15\left(\theta ^{2}+4\theta +4\right)+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Use the distributive property to multiply 15 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta ^{2}+4\theta +4\right)\left(-\frac{5}{18}\right)$$

Multiply 36 and -\frac{5}{18} to get -10.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\left(\theta ^{2}+4\theta +4\right)$$

Use the distributive property to multiply -10 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\theta ^{2}-40\theta -40$$

Combine 15\theta ^{2} and -10\theta ^{2} to get 5\theta ^{2}.

$$36\theta +36=5\theta ^{2}+60\theta +60-40\theta -40$$

Combine 60\theta and -40\theta to get 20\theta .

$$36\theta +36=5\theta ^{2}+20\theta +60-40$$

Subtract 40 from 60 to get 20.

$$36\theta +36=5\theta ^{2}+20\theta +20$$

Subtract 5\theta ^{2} from both sides.

$$36\theta +36-5\theta ^{2}=20\theta +20$$

Subtract 20\theta from both sides.

$$36\theta +36-5\theta ^{2}-20\theta =20$$

Combine 36\theta and -20\theta to get 16\theta .

$$16\theta +36-5\theta ^{2}=20$$

Subtract 20 from both sides.

$$16\theta +36-5\theta ^{2}-20=0$$

Subtract 20 from 36 to get 16.

$$16\theta +16-5\theta ^{2}=0$$

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

$$-5\theta ^{2}+16\theta +16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5\theta ^{2}+a\theta +b\theta +16. To find a and b, set up a system to be solved.

$$a+b=16 ab=-5\times 16=-80$$

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 16.

$$a=20 b=-4$$

Rewrite -5\theta ^{2}+16\theta +16 as \left(-5\theta ^{2}+20\theta \right)+\left(-4\theta +16\right).

$$\left(-5\theta ^{2}+20\theta \right)+\left(-4\theta +16\right)$$

Factor out 5\theta in the first and 4 in the second group.

$$5\theta \left(-\theta +4\right)+4\left(-\theta +4\right)$$

Factor out common term -\theta +4 by using distributive property.

$$\left(-\theta +4\right)\left(5\theta +4\right)$$

To find equation solutions, solve -\theta +4=0 and 5\theta +4=0.

$$\theta =4 \theta =-\frac{4}{5}$$

Steps Using the Quadratic Formula

Variable \theta cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 36\left(\theta +2\right)^{2}, the least common multiple of 2+\theta ,\left(2+\theta \right)^{2},12,18.

$$36\theta +72-36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Subtract 36 from 72 to get 36.

$$36\theta +36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=36\left(\theta ^{2}+4\theta +4\right)\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Multiply 36 and \frac{5}{12} to get 15.

$$36\theta +36=15\left(\theta ^{2}+4\theta +4\right)+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Use the distributive property to multiply 15 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta ^{2}+4\theta +4\right)\left(-\frac{5}{18}\right)$$

Multiply 36 and -\frac{5}{18} to get -10.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\left(\theta ^{2}+4\theta +4\right)$$

Use the distributive property to multiply -10 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\theta ^{2}-40\theta -40$$

Combine 15\theta ^{2} and -10\theta ^{2} to get 5\theta ^{2}.

$$36\theta +36=5\theta ^{2}+60\theta +60-40\theta -40$$

Combine 60\theta and -40\theta to get 20\theta .

$$36\theta +36=5\theta ^{2}+20\theta +60-40$$

Subtract 40 from 60 to get 20.

$$36\theta +36=5\theta ^{2}+20\theta +20$$

Subtract 5\theta ^{2} from both sides.

$$36\theta +36-5\theta ^{2}=20\theta +20$$

Subtract 20\theta from both sides.

$$36\theta +36-5\theta ^{2}-20\theta =20$$

Combine 36\theta and -20\theta to get 16\theta .

$$16\theta +36-5\theta ^{2}=20$$

Subtract 20 from both sides.

$$16\theta +36-5\theta ^{2}-20=0$$

Subtract 20 from 36 to get 16.

$$16\theta +16-5\theta ^{2}=0$$

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.

$$-5\theta ^{2}+16\theta +16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 16 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$\theta =\frac{-16±\sqrt{16^{2}-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Square 16.

$$\theta =\frac{-16±\sqrt{256-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$\theta =\frac{-16±\sqrt{256+20\times 16}}{2\left(-5\right)}$$

Multiply 20 times 16.

$$\theta =\frac{-16±\sqrt{256+320}}{2\left(-5\right)}$$

Add 256 to 320.

$$\theta =\frac{-16±\sqrt{576}}{2\left(-5\right)}$$

Take the square root of 576.

$$\theta =\frac{-16±24}{2\left(-5\right)}$$

Multiply 2 times -5.

$$\theta =\frac{-16±24}{-10}$$

Now solve the equation \theta =\frac{-16±24}{-10} when ± is plus. Add -16 to 24.

$$\theta =\frac{8}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

$$\theta =-\frac{4}{5}$$

Now solve the equation \theta =\frac{-16±24}{-10} when ± is minus. Subtract 24 from -16.

$$\theta =-\frac{40}{-10}$$

Divide -40 by -10.

$$\theta =4$$

The equation is now solved.

$$\theta =-\frac{4}{5} \theta =4$$

Steps for Completing the Square

Variable \theta cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 36\left(\theta +2\right)^{2}, the least common multiple of 2+\theta ,\left(2+\theta \right)^{2},12,18.

$$36\theta +72-36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Subtract 36 from 72 to get 36.

$$36\theta +36=36\left(\theta +2\right)^{2}\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=36\left(\theta ^{2}+4\theta +4\right)\times \frac{5}{12}+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Multiply 36 and \frac{5}{12} to get 15.

$$36\theta +36=15\left(\theta ^{2}+4\theta +4\right)+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

Use the distributive property to multiply 15 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta +2\right)^{2}\left(-\frac{5}{18}\right)$$

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

$$36\theta +36=15\theta ^{2}+60\theta +60+36\left(\theta ^{2}+4\theta +4\right)\left(-\frac{5}{18}\right)$$

Multiply 36 and -\frac{5}{18} to get -10.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\left(\theta ^{2}+4\theta +4\right)$$

Use the distributive property to multiply -10 by \theta ^{2}+4\theta +4.

$$36\theta +36=15\theta ^{2}+60\theta +60-10\theta ^{2}-40\theta -40$$

Combine 15\theta ^{2} and -10\theta ^{2} to get 5\theta ^{2}.

$$36\theta +36=5\theta ^{2}+60\theta +60-40\theta -40$$

Combine 60\theta and -40\theta to get 20\theta .

$$36\theta +36=5\theta ^{2}+20\theta +60-40$$

Subtract 40 from 60 to get 20.

$$36\theta +36=5\theta ^{2}+20\theta +20$$

Subtract 5\theta ^{2} from both sides.

$$36\theta +36-5\theta ^{2}=20\theta +20$$

Subtract 20\theta from both sides.

$$36\theta +36-5\theta ^{2}-20\theta =20$$

Combine 36\theta and -20\theta to get 16\theta .

$$16\theta +36-5\theta ^{2}=20$$

Subtract 36 from both sides.

$$16\theta -5\theta ^{2}=20-36$$

Subtract 36 from 20 to get -16.

$$16\theta -5\theta ^{2}=-16$$

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.

$$-5\theta ^{2}+16\theta =-16$$

Divide both sides by -5.

$$\frac{-5\theta ^{2}+16\theta }{-5}=-\frac{16}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$\theta ^{2}+\frac{16}{-5}\theta =-\frac{16}{-5}$$

Divide 16 by -5.

$$\theta ^{2}-\frac{16}{5}\theta =-\frac{16}{-5}$$

Divide -16 by -5.

$$\theta ^{2}-\frac{16}{5}\theta =\frac{16}{5}$$

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\theta ^{2}-\frac{16}{5}\theta +\left(-\frac{8}{5}\right)^{2}=\frac{16}{5}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$\theta ^{2}-\frac{16}{5}\theta +\frac{64}{25}=\frac{16}{5}+\frac{64}{25}$$

Add \frac{16}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\theta ^{2}-\frac{16}{5}\theta +\frac{64}{25}=\frac{144}{25}$$

Factor \theta ^{2}-\frac{16}{5}\theta +\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(\theta -\frac{8}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\theta -\frac{8}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$\theta -\frac{8}{5}=\frac{12}{5} \theta -\frac{8}{5}=-\frac{12}{5}$$

Add \frac{8}{5} to both sides of the equation.

$$\theta =4 \theta =-\frac{4}{5}$$