In the following exercises, solve.

$1+\frac{9}{p}=\frac{-20}{p^2}$

The given equation is

$1+\frac{9}{p}=\frac{-20}{p^2}$

Some terms of the equation are not defined for the$p=0$

sorole="math" localid="1647788485793" $1+\frac{9}{p}=\frac{-20}{p^2},p\ne 0$

The least common denominator of all denominators in the equation is$p^2$

multiply both sides of the equation by the least common denominator $p^2$

$$\begin{gathered} p^2\left(1+\frac{9}{p}\right)=p^2\left(\frac{-20}{p^2}\right) \\ {p}^2\left(1\right)+p^2\left(\frac{9}{p}\right)=p^2\left(\frac{-20}{p^2}\right) \\ {p}^2+9p=-20 \\ {p}^2+9p+20=0 \end{gathered}$$

Factorize the equation

$$\begin{gathered} p^2+9p+20=0 \\ {p}^2+4p+5p+20=0 \\ p(p+4)+5(p+4)=0 \\ (p+5)(p+4)=0 \end{gathered}$$

Use the Zero Product Property

$$\begin{gathered} p+5=0 \\ p=-5 \end{gathered}$$

or$$\begin{gathered} p+4=0 \\ p=-4 \end{gathered}$$

So solutions of the equation arerole="math" localid="1647788718587" $p=-4\&-5$

Substitute $p=-4$in the equation

$$\begin{gathered} 1+\frac{9}{p}=\frac{-20}{p^2} \\ 1+\frac{9}{-4}\stackrel{?}{=}\frac{-20}{{\left(-4\right)}^2} \\ \frac{4}{4}-\frac{9}{4}\stackrel{?}{=}\frac{-20}{16} \\ \frac{-5}{4}=\frac{-5}{4} \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution $p=-4$is correct.

Substitute $p=-5$in the equation

$$\begin{gathered} 1+\frac{9}{p}=\frac{-20}{p^2} \\ 1+\frac{9}{-5}\stackrel{?}{=}\frac{-20}{{\left(-5\right)}^2} \\ \frac{5}{5}-\frac{9}{5}\stackrel{?}{=}\frac{-20}{25} \\ \frac{-4}{5}=\frac{-4}{5} \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution$p=-5$ is correct.