Q. 296 - Intermediate Algebra

Question

In the following exercises, solve:

$\left(p-5\right)\left(p+3\right)=-7$

Step-by-Step Solution

Answer

On solving the equation, we get $p=4,-2$.

Step 1. Given information.

Consider the given equation,

$\left(p-5\right)\left(p+3\right)=-7$

Write the quadratic equation in standard form,

$$\begin{gathered} p^2+3p-5p-15=-7 \\ {p}^2-2p-8=0 \end{gathered}$$

Step 2. Factor the quadratic expression.

On factoring, we get,

$$\begin{gathered} p^2-4p+2p-8=0 \\ p\left(p-4\right)+2\left(p-4\right)=0 \\ \left(p-4\right)\left(p+2\right)=0 \end{gathered}$$

Use the zero product property,

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

Then,

$$\begin{gathered} p+2=0 \\ p=-2 \end{gathered}$$

Step 3. Check the answers.

Substitute $p=4$ in the given equation,

$$\begin{gathered} \left(4-5\right)\left(4+3\right)=-7 \\ \left(-1\right)\left(7\right)=-7 \\ -7=-7 \end{gathered}$$

This is true.

Substitute $p=-2$ in the given equation,

$$\begin{gathered} \left(-2-5\right)\left(-2+3\right)=-7 \\ \left(-7\right)\left(1\right)=-7 \\ -7=-7 \end{gathered}$$

This is also true.

Hence, the answer is $p=4,-2$.