Chapter 8.5, Problem 73SR

To determine

To find: The value of each equation.

Answer to Problem 73SR

The value is $x=6,16$ .

Explanation of Solution

Given information:

The given information is $22x-x^2=96$ .

Calculation:

Consider the equation:

  $\begin{array}{c}22x-x^2=96\\ -x^2+22x-96=0\\ x^2-22x+96=0\end{array}$

  $\begin{array}{c}{x}^2-22x+96=0\\ x^2-16x-6x+96=0\\ x\left(x-16\right)-6\left(x-16\right)=0\end{array}$

  $\begin{array}{c}\left(x-6\right)\left(x-16\right)=0\\ x=6,16\end{array}$

Insert each value of $x$ into the original equation to check for equivalency.

When $x=6$ ,

  $\begin{array}{c}22x-x^2=96\\ 22\times 6-{\left(6\right)}^2=96\\ 96=96\end{array}$

When $x=16$ ,

  $\begin{array}{c}22x-x^2=96\\ 22\times 16-{\left(16\right)}^2=96\\ 96=96\end{array}$

Therefore, the value is $x=6,16$ .