Chapter 6.1, Problem 53WE

To determine

To calculate: The value of x for which the value of fraction becomes zero $\frac{x^2-2x-15}{x^2+3x-40}$.

Answer to Problem 53WE

The value of x for which the value of fraction $\frac{x^2-2x-15}{x^2+3x-40}$ becomes zero is $-3$.

Explanation of Solution

Given information:

The expression $\frac{x^2-2x-15}{x^2+3x-40}$.

Formula used:

To find the value of the variable that makes the value of fraction zero, equate the fraction to zero and then solve it to find the value(s) of variable that makes it zero and then put the values in the original fraction to check which value makes the fraction zero.

Calculation:

Consider the expression $\frac{x^2-2x-15}{x^2+3x-40}$.

Recall that to find the value of the variable that makes the value of fraction zero, equate the fraction to zero and then solve it to find the value(s) of variable that makes it zero and then put the values in the original fraction to check which value makes the fraction zero.

So, equate the given fraction to zero as,

  $\frac{x^2-2x-15}{x^2+3x-40}=0$

Multiply both sides by $x^2+3x-40$ and then solve it further as,

  $\begin{array}{c}\frac{x^2-2x-15}{x^2+3x-40}\cdot \left(x^2+3x-40\right)=0\cdot \left(x^2+3x-40\right)\\ x^2-2x-15=0\\ x^2-5x+3x-15=0\\ \left(x-5\right)\left(x+3\right)=0\end{array}$

Either $\begin{array}{c}x-5=0\\ x=5\end{array}$ or $\begin{array}{c}x+3=0\\ x=-3\end{array}$.

Put $x=5$ in $\frac{x^2-2x-15}{x^2+3x-40}$ as,

  $\begin{array}{c}\frac{x^2-2x-15}{x^2+3x-40}=\frac{{\left(5\right)}^2-2\left(5\right)-15}{{\left(5\right)}^2+3\left(5\right)-40}\\ =\frac{25-10-15}{25+15-40}\\ =\frac{0}{0}\\ =\infty \end{array}$

Now, put $x=-3$ in $\frac{x^2-2x-15}{x^2+3x-40}$ as,

  $\begin{array}{c}\frac{x^2-2x-15}{x^2+3x-40}=\frac{{\left(-3\right)}^2-2\left(-3\right)-15}{{\left(-3\right)}^2+3\left(-2\right)-40}\\ =\frac{9+6-15}{9-6-40}\\ =\frac{0}{-37}\\ =0\end{array}$

Thus, the value of x for which the value of fraction $\frac{x^2-2x-15}{x^2+3x-40}$ becomes zero is $-3$.