Question

which value(s) of x will be excluded from the domain of the rational function? $y = \frac{x + 5}{2x^{2}+7x + 5}$
-5, $\frac{1}{2}$
$\frac{1}{2}$, 5
-$\frac{1}{2}$, 5
5

Explanation:

Step1: Set denominator equal to 0

$2x^{2}+7x + 5=0$

Step2: Factor the quadratic equation

$2x^{2}+7x + 5=(2x + 5)(x+1)=0$

Step3: Solve for x

For $2x + 5=0$, we have $2x=-5$, so $x =-\frac{5}{2}$. For $x + 1=0$, we have $x=-1$. But if we factor the original denominator $2x^{2}+7x + 5$ in another way:
We can also use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 2$, $b=7$, $c = 5$.
$x=\frac{-7\pm\sqrt{7^{2}-4\times2\times5}}{2\times2}=\frac{-7\pm\sqrt{49 - 40}}{4}=\frac{-7\pm\sqrt{9}}{4}=\frac{-7\pm3}{4}$
$x_1=\frac{-7 + 3}{4}=\frac{-4}{4}=-1$ and $x_2=\frac{-7-3}{4}=\frac{-10}{4}=-\frac{5}{2}$
The values of $x$ that make the denominator 0 are the values to be excluded from the domain.

Answer:

$x=- \frac{5}{2},-1$ (It seems there is a mistake in the provided options as the correct values are not among them. The values that should be excluded from the domain are the roots of the denominator $2x^{2}+7x + 5 = 0$ which are $x=-\frac{5}{2}$ and $x=-1$)