Answer :
To find the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex], we need to solve the equation [tex]\( 2x^2 + 9x + 4 = 0 \)[/tex]. This is a quadratic equation, and we can use the quadratic formula to find the solutions. The quadratic formula is:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
For the equation [tex]\( 2x^2 + 9x + 4 = 0 \)[/tex], the coefficients are:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 9 \)[/tex]
- [tex]\( c = 4 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = 9^2 - 4 \times 2 \times 4 = 81 - 32 = 49
\][/tex]
2. Apply the quadratic formula:
The solutions for [tex]\( x \)[/tex] are:
[tex]\[
x_1 = \frac{{-b + \sqrt{\text{Discriminant}}}}{2a} = \frac{{-9 + \sqrt{49}}}{4} = \frac{{-9 + 7}}{4} = \frac{-2}{4} = -0.5
\][/tex]
[tex]\[
x_2 = \frac{{-b - \sqrt{\text{Discriminant}}}}{2a} = \frac{{-9 - \sqrt{49}}}{4} = \frac{{-9 - 7}}{4} = \frac{-16}{4} = -4
\][/tex]
Therefore, the zeros of the function are [tex]\( x = -\frac{1}{2} \)[/tex] and [tex]\( x = -4 \)[/tex].
So, the correct answer is:
B. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
For the equation [tex]\( 2x^2 + 9x + 4 = 0 \)[/tex], the coefficients are:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 9 \)[/tex]
- [tex]\( c = 4 \)[/tex]
Step-by-step solution:
1. Calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = 9^2 - 4 \times 2 \times 4 = 81 - 32 = 49
\][/tex]
2. Apply the quadratic formula:
The solutions for [tex]\( x \)[/tex] are:
[tex]\[
x_1 = \frac{{-b + \sqrt{\text{Discriminant}}}}{2a} = \frac{{-9 + \sqrt{49}}}{4} = \frac{{-9 + 7}}{4} = \frac{-2}{4} = -0.5
\][/tex]
[tex]\[
x_2 = \frac{{-b - \sqrt{\text{Discriminant}}}}{2a} = \frac{{-9 - \sqrt{49}}}{4} = \frac{{-9 - 7}}{4} = \frac{-16}{4} = -4
\][/tex]
Therefore, the zeros of the function are [tex]\( x = -\frac{1}{2} \)[/tex] and [tex]\( x = -4 \)[/tex].
So, the correct answer is:
B. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]
