College

What are the zeros of the function [tex]y = 2x^2 + 9x + 4[/tex]?

A. [tex]x = -\frac{1}{2}, x = -4[/tex]
B. [tex]x = \frac{1}{2}, x = -4[/tex]
C. [tex]x = -\frac{1}{2}, x = 4[/tex]
D. [tex]x = \frac{1}{2}, x = 4[/tex]

Answer :

To find the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex], we'll use a method called the quadratic formula. This formula is used to find the roots of a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. The quadratic formula is:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

For the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex], the coefficients are:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 9 \)[/tex]
- [tex]\( c = 4 \)[/tex]

First, we calculate the discriminant, which is [tex]\( b^2 - 4ac \)[/tex]:

[tex]\[ b^2 = 9^2 = 81 \][/tex]
[tex]\[ 4ac = 4 \times 2 \times 4 = 32 \][/tex]
[tex]\[ \text{Discriminant} = b^2 - 4ac = 81 - 32 = 49 \][/tex]

Since the discriminant is positive, we have two real and distinct solutions. Now, let's calculate these solutions using the quadratic formula:

1. For the first root:

[tex]\[ x_1 = \frac{-b + \sqrt{49}}{2a} = \frac{-9 + 7}{4} = \frac{-2}{4} = -0.5 \][/tex]

2. For the second root:

[tex]\[ x_2 = \frac{-b - \sqrt{49}}{2a} = \frac{-9 - 7}{4} = \frac{-16}{4} = -4 \][/tex]

Therefore, the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex] are [tex]\( x = -0.5 \)[/tex] and [tex]\( x = -4 \)[/tex].

So, the correct answer is:
A. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]