Answer :
To find the zeros (also called the roots) of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make the function equal to zero. This means we need to solve the equation:
[tex]\[ 2x^2 + 9x + 4 = 0 \][/tex]
The general method for finding the zeros of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is to use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\( a = 2 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = 4 \)[/tex].
1. Calculate the Discriminant:
The discriminant of a quadratic equation is given by [tex]\( b^2 - 4ac \)[/tex].
[tex]\[
b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49
\][/tex]
2. Find the Roots Using the Quadratic Formula:
Since the discriminant is positive (49), there are two real and distinct solutions.
- Root 1:
[tex]\[
x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-9 + \sqrt{49}}{2 \cdot 2} = \frac{-9 + 7}{4} = \frac{-2}{4} = -\frac{1}{2}
\][/tex]
- Root 2:
[tex]\[
x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-9 - \sqrt{49}}{2 \cdot 2} = \frac{-9 - 7}{4} = \frac{-16}{4} = -4
\][/tex]
Thus, the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex] are [tex]\( x = -\frac{1}{2} \)[/tex] and [tex]\( x = -4 \)[/tex]. Therefore, the correct answer is:
D. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]
[tex]\[ 2x^2 + 9x + 4 = 0 \][/tex]
The general method for finding the zeros of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is to use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\( a = 2 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = 4 \)[/tex].
1. Calculate the Discriminant:
The discriminant of a quadratic equation is given by [tex]\( b^2 - 4ac \)[/tex].
[tex]\[
b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49
\][/tex]
2. Find the Roots Using the Quadratic Formula:
Since the discriminant is positive (49), there are two real and distinct solutions.
- Root 1:
[tex]\[
x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{-9 + \sqrt{49}}{2 \cdot 2} = \frac{-9 + 7}{4} = \frac{-2}{4} = -\frac{1}{2}
\][/tex]
- Root 2:
[tex]\[
x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-9 - \sqrt{49}}{2 \cdot 2} = \frac{-9 - 7}{4} = \frac{-16}{4} = -4
\][/tex]
Thus, the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex] are [tex]\( x = -\frac{1}{2} \)[/tex] and [tex]\( x = -4 \)[/tex]. Therefore, the correct answer is:
D. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]
