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Answer :
To find the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex], we can use the quadratic formula. The quadratic formula is given by:
[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 ([tex]\( \Delta \)[/tex]) is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = 9^2 - 4 \times 2 \times 4 = 81 - 32 = 49
\][/tex]
2. Evaluate the Roots:
Since the discriminant is positive ([tex]\( \Delta > 0 \)[/tex]), there are two real roots. We can calculate them by using:
[tex]\[
x_1 = \frac{{-b - \sqrt{\Delta}}}{2a}
\][/tex]
[tex]\[
x_2 = \frac{{-b + \sqrt{\Delta}}}{2a}
\][/tex]
Plugging in the values gives:
[tex]\[
x_1 = \frac{{-9 - \sqrt{49}}}{2 \times 2} = \frac{{-9 - 7}}{4} = \frac{{-16}}{4} = -4
\][/tex]
[tex]\[
x_2 = \frac{{-9 + \sqrt{49}}}{2 \times 2} = \frac{{-9 + 7}}{4} = \frac{{-2}}{4} = -\frac{1}{2}
\][/tex]
Thus, the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex] are [tex]\( x = -4 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/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]
Here, the coefficients are [tex]\( a = 2 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = 4 \)[/tex].
1. Calculate the Discriminant:
The discriminant ([tex]\( \Delta \)[/tex]) is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = 9^2 - 4 \times 2 \times 4 = 81 - 32 = 49
\][/tex]
2. Evaluate the Roots:
Since the discriminant is positive ([tex]\( \Delta > 0 \)[/tex]), there are two real roots. We can calculate them by using:
[tex]\[
x_1 = \frac{{-b - \sqrt{\Delta}}}{2a}
\][/tex]
[tex]\[
x_2 = \frac{{-b + \sqrt{\Delta}}}{2a}
\][/tex]
Plugging in the values gives:
[tex]\[
x_1 = \frac{{-9 - \sqrt{49}}}{2 \times 2} = \frac{{-9 - 7}}{4} = \frac{{-16}}{4} = -4
\][/tex]
[tex]\[
x_2 = \frac{{-9 + \sqrt{49}}}{2 \times 2} = \frac{{-9 + 7}}{4} = \frac{{-2}}{4} = -\frac{1}{2}
\][/tex]
Thus, the zeros of the function [tex]\( y = 2x^2 + 9x + 4 \)[/tex] are [tex]\( x = -4 \)[/tex] and [tex]\( x = -\frac{1}{2} \)[/tex].
So, the correct answer is:
B. [tex]\( x = -\frac{1}{2}, x = -4 \)[/tex]
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