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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 :

We start with the quadratic function

$$
y = 2x^2 + 9x + 4.
$$

To find its zeros, we need to solve the equation

$$
2x^2 + 9x + 4 = 0.
$$

**Step 1. Identify the coefficients.**

The quadratic equation is in the form

$$
ax^2 + bx + c = 0,
$$

where

$$
a = 2,\quad b = 9,\quad c = 4.
$$

**Step 2. Calculate the discriminant.**

The discriminant $\Delta$ is given by

$$
\Delta = b^2 - 4ac.
$$

Substitute the values of $a$, $b$, and $c$:

$$
\Delta = 9^2 - 4(2)(4) = 81 - 32 = 49.
$$

Since the discriminant is positive, we have two real zeros.

**Step 3. Use the quadratic formula.**

The quadratic formula is

$$
x = \frac{-b \pm \sqrt{\Delta}}{2a}.
$$

Substitute the known values:

$$
x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} = \frac{-9 \pm 7}{4}.
$$

**Step 4. Compute the two zeros.**

For the first zero, use the $+$ sign:

$$
x = \frac{-9 + 7}{4} = \frac{-2}{4} = -\frac{1}{2}.
$$

For the second zero, use the $-$ sign:

$$
x = \frac{-9 - 7}{4} = \frac{-16}{4} = -4.
$$

Thus, the zeros of the function are

$$
x = -\frac{1}{2} \quad \text{and} \quad x = -4.
$$

**Answer:** Option B.

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