Question

To determine the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \), we will set the equation equal to zero and solve for \( x \):
\[
2x^2 + 9x + 4 = 0
\]

To solve this quadratic equation, we can use the quadratic formula:

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

In this equation, \( a = 2 \), \( b = 9 \), and \( c = 4 \). Now we will calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 9^2 - 4 \times 2 \times 4
\]
\[
= 81 - 32
\]
\[
= 49
\]

Now, we can substitute the values of \( a \), \( b \), and the discriminant into the quadratic formula:

\[
x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2}
\]
\[
= \frac{-9 \pm 7}{4}
\]

Now we will calculate the two possible values for \( x \):

1. For \( -9 + 7 \):
\[
x = \frac{-2}{4} = -\frac{1}{2}
\]

2. For \( -9 - 7 \):
\[
x = \frac{-16}{4} = -4
\]

Thus, the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \) are:

\[
x = -\frac{1}{2} \quad \text{and} \quad x = -4
\]\
make his like a 6th grader wrote it

Answer 1

To find the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \), we need to make the equation equal to zero and solve for \( x \):

\[
2x^2 + 9x + 4 = 0
\]

We can use the quadratic formula to find \( x \). The quadratic formula is:

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

In our equation, the numbers are:
- \( a = 2 \)
- \( b = 9 \)
- \( c = 4 \)

First, let's find the part called the discriminant, which is \( b^2 - 4ac \):

\[
b^2 - 4ac = 9^2 - 4 \times 2 \times 4
\]
\[
= 81 - 32
\]
\[
= 49
\]

Now we can put our numbers into the quadratic formula:

\[
x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2}
\]
\[
= \frac{-9 \pm 7}{4}
\]

Next, we'll find the two possible values for \( x \):

1. For \( -9 + 7 \):
\[
x = \frac{-2}{4} = -\frac{1}{2}
\]

2. For \( -9 - 7 \):
\[
x = \frac{-16}{4} = -4
\]

So the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \) are:

\[
x = -\frac{1}{2} \quad \text{and} \quad x = -4
\]

That means the points where the graph crosses the x-axis are at \( x = -\frac{1}{2} \) and \( x = -4 \).