Question

Determine the zero(s) of the quadratic function y=6x2+13x+5
. Show all steps.(2 points)

Answer 1

To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:

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

In this case, we have:

- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

### Step 1: Calculate the Discriminant

First, we need to find the discriminant, \( D = b^2 - 4ac \).

\[
D = 13^2 - 4 \cdot 6 \cdot 5
\]
\[
D = 169 - 120
\]
\[
D = 49
\]

### Step 2: Apply the Quadratic Formula

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

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

\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]

### Step 3: Calculate the Square Root

\[
\sqrt{49} = 7
\]

### Step 4: Substitute and Simplify

Now substitute back into the formula:

\[
x = \frac{-13 \pm 7}{12}
\]

This gives us two solutions:

1. **First Solution:**

\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. **Second Solution:**

\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

### Final Result

The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:

\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]