Question

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

Answer 1

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

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

In our function, we identify the coefficients \( a \), \( b \), and \( c \):

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

### Step 1: Calculate the Discriminant

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = 13^2 = 169
\]

\[
4ac = 4 \cdot 6 \cdot 5 = 120
\]

Now, compute the discriminant:

\[
b^2 - 4ac = 169 - 120 = 49
\]

### Step 2: Apply the Quadratic Formula

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

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

Calculating \( \sqrt{49} \):

\[
\sqrt{49} = 7
\]

Now, substitute back into the formula:

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

### Step 3: Calculate the Two Possible Values of \( x \)

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

1. Using the plus sign:

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

2. Using the minus sign:

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

### Final Result

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

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

These values can also be presented as:

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