Question

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

Answer 1

To find the zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \), we need to solve the equation \( 6x^2 + 13x + 5 = 0 \). We can use the quadratic formula:

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

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). For our function:

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

### Step 1: Calculate the Discriminant

The first step is to compute the discriminant \( b^2 - 4ac \):

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

### Step 2: Apply the Quadratic Formula

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

\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
\[
\sqrt{49} = 7
\]
Now substituting back into the formula:

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

### Step 3: Find the Two Solutions

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

1. **Using the positive square root:**

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

2. **Using the negative square root:**

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

### Conclusion

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

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