Question

Determine the zero(s) of the quadratic function y=6x^2+13x+5. Use Algebra (factoring or quadratic formula) to prove your answer. Please write it simply.

Answer 1

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

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

In our quadratic function, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

Now, we first need to calculate the discriminant \( b^2 - 4ac \):

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

Next, we compute the zeros using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-13 \pm \sqrt{49}}{2 \times 6}
\]
\[
x = \frac{-13 \pm 7}{12}
\]

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

1. Using \( -13 + 7 \):
\[
x = \frac{-6}{12} = -\frac{1}{2}
\]

2. Using \( -13 - 7 \):
\[
x = \frac{-20}{12} = -\frac{5}{3}
\]

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

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

Thus, the final answer is:

\[
\boxed{-\frac{1}{2}, -\frac{5}{3}}
\]