Question

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

Answer 1

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

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

where \( a = 6 \), \( b = 13 \), and \( c = 5 \).

1. Calculate the discriminant \( b^2 - 4ac \):

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

So, the discriminant is:

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

2. Now, apply the quadratic formula:

\[
x = \frac{-13 \pm \sqrt{49}}{2 \times 6}
\]
\[
x = \frac{-13 \pm 7}{12}
\]

3. Calculate the two possible values for \( x \):

- For the positive case:

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

- For the negative case:

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

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

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