Question

determine the zeros of y = 6x^2 + 13x + 5

Answer 1

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

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

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

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

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

Now that we have the discriminant, we can proceed to calculate the two possible values of \( x \):

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

Now we calculate the two solutions:

1. Using the plus sign:
\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. Using the minus sign:
\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

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

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