Question

Find the zeros of the quadratic equation y=6x2+13x+5 . You may use any of the following methods: quadratic formula

Answer 1

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

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

In our equation, we have:

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

First, we 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 substitute \( a \), \( b \), and the discriminant into the quadratic formula:

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

Since \( \sqrt{49} = 7 \), we can simplify further:

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

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

1. For the positive case:
\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. For the negative case:
\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

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

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