Question

. Find the zeros of the quadratic equation y=6x2+13x+5
. You may use any of the following methods: factoring, square roots, complete the square, or quadratic formula

Answer 1

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

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

Here, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

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

\[
b^2 = 13^2 = 169
\]

\[
4ac = 4 \cdot 6 \cdot 5 = 120
\]

Now, we find the discriminant:

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

Since the discriminant is positive, there will be two real solutions. Now we apply the quadratic formula:

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

Calculating the square root:

\[
\sqrt{49} = 7
\]

Now we substitute back into the formula:

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

This gives us two solutions:

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

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

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

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