Question

Determine the zero(s) of the quadratic function y=6x2+13x+5 . Show all steps.(2 points)

Answer 1

To determine the zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \), we need to solve the equation \( 6x^2 + 13x + 5 = 0 \).

We can use the quadratic formula:

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

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

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

\[
b^2 - 4ac = 13^2 - 4(6)(5)
\]

Calculating \( 13^2 \):

\[
13^2 = 169
\]

Calculating \( 4 \times 6 \times 5 \):

\[
4 \times 6 \times 5 = 120
\]

Now, plug these values into the discriminant:

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

**Step 2: Calculate the roots using the quadratic formula**

Since the discriminant is positive, there are two real and distinct solutions.

Now substitute \( a \), \( b \), and the discriminant into the formula:

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

Since \( \sqrt{49} = 7 \), we can rewrite it as:

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

Now solve for the two possible values of \( x \):

1. For the positive case:

\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. For the negative case:

\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

**Final Result:**

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

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