Question

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

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 \( c = 5 \).

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

First, we will compute the discriminant:

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

Calculating \( 13^2 \):

\[
13^2 = 169
\]

Now calculating \( 4ac \):

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

Now, substitute these values into the discriminant:

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

**Step 2: Apply the quadratic formula**

Now, substituting the values into the quadratic formula:

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

Calculating \( \sqrt{49} \):

\[
\sqrt{49} = 7
\]

So the equation becomes:

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

**Step 3: Solve for the two possible values of \( x \)**

Now we will find the two values for \( x \):

1. Using the positive sign:

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

2. Using the negative sign:

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

**Step 4: State the zeros**

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

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