Question

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

Answer 1

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

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

In our case, we have:

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

### Step 1: Calculate the discriminant

First, we need to calculate the discriminant \( D \):

\[
D = b^2 - 4ac
\]

Substituting the values of \( a \), \( b \), and \( c \):

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

### Step 2: Apply the quadratic formula

Now, we can substitute the values into the quadratic formula:

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

Since \( \sqrt{49} = 7 \), we have:

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

### Step 3: Calculate the two possible values for \( x \)

Now we will find both values for \( x \):

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

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

### Step 4: Present the final results

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

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

These are the points where the function intersects the x-axis.