Question

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

Answer 1

To determine the zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \), we will use the quadratic formula:

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

where \( a \), \( b \), and \( c \) are the coefficients from the standard form of the quadratic equation \( ax^2 + bx + c \).

In our case, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

### Step 1: Calculate the discriminant

The first step is to compute the discriminant \( D = b^2 - 4ac \).

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

### Step 2: Apply the quadratic formula

Since the discriminant is positive (\( D = 49 \)), we will have two real and distinct solutions. Next, we substitute \( a \), \( b \), and \( D \) into the quadratic formula:

\[
x = \frac{-b \pm \sqrt{D}}{2a}
\]
\[
x = \frac{-13 \pm \sqrt{49}}{2 \times 6}
\]
\[
x = \frac{-13 \pm 7}{12}
\]

### Step 3: Calculate the two solutions

Now, we will calculate the two possible values for \( x \).

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

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

### Conclusion

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

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