Question

Determine the zero of the quadratic function y = 6X^2 + 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. The quadratic formula states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions (or zeros) can be found using:

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

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

### Step 1: Calculate the Discriminant

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

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

Calculating \( 13^2 \):

\[
13^2 = 169
\]

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

\[
4 \cdot 6 = 24 \\
24 \cdot 5 = 120
\]

Now substituting back into the discriminant formula:

\[
D = 169 - 120 = 49
\]

### Step 2: Apply the Quadratic Formula

Since the discriminant \( D \) is positive, we will have two distinct real solutions. Next, we will use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{D}}{2a}
\]

Substituting \( b = 13 \), \( D = 49 \), and \( a = 6 \):

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

Calculating \( \sqrt{49} \):

\[
\sqrt{49} = 7
\]

Now substituting this into the equation:

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

### Step 3: Calculate the Two Solutions

Now we will calculate the two values of \( x \).

**First solution** (\( x_1 \)):

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

**Second solution** (\( x_2 \)):

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

### Conclusion

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

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