Question

Determine the zero(s) of the quadratic function y=6x^2+13x+5
. Show all steps in word form even the numbers and signs.(2 poin

Answer 1

To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we will follow the steps of using the quadratic formula.

1. **Identify the coefficients**:
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \).
For our function, we can identify:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

2. **Write the quadratic formula**:
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

3. **Substitute the values into the formula**:
We will substitute the values of \( a \), \( b \), and \( c \) into the formula:
\[
x = \frac{-13 \pm \sqrt{13^2 - 4 \times 6 \times 5}}{2 \times 6}
\]

4. **Calculate \( b^2 - 4ac \)**:
First, we need to calculate \( 13^2 \):
- \( 13^2 = 169 \)

Next, calculate \( 4ac \):
- \( 4 \times 6 = 24 \)
- Now, multiply by \( c \):
- \( 24 \times 5 = 120 \)

Now we subtract:
\[
b^2 - 4ac = 169 - 120 = 49
\]

5. **Continue substituting back into the formula**:
Now we substitute back into the quadratic formula:
\[
x = \frac{-13 \pm \sqrt{49}}{12}
\]

6. **Calculate the square root**:
The square root of \( 49 \) is \( 7 \):
\[
x = \frac{-13 \pm 7}{12}
\]

7. **Find the two potential solutions**:
We will calculate the two values from the plus and minus:
- Using the plus sign:
\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
- Using the minus sign:
\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

8. **State the zeros of the quadratic function**:
Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
- \( x = -\frac{1}{2} \)
- \( x = -\frac{5}{3} \)

In conclusion, the zeros of the function are \( -\frac{1}{2} \) and \( -\frac{5}{3} \).