Question

Determine the zero(s) of the quadratic function y=6x2+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 can follow these steps:

### Step One: Identify the coefficients
In the quadratic expression \( ax^2 + bx + c \):
- The coefficient \( a \) is six.
- The coefficient \( b \) is thirteen.
- The coefficient \( c \) is five.

### Step Two: Use the quadratic formula
The formula to find the zeros of a quadratic function is given by:
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]
We will substitute the values of \( a \), \( b \), and \( c \) into this formula.

### Step Three: Calculate the discriminant
We need to calculate the discriminant, which is \( b^2 - 4ac \):
- Calculate \( b^2 \):
\[
b^2 = 13^2 = 169
\]

- Calculate \( 4ac \):
\[
4ac = 4 \times 6 \times 5 = 120
\]

- Now, subtract \( 4ac \) from \( b^2 \):
\[
b^2 - 4ac = 169 - 120 = 49
\]

### Step Four: Substitute back into the quadratic formula
Now that we have \( b^2 - 4ac = 49 \), we can find \( \sqrt{49} \):
\[
\sqrt{49} = 7
\]

Substituting these values into the quadratic formula gives:
\[
x = \frac{{-13 \pm 7}}{2 \times 6}
\]
This simplifies to:
\[
x = \frac{{-13 \pm 7}}{12}
\]

### Step Five: Calculate both solutions
Now we will calculate the two possible values for \( x \):

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

2. Second value:
\[
x = \frac{{-13 - 7}}{12} = \frac{{-20}}{12} = -\frac{5}{3}
\]

### Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
- \( x = -\frac{1}{2} \)
- \( x = -\frac{5}{3} \)

Thus, the solutions to the equation are \( x = -0.5 \) and \( x \approx -1.67 \).