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

1 answer

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

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

where \( a = 6 \), \( b = 13 \), and \( c = 5 \).

Step 1: Calculate the discriminant

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

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

Step 2: Use the quadratic formula

Now that we have the discriminant, we can substitute 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 can solve for both possible values of \( x \):

  1. For the positive case:

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

  1. For the negative case:

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

Final Result:

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

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