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

3 answers

To find 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 \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation in the form \( ax^2 + bx + c = 0 \). Here, we have:

  • \( a = 6 \)
  • \( b = 13 \)
  • \( c = 5 \)

Step 1: Calculate the Discriminant

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

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

Step 2: Apply the Quadratic Formula

Now we can substitute the values into the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-13 \pm \sqrt{49}}{2 \times 6} \] \[ = \frac{-13 \pm 7}{12} \]

Step 3: Calculate the Two Possible Values for \( x \)

Now, we will evaluate for both the plus and minus cases:

  1. For the plus case: \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]

  2. For the minus case: \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]

Step 4: Write the Final Answer

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

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

Thus, the zeros are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).

What the answer

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

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

These are the values of \( x \) where the function equals zero.