Find the zeros of the quadratic equation y=6x2+13x+5

. You may use any of the following methods: factoring, square roots, complete the square, or quadratic formula.

1 answer

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

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

In this case, the coefficients are:

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

Now we can calculate the values step by step:

  1. Calculate the discriminant \( b^2 - 4ac \):

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

  1. Calculate the square root of the discriminant:

\[ \sqrt{49} = 7 \]

  1. Now apply the quadratic formula:

\[ x = \frac{{-13 \pm 7}}{{2 \cdot 6}} = \frac{{-13 \pm 7}}{12} \]

  1. Calculate the two possible values for \( x \):
  • For the plus sign: \[ x_1 = \frac{{-13 + 7}}{{12}} = \frac{{-6}}{{12}} = -\frac{1}{2} \]

  • For the minus sign: \[ x_2 = \frac{{-13 - 7}}{{12}} = \frac{{-20}}{{12}} = -\frac{5}{3} \]

Thus, the zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) are:

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