Question

Determine the zero(s) of the quadratic function y=8x^2+14x+3

Answer 1

To determine the zeros of the quadratic function \( y = 8x^2 + 14x + 3 \), we can use the quadratic formula:

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

In our case, the coefficients are:
- \( a = 8 \)
- \( b = 14 \)
- \( c = 3 \)

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

\[
b^2 = 14^2 = 196
\]
\[
4ac = 4 \cdot 8 \cdot 3 = 96
\]
\[
b^2 - 4ac = 196 - 96 = 100
\]

Since the discriminant is positive, there are two distinct real zeros.

Now, we can substitute back into the quadratic formula:

\[
x = \frac{-14 \pm \sqrt{100}}{2 \cdot 8}
\]
\[
x = \frac{-14 \pm 10}{16}
\]

Now we calculate the two potential solutions:

1. \( x_1 = \frac{-14 + 10}{16} = \frac{-4}{16} = -\frac{1}{4} \)
2. \( x_2 = \frac{-14 - 10}{16} = \frac{-24}{16} = -\frac{3}{2} \)

Thus, the zeros of the quadratic function \( y = 8x^2 + 14x + 3 \) are:

\[
x = -\frac{1}{4} \quad \text{and} \quad x = -\frac{3}{2}
\]