Question

find the zeros of 8x^2+14x+3. List the smaller one first

Answer 1

To find the zeros of the quadratic equation \(8x^2 + 14x + 3 = 0\), we can use the quadratic formula:

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

where \(a = 8\), \(b = 14\), and \(c = 3\).

First, we calculate the discriminant:

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

Now we can substitute the values into the quadratic formula:

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

This gives us two possible solutions:

1. For the positive root:
\[
x = \frac{-14 + 10}{16} = \frac{-4}{16} = -\frac{1}{4}
\]

2. For the negative root:
\[
x = \frac{-14 - 10}{16} = \frac{-24}{16} = -\frac{3}{2}
\]

Thus, the zeros of the equation \(8x^2 + 14x + 3 = 0\) are:

\(-\frac{3}{2}\) and \(-\frac{1}{4}\).

Listing the smaller one first, the zeros are:

\[
-\frac{3}{2}, -\frac{1}{4}
\]