Question

According to the Fundamental Theorem of Algebra, 4x2+11x−20=0
has two solutions. How many of those solutions are integers?

Answer 1

To determine the number of integer solutions for the equation \(4x^2 + 11x - 20 = 0\), we start by applying the quadratic formula, which is given by:

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

In our case, the coefficients are:
- \(a = 4\)
- \(b = 11\)
- \(c = -20\)

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

\[
b^2 = 11^2 = 121
\]
\[
4ac = 4 \cdot 4 \cdot (-20) = -320
\]

Now, adding these together:

\[
b^2 - 4ac = 121 - (-320) = 121 + 320 = 441
\]

Next, we check if the discriminant is a perfect square, which will indicate whether the solutions are rational (and possibly integers). We find the square root of \(441\):

\[
\sqrt{441} = 21
\]

Now we substitute back into the quadratic formula:

\[
x = \frac{-11 \pm 21}{2 \cdot 4}
\]

This results in two possible values for \(x\):

1. For the positive case:

\[
x = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4}
\]

2. For the negative case:

\[
x = \frac{-11 - 21}{8} = \frac{-32}{8} = -4
\]

Now we examine the results:

- The first solution \(x = \frac{5}{4}\) is not an integer.
- The second solution \(x = -4\) is an integer.

Thus, among the two solutions of the equation \(4x^2 + 11x - 20 = 0\), only one solution is an integer.

Therefore, the number of integer solutions is:

\[
\boxed{1}
\]