Question

According to the Fundamental Theorem of Algebra, 4x2+11x−20=0 has two solutions. How many of those solutions are integers?(1 point)

Answer 1

To find the integer solutions of the quadratic equation \(4x^2 + 11x - 20 = 0\), we can use the quadratic formula, which is given by:

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

For the equation \(4x^2 + 11x - 20 = 0\), we identify \(a = 4\), \(b = 11\), and \(c = -20\).

First, we need to calculate the discriminant \(D\):

\[
D = b^2 - 4ac
\]

Substituting the values of \(a\), \(b\), and \(c\):

\[
D = 11^2 - 4 \times 4 \times (-20)
\]
\[
D = 121 + 320
\]
\[
D = 441
\]

Next, we compute the square root of the discriminant:

\[
\sqrt{D} = \sqrt{441} = 21
\]

Now we can apply the quadratic formula:

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

This gives us two potential solutions:

1. For the positive case:

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

2. For the negative case:

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

Thus, the two solutions to the equation \(4x^2 + 11x - 20 = 0\) are:

\[
x = \frac{5}{4} \quad \text{and} \quad x = -4
\]

Among these solutions, the only integer solution is:

\(-4\)

Therefore, the number of integer solutions is:

\[
\boxed{1}
\]