Question

According to the Fundamental Theorem of Algebra, 4x^2+11x−20=0 has two solutions. How many of those solutions are integers?

Answer 1

To determine how many solutions of the quadratic equation \( 4x^2 + 11x - 20 = 0 \) are integers, we can use the quadratic formula:

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

Here, \( a = 4 \), \( b = 11 \), and \( c = -20 \).

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

\[
D = b^2 - 4ac = 11^2 - 4(4)(-20)
\]

Calculating \( 11^2 \):

\[
11^2 = 121
\]

Now calculating \( 4(4)(-20) \):

\[
4 \cdot 4 \cdot -20 = -320
\]

So, we find \( D \):

\[
D = 121 - (-320) = 121 + 320 = 441
\]

Next, we check if \( D \) is a perfect square. The square root of \( 441 \) is:

\[
\sqrt{441} = 21
\]

Now substituting \( D \) back into the quadratic formula:

\[
x = \frac{-11 \pm 21}{2 \cdot 4} = \frac{-11 \pm 21}{8}
\]

This gives us two possible solutions:

1.

\[
x_1 = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4}
\]

2.

\[
x_2 = \frac{-11 - 21}{8} = \frac{-32}{8} = -4
\]

Now, we have the two solutions \( x_1 = \frac{5}{4} \) and \( x_2 = -4 \).

Out of these, only \( x_2 = -4 \) is an integer.

Thus, the number of integer solutions for the quadratic equation \( 4x^2 + 11x - 20 = 0 \) is:

\[
\boxed{1}
\]