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

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

In this case, \(a = 4\), \(b = 11\), and \(c = -20\). First, we calculate the discriminant, which is \(b^2 - 4ac\):

\[
b^2 = 11^2 = 121
\]
\[
4ac = 4 \cdot 4 \cdot (-20) = -320
\]
\[
b^2 - 4ac = 121 - (-320) = 121 + 320 = 441
\]

Now, we substitute back into the quadratic formula:

\[
x = \frac{-11 \pm \sqrt{441}}{2 \cdot 4} = \frac{-11 \pm 21}{8}
\]

Calculating the two possible values for \(x\):

1. \(x = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4}\)
2. \(x = \frac{-11 - 21}{8} = \frac{-32}{8} = -4\)

Thus, the solutions to the quadratic equation are \(x = \frac{5}{4}\) and \(x = -4\).

Since one of the solutions, \(-4\), is an integer and the other solution, \(\frac{5}{4}\), is not, there is only **one integer solution**.

Thus, the answer is:

**1**.