According to the Fundamental Theorem of Algebra, 4x2+11x−20=0

Question

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

GPT-4o mini · Answer

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

In our case, the coefficients are:

$a = 4$
$b = 11$
$c = - 20$

First, we need to calculate the discriminant $b^{2} - 4 a c$ :

$b^{2} = 11^{2} = 121$ $4 a c = 4 \cdot 4 \cdot (- 20) = - 320$

Now, adding these together:

$b^{2} - 4 a c = 121 - (- 320) = 121 + 320 = 441$

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 $x$ :

For the positive case:

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

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 $4 x^{2} + 11 x - 20 = 0$ , only one solution is an integer.

Therefore, the number of integer solutions is:

$1$