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

Question

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.