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

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
9 hours ago

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:

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

  1. 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:

\[ \boxed{1} \]