Question

the fundamental theorem of algebra practice
complete this assessment to review what youve learned. it will not count toward your grade.

according to the fundamental theorem of algebra, $4x^2 + 11x - 20 = 0$ has two solutions. how many of those solutions are integers? (1 point)

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Explanation:

Step1: Solve the quadratic equation

We use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). For the equation \(4x^{2}+11x - 20=0\), we have \(a = 4\), \(b=11\), \(c=- 20\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(11)^{2}-4\times4\times(-20)=121 + 320=441\).
Then, \(x=\frac{-11\pm\sqrt{441}}{2\times4}=\frac{-11\pm21}{8}\).

Step2: Find the two solutions

For the plus - sign: \(x_{1}=\frac{-11 + 21}{8}=\frac{10}{8}=\frac{5}{4}\) (not an integer).
For the minus - sign: \(x_{2}=\frac{-11-21}{8}=\frac{-32}{8}=- 4\) (an integer).
So out of the two solutions, only 1 is an integer.

Answer:

1