Problem:
A real number a is chosen randomly and uniformly from the interval [−20,18]. The probability that the roots of the polynomial
x4+2ax3+(2a−2)x2+(−4a+3)x−2
are all real can be written in the form nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let the polynomial be P. Note that
P(x)=(x4−2x2+3x−2)+2a(x3+x2−2x)=(x−1)(x+2)(x2−x+1)+2ax(x−1)(x+2)=(x−1)(x+2)(x2+(2a−1)x+1).
The roots of P(x) will be real precisely when the roots of x2+(2a−1)x+1 are real, which happens if and only if the discriminant (2a−1)2−4≥0. This is equivalent to a≤−21 or a≥23. The desired probability is therefore the probability that a randomly chosen real number from [−20,18] does not lie in (−21,23), which is 18−(−20)18−(−20)−2=3836=1918. The requested sum is 18+19=37.
The problems on this page are the property of the MAA's American Mathematics Competitions