I choose a random integer n between 1 and 10 inclusive. What is the probability that for the n I chose, there exist no real solutions to the equation (x + 4)(x + 5) = n? Express your answer as a common fraction.
+128475 (x + 4)(x + 5) = n simplify as
x^2 + 9x + 20 - n = 0
This will have real solutions if the disxriminant is ≥ 0.....so.......
9^2 - 4 (20 - n) ≥ 0
81 - 80 + 4n ≥ 0
1 + 4n ≥ 0
4n ≥ -1
n ≥ -1/4
In other words.......any positive n will give real solutions....so....the probability of non-real solutions = 0%
