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 $x(x+5) = -n$? Express your answer as a common fraction.
+592 The only way it can equal -n is if x itself is negative. we can expand x(x+5) into x^2+5x=-n. if -n has to be an integer, then so does x.we can count all the possible ways:
x=-1
x^2+5x=-4
x=-2
x^2+5x=-6
x=-3
x^2+5x=-6
x=-4
x^2+5x=-4
x=-5
x^2+5x=0
from this point on, there will be no possible solutions.
so the only numbers that has real solutions that any x(x+5)=-n is n=4 or n=6, which is 2/10, so the answer is 8/10=$\boxed{\dfrac45}$
