+816 I choose a random integer 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.
+128564 x ( x + 5) = -n add n to both sides
x^2 + 5x + n = 0
This will have no real solutions when the disciminant of the quadratic is < 0
So we have
5^2 - 4 (1)(n) < 0
25 - 4n < 0
25 < 4n
25 / 4 < n ⇒ n > 6.5
So n = 7, 8, 9, 10 mean that we have no real solutions for those n's
Then....the probability that there are no real solutions is 4/10
+198 First we find the solution set that results in the equation having no real solutions. We begin by rearranging the equation \(x(x+5) = -n\) to \(x^2 + 5x + n = 0\). If the discriminant \(x^2 + 5x + n = 0\), then there are no real solutions. Thus, we want to solve for \(n\) in the inequality \(25 - 4n < 0\). Adding \(4n\)and dividing by 4, we find \(n>6.25\). The probability that I chose one of the numbers 7, 8, 9, or 10 is \(\boxed{\frac{2}{5}}\) .
