# 2007 iTest Problems/Problem 36

## Problem

Let b be a real number randomly selected from the interval $[-17,17]$. Then, m and n are two relatively prime positive integers such that m/n is the probability that the equation $x^4+25b^2=(4b^2-10b)x^2$ has $\textit{at least}$ two distinct real solutions. Find the value of $m+n$.

## Solution

The equation has quadratic form, so complete the square to solve for x. $$x^4 - (4b^2 - 10b)x^2 + 25b^2 = 0$$ $$x^4 - (4b^2 - 10b)x^2 + (2b^2 - 5b)^2 - 4b^4 + 20b^3 = 0$$ $$(x^2 - (2b^2 - 5b))^2 = 4b^4 - 20b^3$$

In order for the equation to have real solutions, $$16b^4 - 80b^3 \ge 0$$ $$b^3(b - 5) \ge 0$$ $$b \le 0 \text{ or } b \ge 5$$

Note that $2b^2 - 5b = b(2b-5)$ is greater than or equal to $0$ when $b \le 0$ or $b \ge 5$. Also, if $b = 0$, then expression leads to $x^4 = 0$ and only has one unique solution, so discard $b = 0$ as a solution. The rest of the values leads to $b^2$ equalling some positive value, so these values will lead to two distinct real solutions.

Therefore, in interval notation, $b \in [-17,0) \cup [5,17]$, so the probability that the equation has at least two distinct real solutions when $b$ is randomly picked from interval $[-17,17]$ is $\frac{29}{34}$. This means that $m+n = \boxed{63}$.

## See Also

 2007 iTest (Problems) Preceded by:Problem 35 Followed by:Problem 37 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Invalid username
Login to AoPS