There are relatively prime positive integers $m,n$ such that $b-a = \frac{m}{n}$. Find $m+n$ $|x^2 - \frac{5}{2}| \le |x-8|$ if and only if $x$ is in the interval $[a,b]$. There are relatively prime positive integers $m,n$ such that $b-a = \frac{m}{n}$. Find $m+n$.
I have no clue how to approach it.Can anyone help? Thanks in advance.  

Edit 

After Googling on the problem I found that this problem is Purple Comet 2013 HS Problem 9 http://purplecomet.org/home/resource/543/2013HighSchoolProblems.pdf
Here is the solution of problem posted in AOPS
https://artofproblemsolving.com/community/c149h531513_purple_comet_2013_hs_problem_9
$$|x^2 - \frac{5}{2}| \le |x-8|\\ \to \begin{cases}x^2-\frac52=x-8\\x^2-\frac52=-(x-8)\end{cases}\\x=\frac{-1\pm \sqrt{43}}{2}$$
$$x\in [\frac{-1- \sqrt{43}}{2},\frac{-1+ \sqrt{43}}{2}]=[a,b]\\ \to b-a=\frac{-1+ \sqrt{43}}{2}-\frac{-1- \sqrt{43}}{2}$$
$$|x^2 - \frac{5}{2}| \le |x-8|\\ \to \begin{cases}x^2-\frac52=x-8\\x^2-\frac52=-(x-8)\end{cases}\\x=\frac{-1\pm \sqrt{43}}{2}$$
$$x\in [\frac{-1- \sqrt{43}}{2},\frac{-1+ \sqrt{43}}{2}]=[a,b]\\ \to b-a=\frac{-1+ \sqrt{43}}{2}-\frac{-1- \sqrt{43}}{2}$$