June 5th, 2014, 01:28 AM  #1 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0  Equation problem
Hi, can anyone please help me solving this equation: sqrt(x+34sqrt(x1))+sqrt(x+86sqrt(x1))=1 Thanks in advance 
June 5th, 2014, 01:42 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162 
If you substitute x = u² + 1, where u > 0, what do you get?

June 5th, 2014, 02:11 AM  #3 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 
I have done something like that, but when I searched for solutions it goes like this (u2)+(u3)=1 but u2 is in absolute value and so is u1 (I just couldn't find absolute value sign) and it says: 2=<u=<3 and I don't understand why, basically that is the only problem. Last edited by skipjack; June 5th, 2014 at 03:19 PM. 
June 5th, 2014, 05:38 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, wannabemathlete! Quote:
You substituted $x \,=\,u^2+1$ $\quad$and got: $\;u2 + u3 \:=\:1\;\;$ [Correct!] At this point, I reverted to babytalk. $u$ is a number such that its distance from 2 $\quad$plus its distance from 3 is equal to 1. The only such number is: $\:u = 2.5 = \frac{5}{2}$ Therefore: $\:x \:=\:\left(\frac{5}{2}\right)^2 + 1 \:=\:\dfrac{29}{4}$  
June 5th, 2014, 03:31 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162  It's correct! If you think about it, you will see that soroban's geometrical approach is valid, but soroban drew the wrong conclusion. Edit: I rather like this problem. It's actually quit easy, yet cleverly designed so that even an experienced solver may well slip up, which is exactly what happened. Last edited by skipjack; June 5th, 2014 at 03:36 PM. 
June 5th, 2014, 03:35 PM  #6 
Member Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving  
June 5th, 2014, 09:08 PM  #7 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 
Basically, any number from 5 to 10, including those two numbers can be x. Too bad noone actually came up with it, but anyway, if someone solves it please post it here. Thank you guys anyway. Last edited by skipjack; June 5th, 2014 at 11:41 PM. 
June 5th, 2014, 09:24 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
\begin{align*} x &= u^2 + 1 \\ 2 &\le u \le 3 \\ 5 &\le u^2 + 1 \le 10 \\ 5 &\le x \le 10 \\ \end{align*} 
June 5th, 2014, 10:32 PM  #9 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 
Thanks a lot v8archie. Last edited by skipjack; June 5th, 2014 at 11:55 PM. 
June 5th, 2014, 11:53 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162  It was you that first came up with it, albeit adding that you didn't understand why. Then soroban explained why by considering distance. Just replace soroban's mistaken conclusion that u = 2.5 with the conclusion that 2 ≤ u ≤ 3. If you still can't understand soroban's approach, read the first two paragraphs of this article.


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