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 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+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=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 (u-2)+(u-3)=1 but u-2 is in absolute value and so is u-1 (I just couldn't find absolute value sign) and it says: 2=
June 5th, 2014, 05:38 AM   #4
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Hello, wannabemathlete!

Quote:
 Solve for $x\!:\;\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} \:=\:1$

You substituted $x \,=\,u^2+1$

$\quad$and got: $\;|u-2| + |u-3| \:=\:1\;\;$ [Correct!]

At this point, I reverted to baby-talk.

$u$ 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
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Quote:
 Originally Posted by wannabemathlete . . . it says: 2=
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
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Quote:
 Originally Posted by soroban $u$ is a number such that its distance from 2 $\quad$plus its distance from 3 is equal to 1. The only such number is: $\:u = 2.5 = \frac{5}{2}$
Surely any number from 2 - 3 works for this, or am I just being stupid?

 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 no-one 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*} Thanks from wannabemathlete
 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
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Quote:
 Originally Posted by wannabemathlete Too bad no-one actually came up with it
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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