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June 5th, 2014, 01:28 AM   #1
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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
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June 5th, 2014, 01:42 AM   #2
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If you substitute x = u² + 1, where u > 0, what do you get?
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June 5th, 2014, 02:11 AM   #3
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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=<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.
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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: $\:x \:=\:\left(\frac{5}{2}\right)^2 + 1 \:=\:\dfrac{29}{4}$
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June 5th, 2014, 03:31 PM   #5
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Quote:
Originally Posted by wannabemathlete View Post
. . . it says: 2=<u=<3 and I don't understand why
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.
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June 5th, 2014, 03:35 PM   #6
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Quote:
Originally Posted by soroban View Post


$u$ 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?
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June 5th, 2014, 09:08 PM   #7
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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.
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June 5th, 2014, 09:24 PM   #8
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\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*}
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June 5th, 2014, 10:32 PM   #9
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Thanks a lot v8archie.

Last edited by skipjack; June 5th, 2014 at 11:55 PM.
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June 5th, 2014, 11:53 PM   #10
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Quote:
Originally Posted by wannabemathlete View Post
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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