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How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.

$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$

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Please use LaTeX formatting - 5 weeks is enough time to learn! –  nbubis Jul 22 '13 at 8:23
    
@nbubis I'm not sure if it's $\sqrt{x}-1$ or $\sqrt{x-1}$ –  egreg Jul 22 '13 at 8:24
    
@egreg - There were no brackets, so I assumed $\sqrt{x}$. –  nbubis Jul 22 '13 at 8:25
    
its the 2nd one. –  maths lover Jul 22 '13 at 8:25
    
change it to whole root of x-1 –  maths lover Jul 22 '13 at 8:25

3 Answers 3

Straight, you can get the following equation :

$$ \sqrt{(2-\sqrt{x-1})^2} + \sqrt{(3-\sqrt{x-1})^2} =1 $$

which leads to the following equation :

$$ |2-\sqrt{x-1}| + |3-\sqrt{x-1}| =1 $$

Then you will have three cases to discuss :

  • case : $\sqrt{x-1} \leq2$ (equivalent to $x\leq5$) :

$\sqrt{x-1} = 2$ then $x = 5$

  • case : $\sqrt{x-1} >2$ and $\sqrt{x-1} <3$ (equivalent to $5<x<10$) :

The equation below can be written : $$ \sqrt{x-1}-2 + 3-\sqrt{x-1} =1 $$ equivalent to : $ 1=1 $

The solutions belongs to $]5,10[$

  • case : $\sqrt{x-1} \geq3$ (equivalent to $x\geq10$) :

$\sqrt{x-1} = 3$ then $x = 10$

The solutions belongs to $[5,10]$

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In what does this solution differ from mine? –  egreg Jul 22 '13 at 9:25
    
@egreg, Samatix Please, read this (and comments, answers) meta.math.stackexchange.com/questions/8525/… . –  vesszabo Jul 22 '13 at 11:19
    
@vesszabo I don't think you should ping me. I'm not accusing of plagiarism, of course, but my answer appeared before this one. –  egreg Jul 22 '13 at 12:16
    
@egreg, sorry I've written the answer before noticing that you added yours. Your answer was sufficient and complet –  Samatix Jul 22 '13 at 15:34

Put $y=\sqrt{x-1}$, getting $x=y^2+1$; thus the equation becomes $$ \sqrt{y^2-4y+4}+\sqrt{y^2-6y+9}=1 $$ which should ring a bell.

It becomes $|y-2|+|y-3|=1$ that can be treated without resorting to squaring; divide it into cases:
If $y<2$, the equation becomes $2-y+3-y=1$, or $2y=4$, that means $y=2$, absurd.
If $2\le y\le 3$, the equation becomes $y-2+3-y=1$, an identity.
If $y>3$, the equation becomes $y-2+y-3=1$, or $2y=6$ and $y=3$, absurd. Therefore the solutions are all the numbers $x$ such that $2\le\sqrt{x-1}\le3$, that is $4\le x-1\le 9$ or $5\le x\le 10$.

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+1 For using a spoiler tag –  Tobias Kienzler Jul 22 '13 at 12:40

$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
Put x-1 = t^2
so the equation becomes $$\sqrt{t^2+4-4t} + \sqrt{t^2+9-6t} =1$$ complete the squares . EDIT:
My answer was incorrect. Thanks to egreg and Samatix.

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Are you sure that $\sqrt{u^2}=u$? –  egreg Jul 22 '13 at 8:34
    
Hey Simar, if you don't mind but $\sqrt(x^2) = |x|$ not $x$ –  Samatix Jul 22 '13 at 8:37
    
I don't think that you reasoning is correct. –  egreg Jul 22 '13 at 8:37
    
man look at x . It is in sqrt so dont we have to neglect -ve values ? –  Simar Jul 22 '13 at 8:37
    
Corrected it . Sorry ! Now is it alright ? –  Simar Jul 22 '13 at 8:50

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