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[Post New]8 Apr 2008 20:29:26 IST
    Subject: question
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budokai_tenkaichi (62)

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find roots of equation
(x+1)1/2-(x-1)1/2=(4x-1)1/2
Vegeta: While I was with you guys, I became more human. I didn't like that. I even have a family and started to like living on earth.
Vegeta: Don't remind me. I'm mad enough to hurt somebody and pounding you just might be the therapy I need.
Vegeta: You can take control of my mind and my body, but there is one thing a Saiyan always keep... his PRIDE!
[Piccolo and Vegeta sit back to back on a tiny island]
Piccolo: Is it over?
Vegeta: Not until the fish jumps.
[a fish jumps out of the water]
Vegeta: It's over.
    
[Post New]8 Apr 2008 20:31:56 IST
    Subject: Re:question
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sandeepramesh (1090)

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WTH? WHY DO YOU POST THE SAME QUESTION IN ALL THE MATH FORUMS? Huh? Huh? Huh? Huh? Huh? Huh? Huh?
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[Post New]8 Apr 2008 20:34:04 IST
    Subject: Re:question
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budokai_tenkaichi (62)

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whatss ur problem buddy?
Vegeta: While I was with you guys, I became more human. I didn't like that. I even have a family and started to like living on earth.
Vegeta: Don't remind me. I'm mad enough to hurt somebody and pounding you just might be the therapy I need.
Vegeta: You can take control of my mind and my body, but there is one thing a Saiyan always keep... his PRIDE!
[Piccolo and Vegeta sit back to back on a tiny island]
Piccolo: Is it over?
Vegeta: Not until the fish jumps.
[a fish jumps out of the water]
Vegeta: It's over.
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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[Post New]8 Apr 2008 20:35:48 IST
    Subject: Re:question
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sandeepramesh (1090)

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None at all, just the fact that its called ' SPAMMING ' on the net Huh? Huh?
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[Post New]8 Apr 2008 20:38:11 IST
    Subject: Re:question
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budokai_tenkaichi (62)

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why did your yellow faces got down to two??
Vegeta: While I was with you guys, I became more human. I didn't like that. I even have a family and started to like living on earth.
Vegeta: Don't remind me. I'm mad enough to hurt somebody and pounding you just might be the therapy I need.
Vegeta: You can take control of my mind and my body, but there is one thing a Saiyan always keep... his PRIDE!
[Piccolo and Vegeta sit back to back on a tiny island]
Piccolo: Is it over?
Vegeta: Not until the fish jumps.
[a fish jumps out of the water]
Vegeta: It's over.
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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[Post New]8 Apr 2008 20:40:24 IST
    Subject: Re:question
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sandeepramesh (1090)

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Simply. If you want to talk abt this anymore, just nudge me thats all.
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[Post New]8 Apr 2008 20:42:16 IST
    Subject: Re:question
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budokai_tenkaichi (62)

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let the question be a challenge to your skills now
Vegeta: While I was with you guys, I became more human. I didn't like that. I even have a family and started to like living on earth.
Vegeta: Don't remind me. I'm mad enough to hurt somebody and pounding you just might be the therapy I need.
Vegeta: You can take control of my mind and my body, but there is one thing a Saiyan always keep... his PRIDE!
[Piccolo and Vegeta sit back to back on a tiny island]
Piccolo: Is it over?
Vegeta: Not until the fish jumps.
[a fish jumps out of the water]
Vegeta: It's over.
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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[Post New]8 Apr 2008 21:07:57 IST
    Subject: Re:question
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sandeepramesh (1090)

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Ans is clearly no soln. Rewrite the ques as:
 
(x+1)^1/2 = (x-1)^1/2 + (4x-1)^1/2
For 4x-1 >= x+1, there is clearly no soln which is x>= 2/3
But the fn is defined for only x>=1.
 
Q.E.D.
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[Post New]8 Apr 2008 22:17:16 IST
    Subject: Re:question
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hsbhatt (2244)

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edited: something fishy here
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[Post New]8 Apr 2008 22:19:29 IST
    Subject: Re:question
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sandeepramesh (1090)

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how come sir???
 
I think you read it wrongly :)
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[Post New]8 Apr 2008 22:43:19 IST
    Subject: Re:question
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hsbhatt (2244)

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I am not able to understand. Maybe the late hour, but this is what I thought:

\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1} \\ \\ \sqrt{x+1} + \sqrt{x-1} = \frac{2}{\sqrt{4x-1}}

So,

2\sqrt{x+1} = \sqrt{4x-1}+\frac{2}{\sqrt{4x-1}} \\ \\ \text{or} \ 2\sqrt{x+1} = \frac{4x+1}{\sqrt{4x-1}}


4(x+1)(4x-1) = (4x+1)^2

This satisifed by x = \frac{5}{4}

But, it does not satisfy the original equation!!
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[Post New]8 Apr 2008 22:57:45 IST
    Subject: Re:question
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Conjurer (443)

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Even by squaring the terms you get x= 5/4 but since it doesnt satisfy the equation so no roots :)
Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule
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[Post New]8 Apr 2008 23:03:50 IST
    Subject: Re:question
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nadeemoidu (1174)

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nothing fishy.  whenever u square an equation , u should substitute and check in the end . if it does not satisfy the equation , then it is not a root. that's all . u should do that always when u square.
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[Post New]8 Apr 2008 23:05:42 IST
    Subject: Re:question
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sboosy (2860)

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\mbox{Consider the pts.} \ (\sqrt{x+1},0),(\sqrt{x-1},0),(\sqrt{4x-1},0) \ \mbox{and the origin} \ (0,0) \\ \\ \mbox{Along the x-axis,from left to right the order of the pts would be} \\ \\ \ (0,0),(\sqrt{x-1},0),(\sqrt{x+1},0),(\sqrt{4x-1},0) \\ \\ \mbox{The distances of the last three from origin are in order} \\ \\ \sqrt{x-1},\sqrt{x+1},\sqrt{4x-1} \\ \\ \mbox{Coming to the question} \ \sqrt{x+1}-\sqrt{x-1} = \sqrt{4x-1} \\ \\ \mbox{which is nothing but taking distance upto} \ \sqrt{x+1} \\ \\ \mbox{and subtracting} \ \sqrt{x-1} \\ \\ \mbox{which can never equal} \ \sqrt{4x-1} \\ \\ \mbox{because it is rightmost on our numberline} \\ \\ \mbox{Thus no solution}
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[Post New]8 Apr 2008 23:16:32 IST
    Subject: Re:question
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sboosy (2860)

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\sqrt{x+1}-\sqrt{x-1} = \sqrt{4x-1} \\ \\ \mbox{After squaring once we have} \\ \\ -2\sqrt{x^2-1}=\2x-1 ....(1) \\ \\ \mbox{Now consider} \ \sqrt{x+1}+\sqrt{x-1} = \sqrt{4x-1} \\ \\ \mbox{After squaring once we have} \\ \\ 2\sqrt{x^2-1}=\2x-1 ....(2) \\ \\ \mbox{Now we notice that on squaring both (1) and (2) we get the same expression} \\ \\ \mbox{and thus we r also finding solutions of} \\ \\ \sqrt{x+1}+\sqrt{x-1} = \sqrt{4x-1} \ \mbox{instead of} \\ \\ \sqrt{x+1}-\sqrt{x-1} = \sqrt{4x-1} \ \mbox{alone} \\ \\ \mbox{Infact the value of} \ x=\frac{5}{4} \ \mbox{so obtained is solution to} \\ \\ \sqrt{x+1}+\sqrt{x-1} = \sqrt{4x-1}
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