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Fun with Sqrt()	by DennisMV	2003-02-19 19:51:35
Sqrt(x), where x>=0 (principal square root) is defined as a function. Therefore Sqrt(x) is 0 or a positive number I claim that 6 is a solution to Sqrt(x + Sqrt(x-2) )=2 Let's see... Sqrt(6 + Sqrt(6-2) )=2 Sqrt(6 + Sqrt(4) )=2 Sqrt(6 + 2 )=2 Sqrt(8)=2 ... hmm.. nope However, the steps below show that 6 is indeed a solution to the equation. Equation: Sqrt(x + Sqrt(x-2) )=2 [Sqrt(x + Sqrt(x-2) )]^2 = 2^2 x + Sqrt(x-2) = 4 Sqrt(x-2) = 4 - x [ Sqrt(x-2) ]^2 = (4 - x)^2 x - 2 = 16 - 8x +x^2 x^2 - 9x + 18 = 0 (x - 3)(x - 6) = 0 x = 3, 6 so 6 IS a solution to the equation... what happen ? Somebody set up us a six ...
[ Reply ]
M y guess is +/- problems.	by Llyr	2003-02-19 19:53:54
yes, an extraneous solution (n/t)	by DennisMV	2003-02-19 20:05:06
That's the word, thanks. :) (n/t)	by Llyr	2003-02-19 20:06:19
yeah, I think it's +/-	by DennisMV	2003-02-19 20:13:08
x = (+/- sqrt(x)) ^ 2	by LionsPhil	2003-02-19 19:55:56
I didn't look at other answers, but here's mine	by Egaeus	2003-02-19 21:21:25

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