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i am getting two solutions for this problem. but only one is satisfying the equation. could someone pls explain whats going on... http://assets.openstudy.com/updates/attachments/50064e0ae4b0624180677e0a-emily9102-1342590484915-mathproblem5.jpg

Mathematics
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square it and try?
|dw:1342591850277:dw|
x =1 is not satisfying the equation :( not sure why

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Other answers:

hmm
man u have squared it so ofcourse one will not match?
when you square something, you will always get two solutions, and not all of them will be right. substitute the answers back into the original equation to check. in this case, 1 is not the solution as -1 is not equal to 1
im li'l lost.... could u elaborate pls.. ?
i see we are squaring... and losing sign sensitivity is it
u must consider restrictions in every step note that when u have \( x-3=2 \sqrt{x} \) then \( x-3 \ge0 \)
ah that makes sense. thanks mukushla straight to the point. :)
welcome my friend
I suppose this also applies to irrational equations like this? \[\sqrt{x-1}+\sqrt{x+4}=\sqrt{3x+10}\] I got \(x = 5\) and \(x = \frac{-13}{3} \) \(x = \frac{-13}{3} \) is rejected in this case as well because after substituting into the equation, it became undefined.
yeah now i understand thanks higgs :D here, from the first equation itself, we can apply the restriction x >=1. and hence rejected solutions < 1
The sollutions we get after solving , and which does not satisfy the origanal question are called extraneous sollutions. There occurred as we have squared , so there is a chance of getting a extraneous sollution.

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