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ganeshie8
 3 years ago
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/50064e0ae4b0624180677e0aemily91021342590484915mathproblem5.jpg
ganeshie8
 3 years ago
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/50064e0ae4b0624180677e0aemily91021342590484915mathproblem5.jpg

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ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1342591850277:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1x =1 is not satisfying the equation :( not sure why

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man u have squared it so ofcourse one will not match?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when 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

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1im li'l lost.... could u elaborate pls.. ?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1i see we are squaring... and losing sign sensitivity is it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u must consider restrictions in every step note that when u have \( x3=2 \sqrt{x} \) then \( x3 \ge0 \)

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1ah that makes sense. thanks mukushla straight to the point. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I suppose this also applies to irrational equations like this? \[\sqrt{x1}+\sqrt{x+4}=\sqrt{3x+10}\] 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.

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.1yeah now i understand thanks higgs :D here, from the first equation itself, we can apply the restriction x >=1. and hence rejected solutions < 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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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