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ganeshie8
Group Title
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
 2 years ago
 2 years ago
ganeshie8 Group Title
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
 2 years ago
 2 years ago

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f1r1e1s1h1 Group TitleBest ResponseYou've already chosen the best response.0
square it and try?
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
dw:1342591850277:dw
 2 years ago

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

f1r1e1s1h1 Group TitleBest ResponseYou've already chosen the best response.0
man u have squared it so ofcourse one will not match?
 2 years ago

higgs Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
welcome my friend
 2 years ago

higgs Group TitleBest ResponseYou've already chosen the best response.1
I 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.
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago
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