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geoffb Group TitleBest ResponseYou've already chosen the best response.2
\[4 \sqrt{(x+9)} = 20\] Like this? At what part are you getting stuck?
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
well i get x=16 but when i try to see if its true i get 20=20
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
and yes like that
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Yeah, with that 4 outside the square root, I think it's undefined.
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
yeah your probably right. extraneous solution?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
If we defined \(y\) as \(\sqrt{(x+9)}\), we would get \(4y = 20\), with \(y = 5\), and there is no way \(\sqrt{(x+9)} = 5\)
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
hmmm i dont know. but when we divide 4 we get \[\sqrt{x+9}=5\]
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
\(\large{\sqrt{x} = i \sqrt{x}}\) right?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Okay, can we use that somehow?
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Oh wait... We can square both sides, right?
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
and we get x+9=25 because 5x5=25
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Right, so you were right with 16; it is an extraneous root.
 2 years ago

savvers Group TitleBest ResponseYou've already chosen the best response.0
yeah it looks like that then and thanks man
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.2
Hey, no problem. :)
 2 years ago
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