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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?
 one year 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
 one year ago

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

savvers Group TitleBest ResponseYou've already chosen the best response.0
yeah your probably right. extraneous solution?
 one year 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 \(\sqrt{(x+9)} = 5\)
 one year ago

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

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

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

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

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

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

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

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