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vf321Best ResponseYou've already chosen the best response.2
\[\frac{u^2}{(w2)^2}=v\]Implied domain w!=2 \[u^2=v(w2)^2\]Multiplied (w2) on both sides
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Now. What do you think we divide by on both sideS?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
yes. You get: \[u/v = (w2)^2\] Now you need to perform a certain operation to both sides to get rid of the ^2...
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Yes. You get: \[\sqrt{u/v}=\sqrt{(w2)^2}\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
I assume you're in reals domain
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Not quite. what's sqrt((5)^2)?
 one year ago

vf321Best ResponseYou've already chosen the best response.2
yes. And now that you've seen that, what do you think sqrt(x^2) is? it's not just x, since then the answer to sqrt((5)^2) would be 5...
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Well, sort of. For \[x \epsilon R\]\[\sqrt{x^2}=x\]
 one year ago

vf321Best ResponseYou've already chosen the best response.2
But anyways, whats our 'x' in the problem?
 one year ago

storyw4Best ResponseYou've already chosen the best response.0
is the answer w=2v +u√(v)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Well what I got was \[\pm \sqrt{u/v}2\]
 one year ago

storyw4Best ResponseYou've already chosen the best response.0
dw:1344743061685:dw what about this
 one year ago

storyw4Best ResponseYou've already chosen the best response.0
i htought yours was a negative 2
 one year ago

vf321Best ResponseYou've already chosen the best response.2
Oh yeah it was . Typo, my bad.
 one year ago

storyw4Best ResponseYou've already chosen the best response.0
so is my picture correct?
 one year ago
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