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vf321
 2 years ago
Best 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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Now. What do you think we divide by on both sideS?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2yes. You get: \[u/v = (w2)^2\] Now you need to perform a certain operation to both sides to get rid of the ^2...

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Yes. You get: \[\sqrt{u/v}=\sqrt{(w2)^2}\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2I assume you're in reals domain

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Not quite. what's sqrt((5)^2)?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2yes. 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...

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Well, sort of. For \[x \epsilon R\]\[\sqrt{x^2}=x\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2But anyways, whats our 'x' in the problem?

storyw4
 2 years ago
Best ResponseYou've already chosen the best response.0is the answer w=2v +u√(v)

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Well what I got was \[\pm \sqrt{u/v}2\]

storyw4
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1344743061685:dw what about this

storyw4
 2 years ago
Best ResponseYou've already chosen the best response.0i htought yours was a negative 2

vf321
 2 years ago
Best ResponseYou've already chosen the best response.2Oh yeah it was . Typo, my bad.

storyw4
 2 years ago
Best ResponseYou've already chosen the best response.0so is my picture correct?
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