anonymous
  • anonymous
solve for w.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1344742023146:dw|
anonymous
  • anonymous
\[\frac{u^2}{(w-2)^2}=v\]Implied domain w!=2 \[u^2=v(w-2)^2\]Multiplied (w-2) on both sides
anonymous
  • anonymous
Now. What do you think we divide by on both sideS?

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anonymous
  • anonymous
v?
anonymous
  • anonymous
yes. You get: \[u/v = (w-2)^2\] Now you need to perform a certain operation to both sides to get rid of the ^2...
anonymous
  • anonymous
square root
anonymous
  • anonymous
Yes. You get: \[\sqrt{u/v}=\sqrt{(w-2)^2}\]
anonymous
  • anonymous
Now, what's sqrt(x^2)=?
anonymous
  • anonymous
I assume you're in reals domain
anonymous
  • anonymous
yeah i am and just x
anonymous
  • anonymous
Not quite. what's sqrt((-5)^2)?
anonymous
  • anonymous
25
anonymous
  • anonymous
no...
anonymous
  • anonymous
5
anonymous
  • anonymous
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...
anonymous
  • anonymous
+/- x
anonymous
  • anonymous
Well, sort of. For \[x \epsilon R\]\[\sqrt{x^2}=|x|\]
anonymous
  • anonymous
But anyways, whats our 'x' in the problem?
anonymous
  • anonymous
is the answer w=2v +-u√(v)
anonymous
  • anonymous
Well what I got was \[\pm \sqrt{u/v}-2\]
anonymous
  • anonymous
|dw:1344743061685:dw| what about this
anonymous
  • anonymous
That's the same thing.
anonymous
  • anonymous
i htought yours was a negative 2
anonymous
  • anonymous
Oh yeah it was . Typo, my bad.
anonymous
  • anonymous
so is my picture correct?
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
thanks

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