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storyw4

  • 2 years ago

solve for w.

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  1. storyw4
    • 2 years ago
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    |dw:1344742023146:dw|

  2. vf321
    • 2 years ago
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    \[\frac{u^2}{(w-2)^2}=v\]Implied domain w!=2 \[u^2=v(w-2)^2\]Multiplied (w-2) on both sides

  3. vf321
    • 2 years ago
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    Now. What do you think we divide by on both sideS?

  4. storyw4
    • 2 years ago
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    v?

  5. vf321
    • 2 years ago
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    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...

  6. storyw4
    • 2 years ago
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    square root

  7. vf321
    • 2 years ago
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    Yes. You get: \[\sqrt{u/v}=\sqrt{(w-2)^2}\]

  8. vf321
    • 2 years ago
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    Now, what's sqrt(x^2)=?

  9. vf321
    • 2 years ago
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    I assume you're in reals domain

  10. storyw4
    • 2 years ago
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    yeah i am and just x

  11. vf321
    • 2 years ago
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    Not quite. what's sqrt((-5)^2)?

  12. storyw4
    • 2 years ago
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    25

  13. vf321
    • 2 years ago
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    no...

  14. storyw4
    • 2 years ago
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    5

  15. vf321
    • 2 years ago
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    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...

  16. storyw4
    • 2 years ago
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    +/- x

  17. vf321
    • 2 years ago
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    Well, sort of. For \[x \epsilon R\]\[\sqrt{x^2}=|x|\]

  18. vf321
    • 2 years ago
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    But anyways, whats our 'x' in the problem?

  19. storyw4
    • 2 years ago
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    is the answer w=2v +-u√(v)

  20. vf321
    • 2 years ago
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    Well what I got was \[\pm \sqrt{u/v}-2\]

  21. storyw4
    • 2 years ago
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    |dw:1344743061685:dw| what about this

  22. vf321
    • 2 years ago
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    That's the same thing.

  23. storyw4
    • 2 years ago
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    i htought yours was a negative 2

  24. vf321
    • 2 years ago
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    Oh yeah it was . Typo, my bad.

  25. storyw4
    • 2 years ago
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    so is my picture correct?

  26. vf321
    • 2 years ago
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    Yeah.

  27. storyw4
    • 2 years ago
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    thanks

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