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Root equation help

  1. Feb 11, 2007 #1
    Hi, I have the equation

    A = πr² + r²root(k²-1)

    i need to rearrange it to find r

    i go it to

    r³ = (2A/π+root(k²-1)

    to get just r (with no powers) what will the final equation look like and why.

    Thanks
    :biggrin:
     
  2. jcsd
  3. Feb 11, 2007 #2

    arildno

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    Eeeh?

    Your original equation is:
    [tex]A=\pi{r}^{2}+r^{2}\sqrt{k^{2}-1}[/tex]
    Agreed?
     
  4. Feb 11, 2007 #3
    yeh, that's the one
     
  5. Feb 11, 2007 #4

    arildno

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    So, what is a common factor between the two terms on the right-hand side?
     
  6. Feb 11, 2007 #5
    r² is the common factor
     
  7. Feb 11, 2007 #6

    arildno

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    So, you may rewrite your equation as:
    [tex]r^{2}(\pi+\sqrt{k^{2}-1})=A[/tex]
    What would you do next?
     
  8. Feb 11, 2007 #7
    cross multiply?

    [tex]r^{2} = A/(\pi+\sqrt{k^{2}-1})[/tex]
     
  9. Feb 11, 2007 #8

    arildno

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    Which is to divide each side with the factor [itex](\pi+\sqrt{k^{2}-1})[/itex], rather than cross-multiplication.

    1. Now, does this equal what you posted before?

    2. Since you now know the SQUARE of a number, how do we get what the number itself is?
     
  10. Feb 11, 2007 #9
    do you square the whole RHS?
     
  11. Feb 11, 2007 #10

    HallsofIvy

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    You want to get rid of the square on r.


    What is the opposite of squaring?
     
  12. Feb 11, 2007 #11
    sorry, my bad
    square root the whole RHS
     
  13. Feb 11, 2007 #12

    arildno

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    Do you know what an equation is, and what is allowed to do with one???
     
  14. Feb 11, 2007 #13

    arildno

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    Wrong.

    WHAT must you take the square root of?
     
  15. Feb 11, 2007 #14
    isn't the answer:

    [tex]r= \sqrt{A/(\pi+\sqrt{k^{2}-1})}[/tex]
     
  16. Feb 11, 2007 #15

    arildno

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    1. You must take the square root of BOTH sides of the equation, not just of one of the sides as you said. (This you have done)

    2. Put parentheses about the correct radicand.
     
  17. Feb 11, 2007 #16
    cheers, thanks for the help

    So heres another one:

    v² = u² + av² find v

    v² - av² = u²
    v²(1 - a) = u²
    v² = [tex]u²\1-a[/text]
    v = [tex]\sqrt{u²/1-a}[/tex]

    am I right?
     
  18. Feb 11, 2007 #17
    cheers, thanks for the help

    So heres another one:

    v² = u² + av² find v

    v² - av² = u²
    v²(1 - a) = u²
    v² = [tex]u^{2}/1-a[/tex]
    v = [tex]\sqrt{u^{2}/1-a}[/tex]

    am I right?
     
  19. Feb 11, 2007 #18

    arildno

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    Use PARENTHESES ABOUT YOUR DENOMINATOR!!!
    Is it that hard to get?

    Secondly, in the prior exercise I assumed that "r" was a radius, and hence necessarily a non-negative quantity (you didn't say).
    Now, must "v" be a non-negative quantity?
     
  20. Feb 11, 2007 #19
    v is meaningless, i'm just praticing rearranging the formula
     
  21. Feb 11, 2007 #20

    arildno

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    Is it "meaningless"?
    Is it not even a number? :confused:
     
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