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Homework Help: Solving surd equation

  1. Jun 4, 2007 #1
    (urgent)Solving surd equation

    Solve [tex]5\sqrt{x} = \frac{40}{x}[/tex]

    how would I start by solving this?
    Cross multipling seems to be difficul here
    Last edited: Jun 4, 2007
  2. jcsd
  3. Jun 4, 2007 #2
    [tex] x\sqrt{x}=8 [/tex]
    [tex]x^{3/2}=2^3 [/tex]
    Last edited: Jun 4, 2007
  4. Jun 4, 2007 #3


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    Could square both sides, and then cross multiply.
  5. Jun 4, 2007 #4
    substitution ...

    let [tex]u=\sqrt{x}[/tex]

    [tex]5u = \frac{40}{u^2}[/tex]

    [tex]5u - \frac{40}{u^2} = 0[/tex]
    find common denominator and simplify:


    therefore [tex]x=4[/tex]

    ** lol beat me to it
    Last edited: Jun 4, 2007
  6. Jun 4, 2007 #5
    i've seen weird things like this before. where did you learn this? like what country
  7. Jun 4, 2007 #6


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    Whats weird about substitution? Its pretty standard practise anywhere :confused:
  8. Jun 4, 2007 #7
    i was never taught that and
    i don't see the point? [tex]\sqrt{x} = x^\frac{1}{2}[/tex] and then just add exponents of polynomials with like bases when multiplying
  9. Jun 5, 2007 #8
    well ... you weren't taught correctly, lol

    nah, seems simple enough for me.
  10. Jun 5, 2007 #9
    what? how is that incorrect
  11. Jun 5, 2007 #10
    nah im not saying your working is wrong, i'm just saying substitution is a better way, which can be applied; Particularly when doing harder and more complex problems.
  12. Jun 5, 2007 #11
    i seriously don't see how, please show me an example
  13. Jun 5, 2007 #12


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    eventually you will meet problems of that kind ;)
  14. Jun 5, 2007 #13
    ok, this is waaaayy off topic .... but, as per request i shall give multiple:

    perhaps basic:
    [tex] x - 3\sqrt{x} = -2[/tex]
    let [tex]u=\sqrt{x}[/tex]
    [tex] u^2 - 3u + 2 = 0[/tex]
    [tex] (u-1)(u-2) = 0[/tex]
    [tex] u=1 [/tex], [tex] u=2 [/tex]
    [tex]1=\sqrt{x}[/tex] and [tex]x=1[/tex]
    [tex]2=\sqrt{x}[/tex] and [tex]x=4[/tex]

    would you like another ... perhaps trigonometric, involving quotients or exponentials?
  15. Jun 5, 2007 #14


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    Its basically the same as treating some expression (in this case, [tex]\sqrt{x}[/tex] as a variable or symbol itself, and not treating it as an actual expression until the end. So instead of treating [tex]\sqrt{x}[/tex] as the square root of x, its just a symbol, which we treat like any other symbol, such as a, x, y, [tex]\theta[/tex] etc.
  16. Jun 5, 2007 #15

    Gib Z

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    Perhaps a better example of where a substitution would help:

    Solve : [tex]x^8+x^4+\pi =0[/tex]

    Personally I would let u=x^4, what would you do?
  17. Jun 5, 2007 #16


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    I would solve it with my texas ti-83 =P
  18. Mar 20, 2008 #17
    Surds HELP!!

    im having trouble trying to figure out a problem. can someone help me.

    (2/6 -root3)^2 - (2/6+root3)^2

    if that makes sense to anyone help.

    I think that the denominator of both has to be rationalised but do i expand the squared brackets first or later or what?

    any ideas welcome thanks
  19. Mar 20, 2008 #18
    Do you mean

    [tex]\left(\frac{2}{6-\sqrt{3}}\right)^2 - \left(\frac{2}{6+\sqrt{3}}\right)^2[/tex]

    What are you supposed to do with this? Simplify?

    EDIT: Sorry, I didn't see that this question was moved into its separate thread. I'll re-post this there.
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