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Homework Help: Quick Question

  1. Apr 7, 2008 #1
    [SOLVED] Quick Question


    This question may tex ability :shy:

    The Question
    Write [tex]\frac{2\sqrt x+3}{x}[/tex] in the form [tex]2x^p + 3x^q[/tex] where p and q are constants.

    Attempt :eek:

    [tex]\frac{2\sqrt x+3}{x} \times x[/tex]

    [tex]2x + 3x[/tex]

    I basically multiplied the initial equation by x to get rid of the denominator, and now I have what looks like a close to correct answer, but without p and q, either p or q could be one but I think the fact that they have asked for p and q would suggest that they powers must be different.

    Thanks to any helpers =]

  2. jcsd
  3. Apr 7, 2008 #2


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    Can you write [itex]2\sqrt x/x[/itex] in the form [itex]2x^p[/itex]?

    Can you write [itex]3/x[/itex] in the form [itex]3x^q[/itex]?

    Hint: don't multiply by x, this gives something different than what you started with (and are supposed to evaluate).
  4. Apr 7, 2008 #3
    Heed Avodyne's advice! Say you were asked to reduce 64/8 and you decided to multiply by 8 to get rid of the denominator. You now have a different number. Don't confuse this with equations, where you can perform an operation to both sides (like division) and maintain the equality.
  5. Apr 7, 2008 #4


    I don't think I got the first bit correct...

  6. Apr 7, 2008 #5


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    Homework Helper

    Basic algebra:
    \frac{2\sqrt x+3}{x} = \frac{2\sqrt{x}}{x} + \frac{3}{x} = 2 \frac{\sqrt{x}}{x} + 3 \frac{1}{x}
  7. Apr 7, 2008 #6
    I have the 3 but now, but I am still struggling with the root x. Please try and explain how it works.

    Thank you very much for your help so far though, my maths is actually quite good but I have these little cracks I need filling in...

    Thanks again.
  8. Apr 7, 2008 #7


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    You are correct that [itex]\sqrt{3}= 3^{1/2}[/itex]. Now what about the 1/x? What do the "laws of exponents" tell you about [itex]x^m/x^n[/itex]?
  9. Apr 7, 2008 #8
    Okay, I will take your post above as an example HallsofIvy? I tried looking for the root 3 but couldn't find it in my working! :rofl:

    I will take it that my answer is right:


    which gives me...

    [tex]x^\frac{1}{2} - x^{-1} = x^\frac {-1}{2}[/tex]

    Thanks Halls, I hope I am right...
  10. Apr 7, 2008 #9
    Eh, your answer is right but oddly derived.

  11. Apr 8, 2008 #10


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    It was just a typo. He meant
    [tex]x^{\frac{1}{2}}\times x^{-1}= x^{-\frac{1}{2}}[/tex]
  12. Apr 8, 2008 #11
    I appreciate all of your help everyone, thankyou!

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