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Rationalizing the denominator

  1. Apr 18, 2006 #1
    [​IMG]

    You have to multiply the top and bottom by 18 right? From then, i'm confused.
     
  2. jcsd
  3. Apr 18, 2006 #2

    Integral

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    You want to get the square root out of the denominator so you need to multiply by the square root.
     
  4. Apr 18, 2006 #3
    If you want to get the square root then you need to multiply by the square root? Is there two square roots? Of 5 it's originaly 2.24 [Estimated]. But 5x18=90. I don't understand, this is mind twisting but it seems so easy.
     
  5. Apr 18, 2006 #4
    Try multiplying the entire thing by [itex]\frac{\sqrt{18}}{\sqrt{18}}[/itex] and see where you can go from there.
     
  6. Apr 18, 2006 #5

    dav2008

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    Do you understand that [tex]\sqrt{a} \cdot \sqrt{a}=a[/tex]?
     
  7. Apr 23, 2006 #6
    ^Yes i do understand that. I learned that a while ago.

    Yikes i still don't understand how to complete the problem.
    18x5=90
    But how does that occur? Aren't the two denominators suppose to both be the same number?
    2/(Sqrt18)
    ---
    18

    The reason this problem confuses me is becuase I know how to Rationalize the denominator when the extra number like the "5" isn't there.
     
  8. Apr 23, 2006 #7

    dav2008

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    Why should the two denominators both be the same number? 2/4 is the same as 1/2 but they don't have the same denominator.

    If you know how to rationalize the denominator when the extra number isn't there then the procedure is no different here.

    Did you follow ksinclair13's advice?

    The goal is to get rid of any square roots in the denominator. The only thing you can really do to that expression is multiply it by a form of 1 since multiplying it by 1 doesn't change the expression.
     
    Last edited: Apr 23, 2006
  9. Apr 24, 2006 #8

    HallsofIvy

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    Then take it out! How would you rationalize the denominator if the problem were [itex]\frac{2}{\sqrt{18}}[/itex]? After you've done that, just put the 5 back in.
     
  10. Apr 28, 2006 #9
    [tex] 2/sqrt18 \times sqrt18/sqrt18[/tex]

    I understand that part, but can someone please tell me how you would know when to multiply a fraction by its reciprocal opposed to its conjugate.
     
  11. Apr 28, 2006 #10

    HallsofIvy

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    Well, you'd have to tell me what you mean by the "conjugate" of a fraction before I could answer. You multiply by the reciprocal of a fraction when you are dividing by that fraction of course. You are NOT dividing by any fraction here.

    [tex]\frac{2}{\sqrt{18}} \times \frac{\sqrt{18}}{\sqrt{18}}= [/tex] what??

    And what would that be when you put the 5 back in the denominator?
     
  12. Apr 28, 2006 #11

    dav2008

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    I think he means that when you have

    [tex]\frac{2}{1+\sqrt{18}}[/tex]

    you would multiply by [tex]\frac{1-\sqrt{18}}{1-\sqrt{18}}[/tex]

    In a way you're doing that when you're doing the following problem (Obviously this would be doing unnecessary steps and be even more confusing but I'm just showing that you're basically doing the same thing):

    [tex]\frac{2}{\sqrt{18}}[/tex]
    [tex]\frac{2}{0+\sqrt{18}}[/tex]
    [tex]\frac{2}{0+\sqrt{18}} \times \frac{0-\sqrt{18}}{0-\sqrt{18}} [/tex]
    [tex]\frac{2}{\sqrt{18}} \times \frac{-\sqrt{18}}{-\sqrt{18}} [/tex]
    [tex]\frac{2}{\sqrt{18}} \times \frac{\sqrt{18}}{\sqrt{18}} [/tex]

    Like I said, you have to think of what would you have to multiply the denominator by to get rid of the square root. If the denominator is [itex]\sqrt{a}[/itex] then it makes sense to multiply by [itex]\sqrt{a}[/itex]. If the denominator is [itex]b+ \sqrt{a}[/itex] then it makes sense to multiply by [itex]b-\sqrt{a}[/itex] because in general it's true that [itex](c+d)(c-d)=c^2-d^2[/itex] Since you can't change the fraction you have to multiply both numerator and denominator by the same thing. (You have to multiply the original fraction by 1 to not change it)
     
    Last edited: Apr 28, 2006
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