Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to rationalize the numberator?

  1. Sep 24, 2007 #1
    Here is the question: the square root of X minus 3, but the 3 is not under the square root divided by x-9? also another question: the square root of X minus 2, but the 2 is not under the square root divided by 4-x? anyone can help me?
     
  2. jcsd
  3. Sep 24, 2007 #2
    How to rationalize the numerator?

    Here is the question: sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks
     
    Last edited: Sep 24, 2007
  4. Sep 24, 2007 #3
    So you mean sqrt(x)/(x-9) - 3/(x-9)?
     
  5. Sep 24, 2007 #4
    No

    no I mean, sqrt(x) - 3/ x-9 ? also another question: sqrt(x) -2/ 4-x ? anyone can help me? Thanks and I need to rationalize the numerator
     
  6. Sep 24, 2007 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Nope, i guess he means

    [tex]\frac{\sqrt{x} -3}{x-9} [/tex] and [tex] \frac{\sqrt{x} -2}{4-x} [/tex]

    When rationalizing, remember that [itex]a^2 - b^2 \equiv (a+b)(a-b) [/itex]. So take [/itex] a=\sqrt{x} [/itex]. What is 'b' equal to ?
     
  7. Sep 24, 2007 #6
    yeah

    those are the right equations u just put up, but i dont know how to rationalize those equations???
     
  8. Sep 24, 2007 #7
    yeah

    I got x-9/ x-9sqrt(x) + 3x - 27 is that right?
     
  9. Sep 24, 2007 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\frac{\sqrt{x}-3}{x^2- 9}[/tex]
    That it?

    Use (a- b)(a+ b)= a2- b2: multiply both numerator and denominator by [itex]\sqrt{3}+ 3[/itex] and the squareroot will disappear from the numerator. It will, of course, show up in the denominator.

    For the second, assuming you have a fraction with [itex]\sqrt{x}- 2[/itex] in either numerator or denominator, multiply both numerator and denominator by [itex]\sqrt{x}+ 2[/itex]
     
  10. Sep 24, 2007 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is not a "Linear and Abstract Algebra" question and it was also posted in the "General Math" section so I am merging the two threads in that section.
     
  11. Sep 24, 2007 #10
    [tex] \frac {\sqrt{x}-3}{x-9} [/tex]

    Multiply:
    Both [tex] {x-9} [/tex] and [tex] {\sqrt{x}-3}[/tex]

    By the conjugate of [tex]{\sqrt{x}-3}[/tex] which is [tex]{\sqrt{x}+3}[/tex].

    [tex] \frac {\sqrt{x}-3( \sqrt{x}+3)} {x-9( \sqrt{x}+3)}[/tex]

    From here, you should be able to do some algebra, to rationalize the numerator. Do you know what a conjugate is, and why it is used? That might be your issue with not understanding it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to rationalize the numberator?
  1. Rational Numbers (Replies: 6)

Loading...