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Math Help - Radicals and rationalizie

  1. #1
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    Radicals and rationalizie

    Hi All,

    I am doing a Precalculus course and this exercise is from schaum's outlines Precalculus textbook.

    Please can someone help me understand how this guy came to his conclusion?

    Radicals and rationalizie-eq.jpgThe answerRadicals and rationalizie-eq2.jpg
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  2. #2
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    Re: Radicals and rationalizie

    \frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})

    =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)

    =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)\left( \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)

    =\frac{1}{h}\left(\frac{x -x-h}{\sqrt{x(x+h)}}\right)\left( \frac{1}{\sqrt{x}+\sqrt{x+h}} \right)


    Now continue from there....
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  3. #3
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    Re: Radicals and rationalizie

    Quote Originally Posted by SVDM View Post
    Hi All,

    I am doing a Precalculus course and this exercise is from schaum's outlines Precalculus textbook.

    Please can someone help me understand how this guy came to his conclusion?

    Click image for larger version. 

Name:	eq.JPG 
Views:	2 
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ID:	24124The answerClick image for larger version. 

Name:	eq2.JPG 
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Size:	16.6 KB 
ID:	24125
    Multiply numerator and denominator by root(x+h)root(x) Then multiply numerator and denominator by root(x)+root(x+h)
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  4. #4
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    Re: Radicals and rationalizie

    Hallo ou maat!

    My head is spinning!

    Thanks very much as I analyze this for the next 3 hours....

    I never seen this done before! Not in the textbook anyway.

    Stan
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  5. #5
    Member Goku's Avatar
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    Re: Radicals and rationalizie

    Remember that a^2 - b^2 = (a-b)(a+b)...
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    Re: Radicals and rationalizie

    Hi Guys thanks for the help!

    I have another one that I can not solve. I don't find much help in text book and it does not help trying my own dodgy thinks so I hope to learn something from you all.

    This time the stuffed me up with this...

    Radicals and rationalizie-eq3.jpg

    ? I tried to solve this but the answer is so wrong...
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  7. #7
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    Re: Radicals and rationalizie

    Quote Originally Posted by SVDM View Post
    Hi Guys thanks for the help!

    I have another one that I can not solve. I don't find much help in text book and it does not help trying my own dodgy thinks so I hope to learn something from you all.

    This time the stuffed me up with this...

    Click image for larger version. 

Name:	eq3.JPG 
Views:	2 
Size:	17.2 KB 
ID:	24131
    Multiply by numerator and denominator by \left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x}\sqrt[3]{a} + \sqrt[3]{{{a^2}}}} \right)
    Last edited by Plato; June 21st 2012 at 12:23 PM.
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  8. #8
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    Re: Radicals and rationalizie

    Here ia another problem that keep me from sleeping.

    Radicals and rationalizie-eq4.jpg

    what I do next brings me to the answer but i think it is dodgy and Ii would like to see how it is really doubt so Ii can be sure it is right.

    Thanks again.
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  9. #9
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    Re: Radicals and rationalizie

    Quote Originally Posted by SVDM View Post
    what I do next brings me to the answer but i think it is dodgy and Ii would like to see how it is really doubt so Ii can be sure it is right.
    I don't know what's going on in your denominator...

    \frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}

    =\frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}\cdot\frac{\sqrt{x+1} + \sqrt{a+1}}{\sqrt{x+1} + \sqrt{a+1}}

    =\frac{(x+1) - (a+1)}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}

    =\frac{x - a}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}

    =\frac1{\sqrt{x+1} + \sqrt{a+1}}
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  10. #10
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    Re: Radicals and rationalizie

    Hi again,

    I am looking through the text book to find how you came to the conclusion to use a square of a difference in this case. the radical is cubed so I would thought to use differences of two cubes?
    Or am I confused with the radicals somehow? If I disregard manipulating the numerator and just assume the difference of two cubes will remove the radicals my answer is correct. But I am missing something seriously with my understanding of polynomial manipulation because when I actually do the calculation. I get the result below which shows that something is not correct.

    Radicals and rationalizie-radical-polynomial-issue.jpg
    Last edited by SVDM; June 22nd 2012 at 04:20 PM.
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  11. #11
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    Re: Radicals and rationalizie

    to rationalize the numerator of the fraction ...

    \frac{\sqrt[3]{x} - \sqrt[3]{a}}{x-a}


    note that the denominator can be factored as the difference of two cubes ...

    x - a = (\sqrt[3]{x} - \sqrt[3]{a}) \left[(\sqrt[3]{x})^2 + \sqrt[3]{x} \cdot \sqrt[3]{a} + (\sqrt[3]{a})^2 \right]

    can you finish?
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  12. #12
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    Re: Radicals and rationalizie

    Hi thanks,

    So I did misunderstand Pluto above;
    Quote Originally Posted by Plato View Post
    Multiply by numerator and denominator by \left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x}\sqrt[3]{a} + \sqrt[3]{{{a^2}}}} \right)

    Now I see what was done! I feel so stupid! How could I have known? Looks like I have to study the factoring rules some more!

    Thank you very much.
    Last edited by SVDM; June 22nd 2012 at 08:41 PM.
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  13. #13
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    Re: Radicals and rationalizie

    Please can you explain exactly why you came to this conlusion?

    I am trying to understand what to look for to solve the problem myself but getting the answer still does not explain the process it takes to reach the conclusion.

    Thanks again.
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  14. #14
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    Re: Radicals and rationalizie

    Quote Originally Posted by SVDM View Post
    Please can you explain exactly why you came to this conlusion?
    I am trying to understand what to look for to solve the problem myself but getting the answer still does not explain the process it takes to reach the conclusion.
    You must know that x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)
    So when you are faced with \sqrt[3]{x} you use that concept.
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