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Math Help - Please help with squaring booth sides

  1. #1
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    Please help with squaring booth sides

    Hi Please help me to understand how squaring both sides of this equation gets the results on the second line. I have attached the problem. Thanks for any help.
    Attached Thumbnails Attached Thumbnails Please help with squaring booth sides-squaring-both-sides.gif  
    Last edited by coopsterdude; December 3rd 2005 at 11:54 PM.
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  2. #2
    Grand Panjandrum
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    Let:

    \sqrt{x+8}\ =\ \sqrt{x+35}\ -\ 3

    squaring both sides gives:

    (\sqrt{x+8})^2\ =\ (\sqrt{x+35}\ -\ 3)^2

    or:

    x+8\ =\ (\sqrt{x+35}\ -\ 3)^2

    expanding the RHS:

    x+8\ =\ (\sqrt{x+35})^2\ +\ 2.(-3).\sqrt{x+35}\ +\ (-3)^2

    simplifying (a bit):

    x+8\ =\ x+35\ -\ 6.\sqrt{x+35}\ +\ 9,

    finaly:

    x+8\ =\ x\ +\ 49\ -\ 6.\sqrt{x+35},
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  3. #3
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    Quote Originally Posted by coopsterdude
    Hi Please help me to understand how squaring both sides of this equation gets the results on the second line. I have attached the problem. Thanks for any help.
    I really don't know how to quote also the attached message.
    The attached message says:

    >>>
    sqrt(x+8) = sqrt(x+35) -3
    x+8 = x +35 -6sqrt(x+35) +9
    <<<

    Okay.
    Let us forget the radicals first.
    Suppose we instead have:
    y = w -3
    Square both sides,
    y^2 = (w-3)^2
    y^2 = w^2 -3w -3w +9
    y^2 = w^2 -2(3w) +9
    y^2 = w^2 -6w +9 -----------***

    Now, if y were sqrt(x+8), and w = sqrt(x+35), what will happen?
    This will happen:
    [sqrt(x+8)]^2 = [sqrt(x+35)]^2 -6sqrt(x+35) +9
    x+8 = x+35 -6sqrt(x+35) +9
    And that is the same as what was in the attached message.

    ---------
    Squaring a binomial:

    by FOIL,

    (x +y)^2
    = (x+y)(x+y)
    = x^2 +x*y +y*x +y^2
    = x^2 +xy +xy +y^2
    = x^2 +2xy +y^2

    See that 2xy?
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  4. #4
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    Tanks that totally solved my problem. You guys are great.
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