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Math Help - under a square root

  1. #1
    am1
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    under a square root

    if i had squareroot(4+x^2) why can't i simplify it down to (2+x), why isn't that legal?
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  2. #2
    Super Member bigwave's Avatar
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    reason

    Quote Originally Posted by am1 View Post
    if i had squareroot(4+x^2) why can't i simplify it down to (2+x), why isn't that legal?
    \sqrt{4+x^2} is the same as  \left(4 + x^2\right)^{\frac{1}{2}}

    the law is (xy)^a  = x^2y^a

    (x + y)^a \    (x+y) has to be mulitplied by itself a times
    Last edited by bigwave; December 8th 2009 at 09:04 PM. Reason: latex
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  3. #3
    am1
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    ok so <br />
\left(4 + x^2\right)^{\frac{1}{2}}<br />
means to mulitply the inside (4+x^2) by 1/2 if 4 and x are not products of each other?
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  4. #4
    Senior Member Stroodle's Avatar
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    Because that is just the square root of each term, not the square root of both the terms added together; for example:

    \sqrt{3+1}\neq\sqrt{3}+\sqrt{1}

    It equals 2.
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  5. #5
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    Quote Originally Posted by am1 View Post
    if i had squareroot(4+x^2) why can't i simplify it down to (2+x), why isn't that legal?
    It isn't "legal" because it gives the wrong answer! \sqrt{4+x^2} is NOT 2+ x because (2+x)^2= (2+x)(2+x)= 4+ 4x+ x^2 is not equal to 4+ x^2.
    Last edited by mr fantastic; December 9th 2009 at 04:03 AM. Reason: Fixed a latex tag (it's been a while ....!)
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