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Math Help - What can we do with radican the same?

  1. #1
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    What can we do with radican the same?

    Hi all,

    I'm trying to learn about radicals and roots, and what we can do with them.

    I understand the principle is x^2 * x^4 = x^(2+4) = x^6

    Basically if the base is the same and we are multiplying we just add the powers.

    Same powers rules holds true with square roots i.e.

    n\sqrt{x} * n\sqrt{y}= n\sqrt{x * y}

    Now my question is does the power rules still hold true with the radican being the same?

    If not what is it?

    Can we say ...

    a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x} ???

    If not, whats the answer?
    Last edited by Peleus; September 16th 2008 at 02:51 AM.
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  2. #2
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    Quote Originally Posted by Peleus View Post
    n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}
    NO! \sqrt{x} \times \sqrt{y} = \sqrt{xy} \neq \sqrt{x+y}

    \sqrt{x} \times \sqrt{x} = \sqrt{x^2} = x
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  3. #3
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    Just worked out that no that doesn't work, but is there any particular rule for it?

    Cheers

    Edit - Also realised I made a typo in the original math hence it should be been x*y instead of x+y.

    Thanks
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  4. #4
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    Quote Originally Posted by Peleus View Post
    Hi all,

    I'm trying to learn about radicals and roots, and what we can do with them.

    I understand the principle is x^2 * x^4 = x^(2+4) = x^6

    Basically if the base is the same and we are multiplying we just add the powers.

    Same powers rules holds true with square roots i.e.

    {\color{red}{n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}}}

    Now my question is does the power rules still hold true with the radican being the same?

    If not what is it?

    Can we say ...

    a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x} ???

    If not, whats the answer?
    {\color{red}{n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}}} this is wrong it should be {n\sqrt{x} + n\sqrt{y}= n\big(\sqrt{x} + \sqrt{y}\big)} or {n\sqrt{x} * n\sqrt{y}= n^{2}\big(\sqrt{x*y}\big)}
    On the other hand, we can not say a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x}, because this is equal to a\sqrt{4x} * b\sqrt{4x}=a*\sqrt{4x} * b*\sqrt{4x} = a*b\Big(\sqrt{4x}\Big)^{2}=a*b*4*x
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  5. #5
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    For any real numbers a, b, n, m:
    (ab)^n = a^n b^n

    \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

    a^n \times a^m = a^{n+m}

    \frac{a^n}{a^m} = a^{n-m}

    \sqrt[n]{a} = a^{\frac{1}{n}}
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  6. #6
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Chop Suey View Post
    For any real numbers a, b, n, m:
    (ab)^n = a^n b^n

    \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

    a^n \times a^m = a^{n+m}

    \frac{a^n}{a^m} = a^{n-m}

    \sqrt[n]{a} = a^{\frac{1}{n}}
    Also an important one (a^{m})^{n}=a^{mn}=(a^{n})^{m}.
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  7. #7
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    Quote Originally Posted by Peleus View Post
    Hi all,

    I'm trying to learn about radicals and roots, and what we can do with them.

    I understand the principle is x^2 * x^4 = x^(2+4) = x^6

    Basically if the base is the same and we are multiplying we just add the powers.

    Same powers rules holds true with square roots i.e.

    n\sqrt{x} * n\sqrt{y}= n\sqrt{x * y}

    Now my question is does the power rules still hold true with the radican being the same?

    If not what is it?

    Can we say ...

    a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x} ???

    If not, whats the answer?
    The "ath root" of 4x is (4x)^{1/a} and the "bth root" of 4x is (4x)^{1/b}. By the usual formula for multiplying powers, (4x)^{1/a}(4x)^{1/b}= (4x)^{1/a+ 1/b}= (4x)^{\frac{a+b}{ab}}= ^{ab}\sqrt{(4x)^{a+b}}

    Is that what you wanted?
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