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Math Help - What about these types of radicals?

  1. #1
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    What about these types of radicals?

    I have only about fifty more to do at the moment and am struggling on a few more.
    First problem not sure I really get this type of one
    Use rational exponents to simplify each expression.

    8
    √x^4

    Second is to simplify the indicated operation.

    3 3
    √5x^2 * √10y


    Third is:solving this radical equation:
    √x+9 - √x-7=2

    Fourth iserform indicated operation. Write in the form of a + bi
    (5-3i) - (6-9i)
    would I do the foil on this where 30-18i+18i okay no that ones is wrong.
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  2. #2
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    Quote Originally Posted by kbryant05

    8
    √x^4
    Okay,
    What you have in terms of rational exponents is,
    (x^4)^{1/8}=x^{4(1/8)}=x^{1/2}=\sqrt{x}
    Quote Originally Posted by kbryant05
    3 3
    √5x^2 * √10y
    Do you want me to express it as rational exponent?
    Then,
    \sqrt[3]{5x^2}=(5x^2)^{1/3} and,
    \sqrt[3]{10y}=(10y)^{1/3}
    Thus,
    (5x^2)^{1/3}\cdot (10y)^{1/3}=[(5x^2)(10y)]^{1/3}=(50x^2y)^{1/3}

    Quote Originally Posted by kbryant05
    Third is:solving this radical equation:
    √x+9 - √x-7=2
    Do you mean,
    \sqrt{x+9}-\sqrt{x-7}=2
    Or,
    \sqrt{x}+9-\sqrt{x}-7=2
    If it is case #2 then it is always true because the radicals cancel and you are left with 9-7=2 which is always true for any non-negative real number. Thus I presume it is the first one?

    Quote Originally Posted by kbryant05
    Fourth iserform indicated operation. Write in the form of a + bi
    (5-3i) - (6-9i)
    Open parantheses first and watch those signs ,
    5-3i-6+9i
    Add the reals, 5-6=-1
    Add the imaginaries, -3i+9i=6i
    Thus,
    -1+6i
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  3. #3
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    Thumbs up Okay

    Yes on the second one both ways if you can show me,and on the third the first part like you have it is correct not the second part.
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  4. #4
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    Question on my second problem

    The problem is

    3 3
    √5x^2 * √ 10y

    The book is saying 3
    √ 50x^2y as an answer but I see you have 1/3 in there. Where does that come in? I think I am more confused??
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  5. #5
    MHF Contributor Quick's Avatar
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    The 1/3

    I'm back.

    √ 50x^2y as an answer but I see you have 1/3 in there. Where does that come in? I think I am more confused??
    the 1/3 shows  \sqrt[3]{x} and here's why:

     x^1 \cdot x^1 = x^{1+1} = x^2
    therefore,
     x^{1/2} \cdot x^{1/2} = x^{1/2+1/2} = x^1
    notice that the same answer works for  \sqrt {x}
     \sqrt {x} \sqrt {x} = \sqrt {x\cdot x} = \sqrt {x^2} = x
    so  x^{1/2} = \sqrt{x}
    Now we look at 1/3
     x^{1/3} \cdot x^{1/3} \cdot x^{1/3} = x^{(1/3+1/3+1/3)} = x^1
    and the same answer comes around with  \sqrt[3]{x}
     \sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x} = \sqrt[3]{x\cdot x \cdot x} = \sqrt[3]{x^3}=x^1
    so  x^{1/3} = \sqrt[3]{x}

    which makes Hacker's answer,
     (50x^2y)^{1/3} = \sqrt[3]{50x^2y}
    which is the answer in your book.
    Last edited by Quick; June 5th 2006 at 05:48 PM.
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  6. #6
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    Got it

    Good job Quick!!
    Thanks for explaining that. You know the answers are great that they give us in the books but getting to that answer is more than a challenge sometimes for me.
    thanks,
    K
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  7. #7
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    Quote Originally Posted by kbryant05
    Yes on the second one both ways if you can show me,and on the third the first part like you have it is correct not the second part.
    The second case to question 2 has been explained. The solution is any non-negative number.

    The first case:
    \sqrt{x+9}-\sqrt{x-7}=2
    Thus,
    \sqrt{x+9}=2+\sqrt{x-7}
    Square both sides,
    (\sqrt{x+9})^2=(2+\sqrt{x-7})^2
    Thus, use FOIL
    x+9=4+4\sqrt{x-7}+(\sqrt{x-7})^2
    Thus,
    x+9=4+4\sqrt{x-7}+x-7
    Subtract 'x' from both sides and combine numbers,
    12=4\sqrt{x-7}
    Divide by 4,
    3=\sqrt{x-7}
    Square both sides,
    9=x-7
    Thus,
    x=16
    Check this solution to see that it works!
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  8. #8
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    Wink Genius

    Finally understanding these wonderful radicals with real and imaginaries. How fun! Some of the problems start to seem so long and tedious. I think it's easy to get lost into the problems. You guys are great at making it look so simple. Thumbs up to both of you! Thanks for the quick responses also!!
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