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Math Help - Exponents and square roots (once more)

  1. #1
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    Exponents and square roots (once more)

    how is x^2 the answer to this question??

    (\sqrt{x^3})(\sqrt{x})
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by t-dot View Post
    how is x^2 the answer to this question??

    (\sqrt{x^3})(\sqrt{x})
    see post #3 here.

    hence we have: \sqrt {x^3} \sqrt{x} = x^{3/2}x^{1/2} = x^{3/2 + 1/2} = x^2
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  3. #3
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    I like to look at it this way

    \sqrt{x^{3}} \cdot \sqrt{x}

    Then we get

    x \sqrt{x} \cdot \sqrt{x}

    Then we have: x^{2}

    or you could do

    \sqrt{x^{3}} \cdot \sqrt{x}

    and get

    \sqrt{x^{4}}

    and simplify to get x^{2}
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  4. #4
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    Hello, t-dot

    Yet another approach . . .


    How is x^2 the answer to this question? . (\sqrt{x^3})(\sqrt{x})

    We have: . \sqrt{x^3}\cdot\sqrt{x} \;=\;\sqrt{x^3\!\cdot\! x} \;=\;\sqrt{x^4} \;=\;x^2

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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    \sqrt{x^3}\cdot\sqrt{x} \;=\;\sqrt{x^3\!\cdot\! x} \;=\;\sqrt{x^4} \;=\;x^2

    Hmmmm... This is true so long as x is not negative. If it is, then
    \sqrt{x^3}\sqrt{x} = (i\sqrt{-x^3})(i\sqrt{-x}) = i^2\sqrt{x^4} = -x^2

    -Dan
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