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Math Help - Help solving equation

  1. #1
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    Help solving equation

    How is \sqrt{50}y=100 solved? The answer in my book is 2\sqrt{50}
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  2. #2
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    Quote Originally Posted by Mike9182 View Post
    How is \sqrt{50}y=100 solved? The answer in my book is 2\sqrt{50}
    I get

    \sqrt{50}y=100

    5\sqrt{2}y=100

    y=\frac{100}{5\sqrt{2}}

    y=\frac{20}{\sqrt{2}}=10\sqrt{2}
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  3. #3
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    How did \frac{20}{\sqrt{2}} become 10\sqrt{2}?
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  4. #4
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    Quote Originally Posted by Mike9182 View Post
    How did \frac{20}{\sqrt{2}} become 10\sqrt{2}?
    \frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{20 \times \sqrt{2}}{2} = 10 \times \sqrt{2}

    Note: The answer you have written in the question is 2\sqrt{50} = 2 \sqrt{5 \times 5 \times 2} = 2 \times 5 \times \sqrt{2} = 10 \times \sqrt{2}
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  5. #5
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     <br />
\frac{20}{\sqrt{2}}=\frac{20}{\sqrt{2}}\times {\color{red}\frac{\sqrt{2}}{\sqrt{2}}}= \frac{20\sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{20\sqrt{2}}{2}=10\sqrt{2}<br />
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  6. #6
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    Thanks, I understand now.
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  7. #7
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    I have another equation I need help solving, the correct answer is 1/8.

    \frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0
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  8. #8
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    Quote Originally Posted by Mike9182 View Post
    I have another equation I need help solving, the correct answer is 1/8.

    \frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0
    Let  u = x^{-1/3} .

    then the given equation becomes:

    \frac{2u}{3}-\frac{u^2}{3}=0

    \frac{2u - u^2}{3} = 0

    2u - u^{2} = 0

    u^{2} - 2u = 0

    u(u-2) = 0

    So, u = 0 OR u = 2

    substitute u= x^{\frac{-1}{3}}, you get

    x^{\frac{-1}{3}} = 2

    cube both sides to get:

    x^{-1} = 8

    that is

    \frac{1}{x} = 8

    therefore,

    x = \frac{1}{8}
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  9. #9
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    another question

    Quote Originally Posted by harish21 View Post
    Let  u = x^{-1/3} .

    then the given equation becomes:

    \frac{2u}{3}-\frac{u^2}{3}=0

    \frac{2u - u^2}{3} = 0

    2u - u^{2} = 0

    u^{2} - 2u = 0

    u(u-2) = 0

    So, u = 0 OR u = 2

    substitute u= x^{\frac{-1}{3}}, you get

    x^{\frac{-1}{3}} = 2

    cube both sides to get:

    x^{-1} = 8

    that is

    \frac{1}{x} = 8

    therefore,

    x = \frac{1}{8}

    A direct solution if a-b =0 a=b Simplify the equals to find x^1/3 =1/2 so x=1/8


    bjh
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