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Thread: Help solving equation

  1. #1
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    Help solving equation

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

    $\displaystyle \sqrt{50}y=100$

    $\displaystyle 5\sqrt{2}y=100$

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

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

    Note: The answer you have written in the question is $\displaystyle 2\sqrt{50} = 2 \sqrt{5 \times 5 \times 2} = 2 \times 5 \times \sqrt{2} = 10 \times \sqrt{2}$
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  5. #5
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    $\displaystyle
    \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}
    $
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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.

    $\displaystyle \frac{2}{3}x^{-1/3}-\frac{1}{3}x^{-2/3}=0$
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  8. #8
    MHF Contributor harish21's Avatar
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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.

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

    then the given equation becomes:

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

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

    $\displaystyle 2u - u^{2} = 0$

    $\displaystyle u^{2} - 2u = 0$

    $\displaystyle u(u-2) = 0$

    So, $\displaystyle u = 0$ OR $\displaystyle u = 2$

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

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

    cube both sides to get:

    $\displaystyle x^{-1} = 8$

    that is

    $\displaystyle \frac{1}{x} = 8$

    therefore,

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

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

    then the given equation becomes:

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

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

    $\displaystyle 2u - u^{2} = 0$

    $\displaystyle u^{2} - 2u = 0$

    $\displaystyle u(u-2) = 0$

    So, $\displaystyle u = 0$ OR $\displaystyle u = 2$

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

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

    cube both sides to get:

    $\displaystyle x^{-1} = 8$

    that is

    $\displaystyle \frac{1}{x} = 8$

    therefore,

    $\displaystyle x = \frac{1}{8}$

    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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