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Thread: Radicals

  1. #1
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    Radicals

    solve the given equation for x
    (x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2)
    Sorry I didnt write it in radical form, wasnt sure how. btw the -(x)^(1/2) is in the numerator. the answer in the back of the book is 25. Half way thru i got x + (x)^(1/2) - 45 = 0 and then substituted y = (x)^(1/2) and y^2=x for the values. But my end results didnt seem to be right. Sorry if this is cluttered.
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  2. #2
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    Quote Originally Posted by OzzMan View Post
    solve the given equation for x
    (x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2)
    Sorry I didnt write it in radical form, wasnt sure how. btw the -(x)^(1/2) is in the numerator. the answer in the back of the book is 25. Half way thru i got x + (x)^(1/2) - 45 = 0 and then substituted y = (x)^(1/2) and y^2=x for the values. But my end results didnt seem to be right. Sorry if this is cluttered.
    First, if the answer is x = 25, then the equation is NOT (x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2), it's (x-9)^(1/2)=36/(x-9)^(1/2) + (x)^(1/2).

    $\displaystyle \sqrt{x-9} = \frac{36}{\sqrt{x-9} + \sqrt{x}}$


    $\displaystyle \Rightarrow \sqrt{x-9} (\sqrt{x-9} + \sqrt{x})= 36$


    $\displaystyle \Rightarrow x - 9 + \sqrt{x-9} \sqrt{x} = 36$


    $\displaystyle \Rightarrow \sqrt{x-9} \sqrt{x} = 45 - x$


    $\displaystyle \Rightarrow (x - 9)x = (45 - x)^2$


    $\displaystyle \Rightarrow x^2 - 9x = 45^2 - 90x + x^2$


    $\displaystyle \Rightarrow -9x = 45^2 - 90x$


    $\displaystyle \Rightarrow 81x = 45^2 \Rightarrow 9^2 x = 45^2 \Rightarrow x = \left( \frac{45}{9} \right)^2 = 25$.
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  3. #3
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    you wrote it wrong you have + radical x in the denominator its - radical x and its not part of that fraction its over 1
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  4. #4
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    Quote Originally Posted by OzzMan View Post
    you wrote it wrong you have + radical x in the denominator its - radical x and its not part of that fraction its over 1
    Oh I see. The numerator business in you original post and the its over 1 business in your next post are just red herrings. The equation really is as you originally stated:


    $\displaystyle \sqrt{x-9} = \frac{36}{\sqrt{x-9}} - \sqrt{x}$


    $\displaystyle \Rightarrow x - 9 = 36 - \sqrt{x-9} \sqrt{x}$


    from which it follows that:


    Quote Originally Posted by mr fantastic View Post
    [snip]

    $\displaystyle \Rightarrow \sqrt{x-9} \sqrt{x} = 45 - x$


    $\displaystyle \Rightarrow (x - 9)x = (45 - x)^2$


    $\displaystyle \Rightarrow x^2 - 9x = 45^2 - 90x + x^2$


    $\displaystyle \Rightarrow -9x = 45^2 - 90x$


    $\displaystyle \Rightarrow 81x = 45^2 \Rightarrow 9^2 x = 45^2 \Rightarrow x = \left( \frac{45}{9} \right)^2 = 25$.
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  5. #5
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    yeah thats the right equation. Again I'm sorry i didn't use latex. I'm just no good at it. I always get an error.
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  6. #6
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    im a little confused this is probably where i messed up, but i did the cross multiply of radical x-9 and the radicals crossed out like you have but why would radical x-9 stay on the right side also?
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  7. #7
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    Quote Originally Posted by OzzMan View Post
    im a little confused this is probably where i messed up, but i did the cross multiply of radical x-9 and the radicals crossed out like you have but why would radical x-9 stay on the right side also?
    You have to multiply both of the terms on the right hand side by $\displaystyle \sqrt{x - 9}$ ..... So you get a cancelling that leaves you with the 36, AND the $\displaystyle \sqrt{x}$ gets multiplied by $\displaystyle \sqrt{x - 9}$ to leave you with $\displaystyle \sqrt{x - 9} \sqrt{x}$.
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  8. #8
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    yea i thought as much. just didnt see why you had to multipy the radical x by radical x-9. didnt think that applied to cross multiplying. but i see where your coming from.
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