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Thread: square root ques.

  1. #1
    Junior Member shannu82's Avatar
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    Exclamation square root ques.

    Find all the real values of x such that
    1/((sqrt(2x-3))^2) > 1/4

    I just want to know how to do this ques.
    So far i have done the following:
    4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)
    x<19/2 (isolated x...)

    Any help would be greatly appreciated.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by shannu82 View Post
    Find all the real values of x such that
    1/((sqrt(2x-3))^2) > 1/4

    I just want to know how to do this ques.
    So far i have done the following:
    4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)
    x<19/2 (isolated x...)

    Any help would be greatly appreciated.
    Good start, but where did you get 19/2 from? I got 7/2 from adding 3 to both sides to eliminate the -3

    You have: $\displaystyle 4 > 2x-3$

    Add 3 to both sides: $\displaystyle 7 > 2x$

    Can you finish?


    From what I understand you are squaring before taking the square root so the domain is all the real numbers except 3/2
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  3. #3
    Junior Member shannu82's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Good start, but where did you get 19/2 from? I got 7/2 from adding 3 to both sides to eliminate the -3

    You have: $\displaystyle 4 > 2x-3$

    Add 3 to both sides: $\displaystyle 7 > 2x$

    Can you finish?


    From what I understand you are squaring before taking the square root so the domain is all the real numbers except 3/2
    Thanx for replying, ya... i for some reason had squared the 4.... I got 7/2. So I just wanna confirm that the answer is x <7/2 is the final ans? I figure that even if you apply the 3/2 sqrt and sqr cancel each other out!
    ~Priya
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  4. #4
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    e^(i*pi)'s Avatar
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    Yes, that's correct
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by shannu82 View Post
    Find all the real values of x such that
    1/((sqrt(2x-3))^2) > 1/4

    I just want to know how to do this ques.
    So far i have done the following:
    4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)
    $\displaystyle \sqrt{x^2}$ and $\displaystyle \big(\sqrt x\big)^2$ are different, the first one equals $\displaystyle |x|$ which holds for any $\displaystyle x$ and the second one only holds for $\displaystyle x\ge0,$ so on your question, the inequality does actually require that $\displaystyle 2x-3>0,$ and then by algebra it follows that $\displaystyle x<\dfrac72,$ but since $\displaystyle x>\dfrac32,$ the solution for the original inequality is $\displaystyle \dfrac32<x<\dfrac72.$
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  6. #6
    Junior Member shannu82's Avatar
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    I just want to make sure i understand correctly, if the above question was changed from (√(2x-3))^2 to √((2x-3)^2) would it be x<7/2
    Last edited by shannu82; Sep 18th 2010 at 12:12 PM.
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  7. #7
    Math Engineering Student
    Krizalid's Avatar
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    no, it's not correct.

    now in this case we're dealing with $\displaystyle \dfrac1{|2x-3|}>\dfrac14,$ which can be written as $\displaystyle |2x-3|<4,$ we can do this since $\displaystyle |2x-3|>0,$ well it can be zero but makes no sense in this case, so $\displaystyle -4<2x-3<4\implies-\dfrac12<x<\dfrac72.$

    but this is not the solution set since $\displaystyle x\ne\dfrac32,$ so the solution set is actually $\displaystyle \displaystyle\left] -\frac{1}{2},\frac{3}{2} \right[\cup \left] \frac{3}{2},\frac{7}{2} \right[.$
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