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Math Help - equation

  1. #1
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    equation

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  2. #2
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    Re: equation

    Quote Originally Posted by Mhmh96 View Post

    According to wolframalpha your equation does not have real solutions, so you have to find such a way to prove that.
    Last edited by Kmath; June 15th 2012 at 02:49 AM.
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  3. #3
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    Re: equation

    This is tricky equation and simply making algebraic manipulations will require a lot of work.

    In fact this equation doesn't have real roots.
    Let's see why.

    First of all domain of this equation is x>0.
    Next. It is known that for a>0 and b>0 a+b \geq 2\sqrt{ab}
    Proof is straightforward. (\sqrt{a}-\sqrt{b})^2\geq 0 -> a+b- 2\sqrt{ab}\geq 0 -> a+b \geq 2\sqrt{ab}.

    Now,
    \frac {3x+3}{\sqrt{x}}=3\left(\sqrt{x}+\frac{1}{\sqrt{x}  }\right) \geq 3 \left(2\sqrt{\sqrt{x}*\frac{1}{\sqrt{x}}}\right)=6
    Therefore, left side is greater or equal than 6 for all x>0.

    Now, let's see how to handle right part.


    Take expression 5x^2-9x+9 and complete the square: 5x^2-9x+9=\left(\sqrt{5}x-\frac{9}{2\sqrt{5}}\right)^2+9-\frac{81}{20}>0
    From this we have that 5x^2-9x+9>0 for all x.
    Now, divide this expression by 4: 1.25x^2-2.25x+2.25>0 or 2.25x^2-2.25x+2.25>x^2.
    Finally we have that 2.25(x^2-x+1)>x^2
    Dividing by x^2-x+1 (it is positive for all values of x) gives 2.25>\frac{x^2}{x^2-x+1}
    Taking square roots and remembering that x>0 yields
    \sqrt{\frac{x^2}{x^2-x+1}}<1.5 or \frac{x}{\sqrt{x^2-x+1}}<1.5

    Thus, 4+\frac{x}{\sqrt{x^2-x+1}}<4+1.5=5.5

    So, what did we show?
    We showed that left part is at least 6 and right part is less than 5.5. There are no intersections, so equation has no real roots.
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  4. #4
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    Re: equation

    According to WolframAlpha, the RHS attains a maximal value of 4 + \frac{2}{\sqrt{3}}. Still less than 6...

    Proving this using calculus is quite trivial...not a lot of easy algebraic solutions.
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  5. #5
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    Re: equation

    Actually ,i meant ,i dropped 1 so there is no real solution for the previous one .
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  6. #6
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    Re: equation

    Quote Originally Posted by Mhmh96 View Post
    Actually ,i meant ,i dropped 1 so there is no real solution for the previous one .
    Here is WA's solution.
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  7. #7
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    Re: equation

    See, that's why you check what you post before you post it.
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  8. #8
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    Re: equation

    Thats right richard1234 ,i know that the answer is 1,but is there a particular way to solve it ?
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  9. #9
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    Re: equation

    Best bet would be to assume that x = \tan^2 \theta (because if x is real, then x \ge 0).

    Other than that, I don't see any other potentially easy solutions.

    WolframAlpha says the only real solution is x = 1.
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