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Math Help - Simplifying surds

  1. #1
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    Question Simplifying surds

    I have placed this equation into the quadratic formula:

    2x^2 + x(-\sqrt{50} - 2\sqrt{7/2}) + \sqrt{175} = 0

    To get:

    x= \dfrac{-(-\sqrt{50} - 2\sqrt{7/2})\pm\sqrt{(-\sqrt{50} - 2\sqrt{7/2})^2 - 4 \times 2 \times \sqrt{175}}}{2 \times 2}

    Which simplifies to:

    x= \dfrac{\sqrt{50} + \sqrt{14}+\sqrt{\sqrt{2800} - \sqrt{11200}}}{4}

    and

    x= \dfrac{\sqrt{50} + \sqrt{14}-\sqrt{\sqrt{2800} - \sqrt{11200}}}{4}


    Now how do I simplify them to:

    \dfrac{\sqrt{50}}{2} and \sqrt{7/2}

    Can someone show me?
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  2. #2
    A Plied Mathematician
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    I think there is something wrong with your work so far. You're going to get complex numbers, which will never simplify down to the desired results. Double-check your simplification of the discriminant inside the square root, there.

    [EDIT]: I also don't think your results at the end are the roots of the polynomial. Try plugging them in and see if they satisfy the original equation.
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  3. #3
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    The results of \dfrac{\sqrt{50}}{2} and \sqrt{7/2} are the roots of the polynomial. I have checked.

    Maybe I did the expansion of (-\sqrt{50} - 2\sqrt{7/2})^2 wrong somewhere and that has led me to complex numbers...
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Yes, the expansion is wrong.

    (-\sqrt{50} - 2\sqrt{7/2})^2 = 64 + \sqrt{2800}
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  5. #5
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    Then I suggest you check again.

    \frac{\sqrt{50}}{2}= \frac{5}{2}\sqrt{2} is approximately 3.53533, \sqrt{7/2} is approximately 1.87083, and \sqrt{175}= 5\sqrt{3} is approximately 13.22876. 2x^2- (\frac{\sqrt{50}}{2}+ \sqrt{7/2})x+ \sqrt{175}= 0 is 2x^2- 16.76409x+ 13.22876= 0.

    Putting x= 3.53533 into that gives -33.54313 while putting x= 1.87083 into it gives 24.74118. Neither satisifies the equation.

    If a and b are roots of a polynomial of the form (x- a)(x- b)= 0 then we have x^2- (a+ b)x+ ab= 0
    Now, with a= \frac{\sqrt{50}}{2} and b= \sqrt{7/2}, that would be x^2- (\frac{\sqrt{50}}+ \sqrt{7/2})x+ \sqrt{175}= 0. But that is NOT what you have- you have an extra "2" multiplying x^2.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Then I suggest you check again.

    \frac{\sqrt{50}}{2}= \frac{5}{2}\sqrt{2} is approximately 3.53533, \sqrt{7/2} is approximately 1.87083, and \sqrt{175}= 5\sqrt{3} is approximately 13.22876. 2x^2- (\frac{\sqrt{50}}{2}+ \sqrt{7/2})x+ \sqrt{175}= 0 is 2x^2- 16.76409x+ 13.22876= 0.

    Putting x= 3.53533 into that gives -33.54313 while putting x= 1.87083 into it gives 24.74118. Neither satisifies the equation.

    If a and b are roots of a polynomial of the form (x- a)(x- b)= 0 then we have x^2- (a+ b)x+ ab= 0
    Now, with a= \frac{\sqrt{50}}{2} and b= \sqrt{7/2}, that would be x^2- (\frac{\sqrt{50}}+ \sqrt{7/2})x+ \sqrt{175}= 0. But that is NOT what you have- you have an extra "2" multiplying x^2.
    Actually this becomes:

    x^2 - (\frac{\sqrt{50}}{2} + \sqrt{7/2})x + \frac{\sqrt{175}}{2} = 0

    Multiplying throughout by 2 then gives:

    2x^2 - (\sqrt{50}} + 2\sqrt{7/2})x + \sqrt{175} = 0
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  7. #7
    Senior Member Educated's Avatar
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    Here:

    Quadratic formula solve equation - Wolfram|Alpha

    I have checked 4 times now and the roots are correct.

    I just want to know how to simplify it now.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    x= \dfrac{\sqrt{50} + \sqrt{14}\pm\sqrt{64+\sqrt{2800} - \sqrt{11200}}}{4}

    x= \dfrac{5\sqrt{2} + \sqrt{14}\pm\sqrt{64+20\sqrt{7} - 40\sqrt{7}}}{4}

    x= \dfrac{5\sqrt{2} + \sqrt{14}\pm\sqrt{64-20\sqrt{7}}}{4}

    x= \dfrac{5\sqrt{2} + \sqrt{14}}{4} \pm \dfrac{\sqrt{64-20\sqrt{7}}}{4}

    x= \dfrac{5\sqrt{2} + \sqrt{14}}{4} \pm \sqrt{\dfrac{64-20\sqrt{7}}{16}}

    x= \dfrac{5\sqrt{2} + \sqrt{14}}{4} \pm \sqrt{4-\dfrac{5}{4}\sqrt{7}}

    This is What I got so far
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