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Math Help - Roots of polynomial

  1. #1
    Senior Member pankaj's Avatar
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    Roots of polynomial

    Consider
     <br />
x^5+ax^4+bx^3+cx^2+dx+e=0,x\in R<br />

    1.If all roots are positive then which of the following is true
    (A) a^5<5^5 e
    (B) a^5>5^5 e
    (C) a^4=5^4 e
    (D) a^3=5^3 e


    2.If roots can take any sign
    (A )(a^2-2b)^5>5e^2
    (B) (a^2+2b^2)>5e^2
    (C) (2a^2-b)>5e^2
    (D)None of these

    3.If one root is zero
    (a) 9a^2<24b
    (B) 9a^2\geq 24b
    (C) 9a^2+24b>0
    (d)None of these


    4.If all coefficients of the equation are positive then which of the following is true
    (A) a^5<5^5 e
    (B) a^5<5e
    (C) a^5<3^5 e
    (d)None of these
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  2. #2
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    I seem to have got some questions hopefully though

    1)Let roots be x_{1},x_{2},x_{3},x_{4},x_{5}

    x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=-a

    x_{1}x_{2}x_{3}x_{4}x_{5}=-e

    A.M\geq G.M

     <br />
\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}\geq (x_{1}x_{2}x_{3}x_{4}x_{5})^{\frac{1}{5}}<br />

     <br />
-a\geq 5(-e)^{\frac{1}{5}}<br />

     <br />
a^5\leq 5^5e<br />


    3)If one root is zero,then let f(x)=x^4+ax^3+bx^2+cx+d.
    Now if f(x)=0 has 4 real roots then f'(x)=0 will have 3 roots(By Rolle's theorem) and consequently f"(x)=0 will have 2 roots for which the condition is that the equation
    6x^2+3ax+b=0
    must have its discriminant greater than zero.
    i.e. 9a^2>24b


    4)Since all the coefficients are positive then by Descartes rule of signs the equation will have atmost zero positive roots.Therefore all the roots of the equation are negative.

    A.M\geq G.M


     <br />
\frac{(-x_{1})+(-x_{2})+(-x_{3})+(-x_{4})+(-x_{5})}{5}\geq (-x_{1}x_{2}x_{3}x_{4}x_{5})^{\frac{1}{5}}<br />

     <br />
a\geq 5e^{\frac{1}{5}}<br />

    a^5 \geq 5^5 e


    It is the 2) that I am not getting
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