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Thread: Roots of polynomial

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

    Consider
    $\displaystyle
    x^5+ax^4+bx^3+cx^2+dx+e=0,x\in R
    $

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


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

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


    4.If all coefficients of the equation are positive then which of the following is true
    (A)$\displaystyle a^5<5^5 e$
    (B)$\displaystyle a^5<5e$
    (C)$\displaystyle a^5<3^5 e$
    (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 $\displaystyle x_{1},x_{2},x_{3},x_{4},x_{5}$

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

    $\displaystyle x_{1}x_{2}x_{3}x_{4}x_{5}=-e$

    $\displaystyle A.M\geq G.M$

    $\displaystyle
    \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}}
    $

    $\displaystyle
    -a\geq 5(-e)^{\frac{1}{5}}
    $

    $\displaystyle
    a^5\leq 5^5e
    $


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


    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.

    $\displaystyle A.M\geq G.M$


    $\displaystyle
    \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}}
    $

    $\displaystyle
    a\geq 5e^{\frac{1}{5}}
    $

    $\displaystyle a^5 \geq 5^5 e$


    It is the 2) that I am not getting
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