1. ## 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

2. 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