Thread: Roots of polynomial

1. Roots of polynomial

Consider
$
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) $a^5<5^5 e$
(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

2. I seem to have got some questions hopefully though

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

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

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

$A.M\geq G.M$

$
\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}}
$

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

$
a^5\leq 5^5e
$

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.

$A.M\geq G.M$

$
\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}}
$

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

$a^5 \geq 5^5 e$

It is the 2) that I am not getting