Question

If $a_1,a_2,a_3,a_4,a_5$ are the roots of the equation $6x^5-41x^4+97x^3-97x^2+41x-6=0,$ such that $|a_1|\le |a_2|\le |a_3|\le |a_4|\le |a_5|,$ then which of the following is/are correct?

• A. $a_1,a_2,a_3$ are in G.P.
• B. $a_1,a_2,a_3$ are in H.P.
• C. $a_3,a_4,a_5$ are in A.P.
• D. the equation has three real roots and two imaginary roots.

Answer 1

Answer: (B), (C)

$a_3,a_4,a_5$ are in A.P.

This is a reciprocal equation of odd degree with opposite signs.
$\therefore x=1$ is a root.
Dividing L.H.S. by $x-1$, the given equation reduces to
$6x^4-35x^3+62x^2-35x+6=0$
Divide above equation by $x^2$,
$6(x^2+\frac{1}{x^2})-35(x+\frac{1}{x})+62=0$
Put $x+\frac{1}{x}=y$
$\Rightarrow (x^2+\frac{1}{x^2})=y^2-2$
$\Rightarrow 6(y^2-2)-35y+62=0\Rightarrow (3y-10)(2y-5)=0\Rightarrow y=x+\frac{1}{x}=\frac{10}{3}\text{or}\frac{5}{2}$
$\Rightarrow 3x^2-10x+3=0\text{or}2x^2-5x+2=0\Rightarrow (3x-1)(x-3)=0\text{or}(2x-1)(x-2)=0\Rightarrow x=\frac{1}{3},3\text{or}x=\frac{1}{2},2$
$\therefore (a_1,a_2,a_3,a_4,a_5)=(\frac{1}{3},\frac{1}{2},1,2,3)$
$\Rightarrow a_3,a_4,a_5$ are in A.P.
$\Rightarrow a_1,a_2,a_3$ are in H.P.