Question

The equation $a{x}^2-2bx+c=0,b{x}^2-2cx+a=0$ and $c{x}^2-2ax+b=0$ will have only positive roots provided

• A. $a,b,c$ are all positive
• B. $a,b,c$ are all negative
• C. $a>b>c$ or $a<b<c$
• D. $a=b=c$

Answer 1

The correct option is D $a=b=c$

Let $({\alpha }_1,{\beta }_1);({\alpha }_2,{\beta }_2);({\alpha }_3,{\beta }_3)$ be roots of the three equations repectively
$\Rightarrow {\alpha }_1+{\beta }_1=\frac{2b}{a};{\alpha }_1{\beta }_1=\frac{c}{a}$
${\alpha }_2+{\beta }_2=\frac{2c}{b};{\alpha }_2{\beta }_2=\frac{a}{b}$
${\alpha }_3+{\beta }_3=\frac{2a}{c};{\alpha }_3{\beta }_3=\frac{b}{c}.$
This implies
$a,b,c$ are all of the same sign.
$\therefore ab>0,bc>0,ca>0$
But for roots to exist,$4b^2-4ac\ge 0$ ; $4c^2-4ab\ge 0$ ; $4a^2-4bc\ge 0$
$\Rightarrow a^2+b^2+c^2<ab+bc+ca\dots \dots$
Adding the three inequalities$=\frac{1}{2}\left(\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)\ge 0$
$\Rightarrow a=b;b=c;c=a$
$a=b=c$