Question

Suppose that the three quadratic equations $a{x}^2-2bx+c=0,b{x}^2-2cx+a=0$ and $c{x}^2-2ax+b=0$ all have only positive roots. Then,

• A. $b^2=ca$
• B. $c^2=ab$
• C. $a^2=bc$
• D. $a=b=c$

Answer 1

Answer: (A), (B), (C), (D)

All the options are correct.
Given: $a{x}^2-2bx+c=\mathrm{0...}\left(i\right),b{x}^2-2cx+a=\mathrm{0...}\left(ii\right)$ and $c{x}^2-2ax+b=\mathrm{0...}\left(iii\right)$ all have positive roots only.
Product of roots $>0$, therefore,
$\Rightarrow \frac{c}{a}>0,\frac{a}{b}>0,\frac{b}{c}>0$
$\Rightarrow a,b,c$ have same sign.

Also, Discriminant $D=B^2-4AC\ge 0$ when roots are positive for standard quadratic equation $A{x}^2+Bx+C=0$.

Therefore,

$\Rightarrow 4b^2-4ac\ge 0$ for equation (iv)

$\Rightarrow 4c^2-4ab\ge 0$ for equation (v)

$\Rightarrow 4a^2-4cb\ge 0$ for equation (vi)

Adding (iv),(v) and (vi), we get,

$\Rightarrow (4b^2-4ac)+(4c^2-4ab)+(4a^2-4cb)\ge 0$

$\Rightarrow (b^2+a^2+c^2)-ac-ab-cb\ge 0$

$\Rightarrow \frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]\ge 0$

$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

$\Rightarrow (a-b{)}^2=0,(b-c{)}^2=0$ and $(c-a{)}^2=0$

$\Rightarrow a=b=c$

Also, from (iv),(v) and (vi), we get,

$\Rightarrow b^2=ac,c^2=ab,a^2=cb$