Let three quadratic equations ` ax^(2) - 2bx + c = 0, bx^(2) - 2 cx + a = 0`
and `cx^(2) - ax + b = 0 `, all have only positive roots. Then ltbr. Which of these are always ture?

Options

A. `b^(2) = ac `
B. `c^(2) = ab `
C. each pair of equations has exactly one root common
D. each pair of equations has two roots common

Answer

1,2,4

Explanation

For each equation , required condition
(i) Dicriminant ` ge 0 rArr b^(2) le ac, c^(2) ge ab , a^(2) ge bc `
(ii) ` f(0) gt 0 rArr c,a,b ge 0 `
(iii) Abscissa of vertex ` gt 0 rArr (b)/(a) gt 0 , (c)/(b) gt 0, (a)/(c) gt0 ` .
Now, `b^(2) ac , c^(2) ge ab rArr b^(2) c^(2) ge a^(2) bc rArr bc ge a^(2).`
But ` a^(2) ge bc. So, a^(2) = bc `
Similarly, ` b^(2) = ac and c^(2) = ab `
Therefore , a = b= c .