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

Options

A. `b^(2) = ca`
B. `c^(2) = ab`
C. `a^(2) = bc`
D. `a = b = c`

Answer

A, B, C, D

Explanation

`ax^(2) - 2bx + c = 0`
`x^(2) - 2cx + a = 0` and `cx^(2) - 2ax + b = 0`
All three equation have positive roots only `(1)`
`rArr "vertex" (x-"coordinate") gt 0`
`rArr (b)/(a) gt0`
`(II)`
`a. f(0) gt 0 rArr a.c gt 0`
(A) If `a gt 0`
`I^(st)b gt 0 , c gt 0II^(nd)`
(B) If `a lt 0`
`I^(st)b gt 0 , c gt 0II^(nd)`
`rArr` either `a gt 0, b gt 0, cgt 0`
Or `a lt 0, b lt 0 , c lt 0`
Case No. `1`
`a gt 0, b gt 0` and `c gt 0`
`rArr` Discriminant `ge 0`
`a^(2) ge bc`........(I) , `a^(4) ge b^(2)c^(2)` .......(1)
`b^(2) ge ac` ....(II) , `b^(4)gea^(2)c^(2)`
`c^(2) ge ab`.....(III) , `c^(4)ge a^(2)b^(2)`......(3)
Put `b^(2)` from equation `(II)` to `(I)`
`rArr a^(4) ge ac^(3) rArr a^(3) ge c^(3) "......"(4)`
Now put value of `b^(2)` from equation `n(II)`
`rArr c^(4) ge a^(3)c`
`rArr c^(3) ge a^(3)`.......(5)
So form equation no `(4)` and `(5)`
`c^(3) = a^(3) rArr c = a`
Similarly we can do `b = a c = c`
So `a = b = c`
Hence `a^(2) = bc`
`b^(2) = ac`
`a^(2) = bc` and `c^(2) = a^(2)`
Similarly we can do in case of `a lt 0, b lt 0` and `c lt 0` so option no `A, B, C, D` all are correct