If `a,b,c in R - {0}` and `a^(2) = bc` and `a + b +c = abc` then the least possible value of `a^(2)` is

Answer

`3`

Explanation

Given `bc = a^(2) …(i)`
`b + c = abc - a`
`b + c = a^(3) - a…..(ii)`
`:.` equation `x^(2) - (b + c)x + bc = 0overset(b)underset(c)(lt)`
`x^(2) - (a^(3) - a) x + a^(2) = 0` has two roots `b` and `c`.
`b` and `c` are equal so
`D ge 0`
`(a^(3) - a)^(2) - 4a^(2) ge 0`
`a^(2)(a^(4) -2a^(2) - 3) ge 0`
`a^(2)(a^(2) - 3)(a^(2) + 1) ge 0`
`a^(2) - 3 ge 0`
`:. a^(2) ge 3`
`:. (a^(2))_(min) = 3`