The real numbers `x_1, x_2, x_3` satisfying the equation `x^3-x^2+b x+gamma=0` ar ein A.P. Find the intervals in which `betaa n dgamma` lie.

Answer

`beta in [-oo, (1)/(3)] and gamma in [-(1)/(27), oo]`

Explanation

Since, `x_(1), x_(2), x_(3)` are in an AP. Let `x_(1) = a -d, x_(2) = a and x_(3) a +d and x_(1), x_(2), x_(3)` be the roots of `x^(3) -x^(2) + beta x + gamma = 0`
`:. Sigma alpha = a -d + a + a + d = 1`
`rArr a = 1//3`...(i)
`Sigma alpha beta = (a -d) a + a ( a+d) + (a -d) (a +d) = beta`....(ii)
and `alpha beta gamma = (a -d) a (a +d) = - gamma`....(iii)
From Eq. (i),
`3a = 1 rArr a = 1//3`
From Eq. (ii), `3a^(2) -d^(2) = beta`
`rArr 3(1//3)^(2) - d^(2) = beta` [from Eq. (i)]
`rArr 1//3 - beta = d^(2)`
Note In this equation, we have two variables `beta and gamma` but we have only one equation. So at first sight it looks that this equation cannot solve but we know that `d^(2) ge 0, AA d in R`, then `beta` can be solved. This trick is frequently asked in IIT examples.
`rArr (1)/(3) - beta ge 0`
`rArr beta le (1)/(3) rArr beta in [-oo, 1//3]`
From Eq. (iii), `a(a^(2) -d^(2)) = - gamma`
`rArr (1)/(3) ((1)/(9) -d^(2)) = - gamma rArr (1)/(27) - (1)/(3) d^(2) = - gamma`
`rArr gamma + (1)/(27) = (1)/(3) d^(2) rArr gamma + (1)/(27) ge 0`
`rArr gamma ge - 1//27`
`rArr gamma in [-(1)/(27),oo)`
Hence, `beta in (-oo, 1//3] and gamma in [-1//27, oo)`