Let `f(x) = x2 + b_1x + c_1. g(x) = x^2 + b_2x + c_2`. Real roots of `f(x) = 0` be `alpha, beta` and real roots of `g(x) = 0` be `alpha+gamma, beta+gamma`. Least values of `f(x)` be `- 1/4`Least value of `g(x)` occurs at `x=7/2`

Options

A. `3,-4`
B. `-3,4`
C. `3,-4`
D. `-3,-4`

Answer

3

Explanation

`(beta - alpha) = ((beta + h)-(alpha + h))`
`(beta + alpha)^(2) -4alphabeta = [(beta + h) +(alpha + h)]^(2) - 4(beta + h) (alpha + h)`
`(-b_(1))^(2) - 4c_(1) = (-b_(2))^(2) - 4c_(2)`
`D_(1) = D_(2)`
The least value of f(x) is .
`-(D_(1))/(4) = - (1)/(4) and D_(2) = 1`
Therefore, the least value of
`g(x)` is `-(D_(2))/(4) = -(1)/(4)`
The least value of the of g(x) occurs at
`-(b_(2))/(2)=(7)/(2)` or `b_(2) = -7`
Now, `b_(2)^(2)-4c_(2) = D_(2)`
or ` 49-4c_(2) = 1` or `(48)/(4) = c_(2) or c_(2) = 12`
`rArr x^(2) - 7x + 12 = 0` or `x = 3,4`