Let `f(x)=x^(2)+bx+c` and `g(x)=x^(2)+b_(1)x+c_(1)`
Let the real roots of `f(x)=0` be `alpha, beta` and real roots of `g(x)=0` be `alpha +k, beta+k` fro same constant `k`. The least value fo `f(x)` is `-1/4` and least value of `g(x)` occurs at `x=7/2`
The valueof `b_(1)` is

Options

A. `-5`
B. 9
C. `-8`
D. `-7`

Answer

4

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`