if `alpha,beta` be the roots of the equation `(x-a)(x-b)+x=0(c ne0),` then the roots of the equation `(x-c-alpha)(x-c-beta)=c `are

Options

A. a and b + c
B. a + c and b
C. a + c and b + c
D. a - b and b - c

Answer

C

Explanation

We have, `(x-a) (x-b) + c = 0`
`rArr" "x^(2) - x(a+b) + ab + c = 0`
Since `alpha, beta` are roots of this equation.
`therefore" "alpha + beta = a + b and alpha beta = ab + c`.
Now, `(x-c-alpha)(x - c- beta) = 0`
`rArr" "(x-c)^(2)-(x-c)(alpha+beta)+alpha beta -c = 0`
`rArr" "(x-c)^(2)-(x-c)(a+b)+a b+c-c = 0`
`rArr" "(x-c)^(2)-(x-c)(a+b)+ab = 0`
`rArr" "((x-c)-a)((x-c)-b) = 0 rArr x = c + a and x = c + b`
Thus, the given equation has c + a and c + b its roots.