Let `alpha,beta` be the roots of the equation `(x-a)(x-b)=c ,c!=0` Then the roots of the equation `(x-alpha)(x-beta)+c=0` are `a ,c` (b) `b ,c` `a ,b` (d) `a+c ,b+c`

Options

A. `a,c`
B. `b,c`
C. `a,b`
D. `a+b,a+c`

Answer

C

Explanation

Given equation is
` (x-a) (x-b) = c, c!= 0`
`rArr x^(2) - (a+b) x + ab - c = 0 `
Let `alpha,beta ` be the roots of this equation.
`:. alpha + beta = a+ b, alpha beta = ab - c `
Consider `(x-alpha) (x-beta)+ c =0`
` x^(2) - ( alpha + beta) x + alpha beta + c = 0 `
Roots of the eqution is
` x= ((alpha+ beta) pm sqrt((alpha + beta)^(2)- 4 ( alpha beta + c) ))/2`
`((a+b) pm sqrt((a+b)^(2)-4 (ab-c+c)))/2`
`=( (a+b) pm sqrt(a^(2) + b^(2) -2ab))/2`
` ((a+b) pm sqrt((a-b))^(2))/2= ((a+b) pm (a-b))/2`
`= (a+b+a-b)/2,(a+b-a+b)/2 = a, b.`
Hence, roots of the equation
` (x-alpha) (x-beta) + c = 0 ` are a and b .