Question 31494: Let p,q be the roots of the equation (x-a)(x-b)=c with c<>0.Then the roots of the equation (x-p)(x-q)+c=0 are which of the following:-
A.a,b
B.a,c
C.b,c
D.a,b,c
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! None of the given options hold true here.
Let us consider a=b=2; and c=9 in the given equation (x-a)(x-b)=c
So, (x-2)^2=9=>(x-2)=3 or (x-2)=-3
=>x=5 or x=-1; same as p=5 and q=-1
This comes down to 5(value of p) and -1(value of q) are the roots of the equation (x-a)(x-b)=c where a=2, b=2 and c=9;
Now,
(x-p)(x-q)+c=0 gives
(x-5)(x+1)+9=0
=>x^2-4x-5+9=0
=>x^2-4x+4=0
=>(x-2)^2=0
=>x=2 or x=-2
The roots of the given equation are 2 and -2
The roots contradict with all given options in terms of a, b, or c
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