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Q.
If $\alpha, \beta$ are the roots of $(x-a)(x-b)=c, c \neq 0$, then the roots of $(x-\alpha)(x-\beta)+c=0$ shall be
Complex Numbers and Quadratic Equations
Solution:
Given $(x-a)(x-b)=c$
$\therefore \alpha+\beta=a+b$ and $\alpha \beta=a b-c$
Now. given equation $(x-\alpha)(x-\beta)+c=0$
$\Rightarrow x^{2}-(\alpha+\beta) x+\alpha \beta+c=0$
If its roots be $p$ and $q$, then
$p + q =(\alpha+\beta)= a + b$
$pq =\alpha \beta+ c = ab - c + c =a b$
So, it can be given by $x^{2}-(a+b) x+a b=0$
So, its roots will be $a$ and $b$