Theory of Equations 3 Question 1
1. Let $\alpha, \beta$ be the roots of the equation, $(x-a)(x-b)=c, c \neq 0$. Then the roots of the equation $(x-\alpha)(x-\beta)+c=0$ are
(a) $a, c$
(b) $b, c$
(c) $a, b$
(d) $a+c, b+c(1992,2 M)$
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Solution:
- Given, $\alpha, \beta$ are the roots of $(x-a)(x-b)-c=0$
$\Rightarrow \quad(x-a)(x-b)-c=(x-\alpha)(x-\beta)$
$\Rightarrow \quad(x-a)(x-b)=(x-\alpha)(x-\beta)+c$
$\Rightarrow a, b$ are the roots of equation $(x-\alpha)(x-\beta)+c=0$
