• question_answer Let a, b be the roots of the equations $(x-a)(x-b)+c,c\ne 0,$ Then the roots of the equations - $\left( \mathbf{x}-\mathbf{a} \right)\left( \mathbf{x}-\mathbf{b} \right)+\mathbf{c}=\mathbf{0}$are:- A) $\left( a,c \right)$                       B) $\left( b,c \right)$          C) $\left( a+c,b+c \right)$              D) $a,b$

[d]   $\because \left( x-a \right)\left( x-b \right)=c$ $\Rightarrow {{x}^{2}}-\left( a+b \right)x+a.b-c=0$              ?..(i) $\because \alpha \And \beta$be the root of the equ.(i), we have $\alpha +\beta =\frac{-b}{a}=a+b$ $\alpha -\beta =\frac{c}{a}=ab-c$ Now, $\because \left( x-\alpha \right)\left( x-\beta \right)+c=0$ ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha .\beta +c=0$ ${{x}^{2}}-\left( a+b \right)x+ab-c+c=0$ ${{x}^{2}}-\left( a+b \right)x+a.b=0$ $\left( x-a \right)\left( x-b \right)=0$ If $\left( x-a \right)=0$                       $~\therefore x=a$ If $\left( x-b \right)=0$                       $~\therefore x=b$ So, $~x=a,b$. Hence option [d] is correct.