A) \[\left( a,c \right)\]
B) \[\left( b,c \right)\]
C) \[\left( a+c,b+c \right)\]
D) \[a,b\]
Correct Answer: D
Solution :
[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.
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