A) \[(x-\,2)\]
B) \[{{k}_{1}}=1,\,{{k}_{2}}=-\,6\]
C) \[{{k}_{1}}=1,\,{{k}_{2}}=-\,6\]
D) \[{{k}_{1}}=6,\,{{k}_{2}}=-\,1\]
Correct Answer: B
Solution :
[b] \[\therefore \,\,\,If\,x-\,2=0\] \[\therefore \,\,\,x=2\] \[then,\] \[{{x}^{2}}+{{k}_{1}}x+{{k}_{2}}=0\] \[{{(2)}^{2}}+{{k}_{1}}\times 2+{{k}_{2}}=0\] \[2{{k}_{1}}+{{k}_{2}}=-4\] ?. (i) \[If\,\,x+3=0\] \[\therefore x=-\,3\] \[then,\,\] \[{{x}^{2}}+{{k}_{1}}x+{{k}_{2}}=0\] \[{{(-\,3)}^{2}}+{{k}_{1}}x-\,3+{{k}_{2}}=0\] \[\therefore 3{{k}_{1}}-\,{{k}_{2}}=9\] ?. (ii) From equation (i) and (ii), We get \[{{k}_{1}}\]\[=1\] and \[{{k}_{2}}\]
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