A) 3
B) 1
C) - 2
D) 2
Correct Answer: D
Solution :
Given equations are \[{{x}^{2}}-3x+a=0\] ??(i) and \[{{x}^{2}}+ax-3=0\] ??(ii) Subtracting (ii) from (i), we get Þ \[-3x-ax+a+3=0\] \[\Rightarrow (a+3)(-x+1)=0\] Þ either \[a=-3\] or \[x=1\] When \[a=-3,\]the two equations are identical. So, we take \[x=1,\] which is the common root of the two equations. Substituting \[x=1\] in (i), we get \[1+a-3=0\Rightarrow a=2.\]You need to login to perform this action.
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