A) 10
B) 12
C) 14
D) 16
Correct Answer: A
Solution :
We have, \[{{x}^{2}}+px+3=0\] ?(i) \[{{x}^{2}}+qx+5=0\] ?(ii) And \[{{x}^{2}}+(p+q)x+24=0\] ?(iii) \[\therefore (1)+(2)-(3)\Rightarrow {{x}^{2}}=16\Rightarrow x=-4,\,\,4\] So, put common negative root is\[-4\] \[\therefore \] Put \[x=-4\]in (10, we get \[p=\frac{19}{4}\] and put \[x=-4\] in (2), we get\[q=\frac{21}{4}\]. Hence, \[(p+q)=\frac{19}{4}+\frac{21}{4}=10\]You need to login to perform this action.
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