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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    If the equations \[{{x}^{2}}+px+3=0,{{x}^{2}}+qx+5=0\] and \[{{x}^{2}}+(p+q)x+24=0\] have a common negative root, then the value of \[(p+q)\], is

    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, \[(p+q)=\frac{19}{4}+\frac{21}{4}=10\]


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