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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Self Evaluation Test - Complex Numbers and Quadratic Equations

  • question_answer
    If both the roots of \[k(6{{x}^{2}}+3)+rx+2{{x}^{2}}-1=0\]and \[6k(2{{x}^{2}}+1)+px+4{{x}^{2}}-2=0\] are common, then \[2r-p\]is equal to

    A) \[-1\]    

    B) 0   

    C) 1                     

    D) 2

    Correct Answer: B

    Solution :

    Given equation can be written as \[(6k+2){{x}^{2}}+rx+3k\,-1=0\]                    ... (i) and \[2\left( 6k + 2 \right){{x}^{2}} +px+ 2\left( 3k -1 \right) = 0\]     ... (ii) Condition for common root is \[\frac{12\,k+4}{6\,k+2}=\frac{p}{r}=\frac{6k-2}{3k-1}=2\,\,\,or\,\,2r-p=0\]


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