JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer 10) If the roots of equation \[\frac{{{x}^{2}}-bx}{ax-c}=\frac{m-1}{m+1}\]are equal but opposite in sign, then the value of \[m\] will be [RPET 1988, 2001; MP PET 1996, 2002; Pb. CET 2000]

    A) \[\frac{a-b}{a+b}\]

    B) \[\frac{b-a}{a+b}\]

    C) \[\frac{a+b}{a-b}\]

    D) \[\frac{b+a}{b-a}\]

    Correct Answer: A

    Solution :

    Given equation can be written as \[(m+1){{x}^{2}}-\{m(a+b)-(a-b)\}x+c(m-1)=0\].   Roots are equal and of opposite sign. So sum of roots is equal to zero. Þ \[0=m(a+b)-(a-b)\]Þ \[m=\frac{a-b}{a+b}\].

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