JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer If one root of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is equal to the nth power of the other root, then the value of \[{{(a{{c}^{n}})}^{\frac{1}{n+1}}}+{{({{a}^{n}}c)}^{\frac{1}{n+1}}}=\] [IIT 1983]

    A) \[b\]

    B) - b

    C) \[{{b}^{\frac{1}{n+1}}}\]

    D) \[-{{b}^{\frac{1}{n+1}}}\]

    Correct Answer: B

    Solution :

    Let \[\alpha ,{{\alpha }^{n}}\] be the two roots. Then \[\alpha +{{\alpha }^{n}}=-b/a,\alpha {{\alpha }^{n}}=c/a\] Eliminating\[\alpha \], we get \[{{\left( \frac{c}{a} \right)}^{\frac{1}{n+1}}}+{{\left( \frac{c}{a} \right)}^{\frac{n}{n+1}}}=-\frac{b}{a}\] Þ \[a.{{a}^{-\ \frac{1}{n+1}}}.{{c}^{\frac{1}{n+1}}}+a.{{a}^{-\frac{n}{n+1}}}.{{c}^{\frac{n}{n+1}}}=-b\] or\[{{({{a}^{n}}c)}^{\frac{1}{n+1}}}+{{(a{{c}^{n}})}^{\frac{1}{n+1}}}=-b\].

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