JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    The roots a and P of the quadratic equation\[p{{x}^{2}}+qx+r=0\]are real and of opposite signs. The roots of \[\alpha {{(x-\beta )}^{2}}+\beta {{(x-\alpha )}^{2}}=0\]

    A)  positive                              

    B)  negative

    C)  of opposite signs            

    D)  non real

    Correct Answer: C

    Solution :

     We have\[\alpha +\beta =-\frac{q}{p},\alpha \beta =\frac{r}{p}\]Now the given equation \[\alpha {{(x-\beta )}^{2}}+\beta {{(x-\alpha )}^{2}}=0\] \[\Rightarrow \]\[(\alpha +\beta ){{x}^{2}}-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0\] \[\Rightarrow \]\[\left( -\frac{q}{p} \right){{x}^{2}}-4\frac{r}{p}x+\frac{r}{p}\left( -\frac{q}{p} \right)=0\] \[\Rightarrow \]\[pq{{x}^{2}}+4prx+rq=0\]??(i) Since \[\alpha \] and \[\beta \] have opposite signs, therefore p and r must have opposite signs. \[\Rightarrow \]pq and rq must have opposite signs \[\Rightarrow \] roots of equation (i) have opposite signs


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