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  • The value of a for which one root of the quadratic equation (a^2-5a+3)x^2+(3a-1)x+2=0 is twice as large as the other is

The value of a for which one root of the quadratic equation (a^2-5a+3)x^2+(3a-1)x+2=0 is twice as large as the other is

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Solution:  We have ,      

              (a^2-5a+3)x^2+(3a-1)x+2=0

         Let the roots are  alpha and 2alpha

      \ \ Rightarrow alpha+2alpha=frac1-3aa^2-5a+3 hspace0.2cmand hspace0.2cmalpha cdot 2alpha =frac2a^2-5a+3\ \ Rightarrow 2[frac19frac(1-3a)^2(a^2-5a+3)^2]=frac2a^2-5a+3\ \ Rightarrow 9a^2-6a+1=9a^2-45a+27

 Hence ,     39a=26Rightarrow a=frac23.

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Deependra Verma

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