The value of ‘a’ for which one root of the quadratic equation (a2-5a+3)x2+(3a-1)x+2=0 is twice as large as the other, is

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-23

-13

answer is A.

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Detailed Solution

Let the roots are α and 2α⇒α+2α=1-3aa2-5a+3 and α.2α=2a2-5a+3⇒219 (1-3a)2(a2-5a+3)2=2a2-5a+3⇒(1-3a)2(a2-5a+3)=9⇒9a2-6a+1=9a2-45a+27⇒39a=26⇒a=23

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