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If α & β be the roots of ax2+bx+c = 0, then limx→α( 1+ax2+bx+c)1/(x-α) is :-
Hi
As α and β are roots
so,
ax2 + bx + c = a(x - α) (x - β)
Let
y = limx->α {1+ a(x - α)(x - β)}1/(x - α)
=> log y = limx->α log [1+ a (x - α)(x - β)] / (x - α)
= limx->α [a(x - α) + a (x - β)] / 1+a(x - β)(x-α)
= a(α - β)
=> y = ea(α - β)
Thanks
Anurag Kishore
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