limit

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1271

. Let

and

be the roots of the equation $(( 1+ a)^\frac{1}{3} -1) x^2 +( ( 1+ a )^\frac{1}{2}-1)x + ( (1+a)^\frac{1}{6} -1) = 0$

approaches to

.

My work i try to find the roots by using quadratic formula and then apply limit . (I have solved this question by taking sum of roots and product ) i want to solve this question by quadratic formula only . irrational part in discriminant is quite cumbersome .how to handle discriminant?

Shadow · **Posted:** Tue, 17 Apr 2012 21:47:36 UTC

mun wrote:

. Let

and

be the roots of the equation $(( 1+ a)^\frac{1}{3} -1) x^2 +( ( 1+ a )^\frac{1}{2}-1)x + ( (1+a)^\frac{1}{6} -1) = 0$

approaches to

.

My work i try to find the roots by using quadratic formula and then apply limit . (I have solved this question by taking sum of roots and product ) i want to solve this question by quadratic formula only . irrational part in discriminant is quite cumbersome .how to handle discriminant?

Why not use L'Hopital's rule on the polynomial as a goes to 0 (divide by the leading term first), then use the quadratic formula on the resulting polynomial? iirc that works.

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1271

sir why direct quadratic formula is not working. i have used binomial theorem in discriminant(on every radical expression) and it is creating problem.

Shadow · **Posted:** Wed, 18 Apr 2012 02:16:46 UTC

mun wrote:

sir why direct quadratic formula is not working. i have used binomial theorem in discriminant(on every radical expression) and it is creating problem.

Well that's probably why, that sounds terrible to deal with. I mean, if you want to use the quadratic formula, that's fine, but then use L'Hopital to calculate the limit.

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