Cubic with Vieta's formula

I solve the problem using Vieta's Formula

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[PreMath]

Great presentation! Thanks for sharing. Keep rocking 👍

[davidblauyoutube]

Great video!

I was taught a shortcut in situations like this. Since you are looking for the sum of the reciprocals of the roots (and no root is zero), let x = 1/y. Then the desired number is the sum of the /roots/ of the polynomial expressed in y, which simply has the coefficients reversed, i.e. 2y^3 + 6y^2 + 3y + 1 = 0. Using the sum-of-roots formula now gives the answer -6/2 = -3.

This lets you quickly calculate e.g. 1/ab + 1/ac + 1/bc (in x) as ab+ac+bc (in y), which is 3/2 by inspection. 🙂

[siddharthsoni]

Bhai itne easy, thode mushkil lao na bhai