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How to prove that a particular cubic equation has three real and distinct roots without finding its discriminant via calculus method?

Please do not use mathematical concepts beyond high school/IIT-JEE. Well and if someone is using the method of finding stationary points please explain the logic behind the technique.

1 Answers1

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I suppose you know derivatives.

Given $y(x)=ax^3+bx^2+cx+d$, find the derivative $ y'(x)=3ax^2+2bx+c$.

To find stationary points you have to solve the equation: $$ 3ax^2+2bx+c=0 $$ that is second degree. So: if this equation has no real roots then the cubic has only one real root. If you find two real solutions $x_1,x_2$ then:

if $ y(x_1)y(x_2) <0$ the cubic has three distinct real roots,

if $ y(x_1)y(x_2) =0$ the cubic has two distinct real roots, one of them double, or three coincident real roots.

if $ y(x_1)y(x_2) >0$ the cubic has only one real root.

Emilio Novati
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  • The first option should be y(x1)(x2)<0 right?(for 3 distinct real roots)Thanks a lot.I understood the technique but can someone explain me the concept behind the technique please?What are stationary points? –  Mar 09 '15 at 16:07
  • Yes! I correct the typo. And added a more exact solution. – Emilio Novati Mar 09 '15 at 16:08
  • Stationary points are points of local maximum or minimum of the function or point of horizontal flexion of his graph. – Emilio Novati Mar 09 '15 at 16:20
  • Got it.This helped.https://www.youtube.com/watch?v=h1nG9i_eAjM –  Mar 09 '15 at 16:25
  • Thank's to Youtube!!! Have you understood ? Don't forget to accept if you are satisfied :) – Emilio Novati Mar 09 '15 at 17:55