cubics
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.
I'll give you some hints.
1) Transform the equation to the form $x^3 + 3bx + c$. (The 3 just comes from a simple change of variable, but will come in handy). Your transformation method may come in handy here.
2) Now for the clever trick. Let $x = w - \frac{b}{w}$. This transforms the equation into $w^3 + c - \frac{b^3}{w^3} = 0$... which is a quadratic in $w^3$. So solve this
3) You now have a value for $w$. Can you now find a root of the equation?
Consider $g(x)=x^{3}-3x^{2}=x^{2}(x-3)$, so that $f(x)=g(x)+(2-k)$. The plot of $g(x)$ is as follows;
So we see that $g(x)$ has $2$ roots, one at $x=0$ and the other at $x=3$. It also has a local minimum at $x=2$, with value $f(2)=-4$ and a local maximum of at $x=0$.
Now, $f(x)$ is just a vertical translation of $g(x)$ by $(2-k)$. So for $(2-k)<0$ we have one root, as $g(x)$ is being translated down so its local maximum is below the $x$ axis. For $(2-k)=0$ we have two roots, as $g(x)=f(x)$. For $0<(2-k)<4$ we have three roots, as $g(x)$ has been translated up so that the $x$ axis is between its local minimum and local maximum. For $(2-k)=4$ we have two roots, as $g(x)$ has been translated up until its local minimum is sitting right on the $x$ axis, and finally for $(2-k)>4$ we have one root, as $g(x)$ has been translated until it's local minimum is above the $x$ axis.
As we have $3$ roots for $0<(2-k)<4$, we have by solving the inequality that there are three real roots for $-2<k<2$, i.e. for $k\in(-2,2)$.
As we have a single real, root for $(2-k)<0$ and for $(2-k)>4$, we have upon solving the inequalities that there is one real root for $k>2$ or $x<-2$, i.e. for $k\in (-2,\infty)\cup(2,\infty)$
Best Answer
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.