Using Rolle's theorem to prove for roots (part 2)

[Joe20]

Hi, I have done up the proof for the question below. Please correct me if I have done wrong for the proof. Thanks in advanced!Question: Prove that if ab < 0 then the equation ax^3 + bx + c = 0 has at most three real roots.Proof:

Let f(x) = ax^3 + bx + c.

Assume that f(x) has 4 distinct roots, f(p) = f(q) = f(r) = f(s) = 0, there is a point x1 \in (p,q) such that f'(x1) = 0 ; x2 \in (q, r) such that f'(x2) = 0 ; x3 \in (r,s) such that f'(x3) = 0.

Since ab < 0 then there are two possibilities where a>0 and b<0 or a <0 , b > 0.

f'(x) = 3ax^2+b

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 > 0 and b < 0, then f'(x) = 0

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 < 0 and b > 0, then f'(x) = 0

This is not true because the equation f'(x) = 0 has only two roots.

Hence the given equation has at most three real roots when ab < 0.

 

[Olinguito]

But all cubic polynomial equations have at most three real roots! (Nerd)

 

[Joe20]

But all cubic polynomial equations have at most three real roots! (Nerd)

Here is the revised proof:

Proof:

Let f(x) = ax^3 + bx + c.

Assume that f(x) has 3 distinct roots, f(p) = f(q) = f(r) = 0, there is a point x1 element of (p,q) such that f'(x1) = 0 ; x2 element of (q, r) such that f'(x2) = 0.

Since ab < 0 then there are two possibilities where a>0 and b<0 or a <0 , b > 0.

f'(x) = 3ax^2+b

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 > 0 and b < 0, then f'(x) = 0

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 < 0 and b > 0, then f'(x) = 0

f'(x) = 0 has two roots.

Hence the given equation has at most three real roots when ab < 0.

Will this be ok? Or need further improvement? If so, how can it be improved?