- #1

Joe20

- 53

- 1

Question:

Prove that if ab > 0 then the equation ax^3 + bx + c = 0 has exactly one root by rolle's theoremProof:

Let f(x) = ax^3+bx+c = 0. f(x) is continuous and differentiable since it is a polynomial.

Assume f(x) has 2 roots, f(a) = 0 and f(b) = 0, there is a point d element of (a,b) such that f'(d) = 0.

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

Since ab>0, a and b must be both positive or both negative.

f'(d) = 3a(d)^2+b = 3ad^2+b not equal to 0 instead >0 since any values of d for d^2 will be positive.

Likewise f'(d) = 3a(d)^2 + b will not equal to 0 instead <0 for all negative values of a and b.

Hence, a contradiction, f has exactly one root.