Question

If b² < 2ac, then prove that ax³ + bx² + cx + d = 0 has exactly one real root.

Answer 1

Let all the roots of f(x) = ax³ + bx² + cx + d and f(x) = 0 be real. So 

f ′(x) = 3ax² +2bx + c = 0 has two real roots.

But its discriminant is (2b) 2 − 4⋅3⋅ac = (b² − 2ac) − ac < 0 (as b² < 2ac) which is a contradiction. So f(x) = 0 will not have all the roots real.