Question

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

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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.