If `b^2<2a c` , then prove that `a x^2+b x^2+c x+d=0` has exactly one real root.

Explanation

Let `alpha , beta, gamma` be the roots of `ax^(3) + bx^(2) + cx + d = 0` Then
`alpha + beta + gamma = -(b)/(a)`
`alpha beta + beta gamma + gammaalpha = (c)/(a)`
`alpha beta gamma = (-d)/(a)`
`therefore alpha ^(2) + beta ^(2) +gamma^(2) = (alpha + beta +gamma)^(2) - 2(alpha beta + beta gamma + gamma alpha)`
`(b^(2))/(a^(2)) - (2c)/(a) = (b^(2) - 2ac)/(a^(2))`
`rArr alpha^(2) + beta^(2) + gamma^(2) lt 0 (because b^(2) lt 2ac)`
Which is not possible if all `alpha, beta, gamma` are real. So at least one root is nonreal, but complex roots occur in pair. Hence, given cubic equattion has tow nonreal roots and one real root.