Question

If \(b^{2}<2ac\) and \(a,b,c,d \in \mathbb{R},\) then the number of real roots of the equation \(ax^{3}+bx^{2}+cx+d=0\) are

Answer 1

Given: \(ax^{3}+bx^{2}+cx+d=0,\) \(b^{2}<2ac\) and \(a,b,c,d \in \mathbb{R}\)

Let \(\alpha, \beta, \gamma\) be the roots of \(ax^{3}+bx^{2}+cx+d=0 \)

\(\therefore ~\text{sum of roots}~ =\alpha+ \beta+ \gamma = -\dfrac{b}{a}\)
\(\text{sum of roots taken two at a time}~ =\alpha \beta+ \beta\gamma+ \gamma \alpha = \dfrac{c}{a}\)
\(\text{product of roots }~ =\alpha.\beta.\gamma = \dfrac{-d}{a}\)

Now, \(\alpha^{2}+ \beta^{2}+ \gamma^{2} = (\alpha+ \beta+ \gamma)^{2} -2(\alpha \beta+ \beta\gamma+ \gamma \alpha)\)

\(\Rightarrow \alpha^{2}+ \beta^{2}+ \gamma^{2} = (-\dfrac{b}{a})^{2} -2(\dfrac{c}{a})\)
\(\Rightarrow \alpha^{2}+ \beta^{2}+ \gamma^{2} = \dfrac{b^{2} -2 ac } {a^{2} }\)
We have given that \(b^{2} -2 ac \) is negative,
\(\therefore \alpha^{2}+ \beta^{2}+ \gamma^{2} <0,\) which is not possible if all \(\alpha, \beta, \gamma\) are real. Therefore atleast one roots is non-real, but complex roots occurs in pair. Hence given cubic equation has one real root and two non real roots.