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Consider the cubic equation $x^{3}+a x^{2}+b x+c=0$, where $a, b, c$ are real numbers. Which of the following statements is correct?
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Verified Answer
The correct answer is:
If $a^{2}-2 b < 0$, then the equation has one real and two imaginary roots
$\begin{array}{l}
f(x)=x^{3}+a x^{2}+b x+c \\
f^{\prime}(x)=3 x^{2}+2 a x+b \\
D=4 a^{2}-4.3 \quad b=4\left(a^{2}-3 b\right)
\end{array}$
If $a^{2} < 2 b \Rightarrow a^{2} < 3 b \Rightarrow D < 0 \Rightarrow f^{\prime}(x)=0$ has non real roots
Hence $\mathrm{f}(\mathrm{x})=0$ has 1 real and two imaginary roots
f(x)=x^{3}+a x^{2}+b x+c \\
f^{\prime}(x)=3 x^{2}+2 a x+b \\
D=4 a^{2}-4.3 \quad b=4\left(a^{2}-3 b\right)
\end{array}$
If $a^{2} < 2 b \Rightarrow a^{2} < 3 b \Rightarrow D < 0 \Rightarrow f^{\prime}(x)=0$ has non real roots
Hence $\mathrm{f}(\mathrm{x})=0$ has 1 real and two imaginary roots
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