Three real roots of a cubic

Question

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Question: If the equation $z^3-mz^2+lz-k=0$ has three real roots, then necessary condition must be _______

• $l=1$
• $ l \neq 1$
• $ m = 1$
• $ m \neq 1$

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I know there is a question here on stack about discriminant and all for these three distinct real roots but I do not feel myself competent to use that. However I tried differentiating the function and for real roots I ended up with $m^2 \geq 3l$

Answer 1

Let's suppose that $z^3-mz^2+lz-k$ has three real roots $a$, $b$, and $c$. Then $$z^3-mz^2+lz-k = (z-a)(z-b)(z-c)$$

so $$z^3-mz^2+lz-k = z^3-(a+b+c)z^2+(ab+bc+ca)z-abc$$

which leads to $$m=a+b+c \quad \text{and} \quad l=ab+bc+ca$$

This does not give any condition on $m$ being equal to $1$ or not : for example, for $a=-1$, $b=0$, $c=2$, then $m=1$ ; but for $a=1$, $b=2$, $c=3$, then $m \neq 1$.

Moreover, this does not give any condition on $l$ either : for example, for $a=1/2$, $b=2$ and $c=0$, then $l=1$ ; but for $a=1$, $b=2$, $c=3$, then $l\neq 1$.

Finally, neither of the given condition is necessary.