Prove $m^2 \ge 4kx$ for Real Solutions of $x^3+mx+k=0$

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The inequality $m^2 \ge 4kx$ holds for any real solution of the cubic equation $x^3 + mx + k = 0$. This conclusion is derived from analyzing the relationship between the coefficients and the roots of the polynomial. The discussion highlights the importance of understanding the properties of cubic equations and their implications on real solutions.

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Prove that any real solution of $x^3+mx+k=0$ satisfies the inequality $m^2 \ge 4kx$.
 
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If $x$ is a real root of the cubic equation, then the quadratic equation $a^2 + ma + kx = 0$ has a real root, namely $a = x^2$. Thus, the discriminant of the quadratic is nonnegative. This implies $m^2 \ge 4kx$.
 
Euge said:
If $x$ is a real root of the cubic equation, then the quadratic equation $a^2 + ma + kx = 0$ has a real root, namely $a = x^2$. Thus, the discriminant of the quadratic is nonnegative. This implies $m^2 \ge 4kx$.

Awesome, Euge! Thanks for participating and thanks for your great solution too! :)
 

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