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

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The discussion focuses on proving the inequality $m^2 \ge 4kx$ for real solutions of the cubic equation $x^3 + mx + k = 0$. Participants engage in exploring the conditions under which this inequality holds true, emphasizing the relationship between the coefficients and the roots of the polynomial. The conversation highlights the significance of this inequality in understanding the nature of the solutions. A successful proof would demonstrate that all real solutions adhere to this condition, reinforcing the connection between the cubic's parameters and its roots. The thread concludes with appreciation for contributions that clarify the proof.
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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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