If, a, b, c, d and e are all real numbers how could I prove that the 5 solutions of the equation: $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e == 0$ cannot all be real valued if: $2a^2 < 5b$
Any assistance is appreciated.
If, a, b, c, d and e are all real numbers how could I prove that the 5 solutions of the equation: $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e == 0$ cannot all be real valued if: $2a^2 < 5b$
Any assistance is appreciated.
If there are 5 zeros, then the first derivative has 4 zeros, the second has 3 zeros, and the third has 2 zeros, all counted with multiplicity. The condition on $a$ and $b$ is exactly the condition that the third derivative has no real zeros.