Real solutions of $x^5+ax^4+bx^3+cx^2+dx+e=0$

Question

If $2a^2＜5b$,prove that the equation $x^5+ax^4+bx^3+cx^2+dx+e = 0$ has at least one complex root。Thanks.

Answer 1

I am assuming all coefficients are real.

If $f(x) = x^5+ax^4+bx^3+cx^2+dx+e $, then $f'(x) = 5x^4+4ax^3+3bx^2+2cx+d $, $f''(x) = 20x^3+12ax^2+6bx+2c $, $f'''(x) = 60x^2+24ax+6b $.

If $f$ has 5 real roots, then $f'$ has 4 real roots, $f''$ has 3 real roots, and $f'''$ has 2 real roots.

The roots of $f'''$ are the roots of $0 = 10x^2+4ax+b $ and the discriminant of this is $(4a)^2-4\cdot10b =16a^2-40b =8(2a^2-5b) $. If this has real roots, then $2a^2-5b \ge 0 $.

Therefore, if $2a^2< 5b$, $f'''$ has complex roots so $f$ can not have all real roots.