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

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.
