maybe a silly doubt

can a cubic eqn. have three complex roots...!!

I mean if there's 1 so there will be 2
can it be 3?

if it can be...

then look at this question,

if b² < 2ac, then prove that ax³ + bx² + cx + d = 0 has exactly one real root.

if we assume the roots to be x1, x2 and x3 then we come to,

x₁² + x₂² + x₃² < 0

-> Which on earth is not possible if all of the three are real...
so atleast one is complex
now, complex roots occurr in pairs so,
it means the eqn. has two complex roots..!!
so it has exactly one real and two complex roots...

this is what the bookish soln. says..!!

but can't all the three roots be complex...?

Let us suppose it can't be then check out the eqn. below

f(x) = x⁶ + x⁴ + x² + x + 3

clearly, f(+x) has no changes in signs of the coeff..
and, f(-x) has one change in sign of the coeff..

so by Descartes' law,
it can have atmost 1 negative real root and 0 positive real root..
or, it can have only 1 real root
-> it has five complex roots...... How????????