Question

The number of distinct real roots of the equation x^9+ x^7+ x^5+ x^3+ x+ 1=0 is?

Answer 1

See diagram.

$y = 1+x+x^3+x^5+x^7+x^9 \\ \\ y = -4\ \ at\ \ \ x = -1 \\ \\ y = 1\ at\ x=0$

So  y = 0 when x is in between -1 and 0.  for x >0 , y >0 and increasing. For x < -1 , y is decreasing more and more. So there is only one real root  -1 < α < 0.

Slope of y : 1+3x^2+5x^4+7x^6+9x^8.  it is always positive. So y is always increasing as x increases. So y = 0 is possible only once.

You can find that approximately as: - 0.63

Answer 2

x^9+ x^7+ x^5+ x^3+ x+ 1=0From the plot of the curve, it is clear that it has only one real solution and 8 imaginary solutions.