The number of real roots of x8-x5+x2-x+1=0 is?
a] 2
b] 4
c] 6
d] 0
2305\left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}x^2-x+1
= \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x^2-\frac{4}{3}x+\frac{4}{3}\right)
= \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x-\frac{2}{3}\right)^2+1-\frac{4}{9}>0\quad \forall x\in \mathbb{R}
Hence number of real roots = 0.(d)
383d] zero
Rishabh Upadhyay whats your explanation?
Soumyabrata Mondal i jst put some values and any of that does nt satisfied the equation....i guess it suld be zero....for confirmation plot the graph.. 
Hardik Sheth is it so easy to plot the graph widout knowing the roots ,if any..??!!
2305I had answered the same question over here (using AM-GM inequality):
http://math.stackexchange.com/questions/366093/calculate-the-number-of-real-roots-of-x8-x5x2-x1-0/366467#366467
So you might want to check that out too.