no. of real roots.

NO OF REAL OOTS I THINK ITS GONNA 6....... ALL ARE POSITIVE REAL ROOTS, NO NEGATIVE REAL ROOTS..... YOUR QUESTION IS NOT COMPLETE I THINK. THE QUESTION SHOULD BE THE MAX NUMBER OF REAL ROOTS IN THIS EQUATION....... BECAUSE MIN NO OF REAL ROOTS IN THIS EQUATION IS ZERO. A

THANKS.... SOUARDEEP MAJUMDER

x^7+ 1/ x+1 = 1/4

so x not equal to -1

i think no real root

Its obvious that for x≤0, the expression is positive.

For x≥1, since xⁿ⁺¹â‰¥xⁿ, we have

(x^6-x^5)+(x^4-x^3)+(x^2-x)+\frac{3}{4} \ge \frac{3}{4}>0

For 0<x<1, we note that

x^n-x^{n+1} = x^n(1-x)<x(1-x) \le \frac{(x+1-x)^2}{4} = \frac{1}{4} \Rightarrow x^{n+1}-x^n > -\frac{1}{4}

Hence for 0<x<1, we have

(x^6-x^5)+(x^4-x^3)+(x^2-x)+\frac{3}{4} > -\frac{3}{4}+ \frac{3}{4}=0

Hence, there are no real roots

$Thanks Nishant Sir,hsbhatt Sir for Nice Solution.\\\\ Same as hsbhatt Sir.\\\\ The Given expression is always valid for $x\leq0$\\\\ Now for $0<x<1$\\\\ $x^6-x^5+x^4-x^3+x^2-x+\frac{3}{4}$\\\\ \Rightarrow x^4(x^2-x)+x^2(x^2-x)+(x^2-x)+\frac{3}{4}\Leftrightarrow(x^4+x^2+1).(x^2-x)+\frac{3}{4}.\\\\ $(x^4+x^2+1).(x^2-x)+\frac{3}{4}\geq (x^4+x^2+1)(-\frac{1}{4})+\frac{3}{4}\geq \frac{1}{4}\left \{ (1-x^4)+(1-x^2) \right \}>0$\\\\ Now The expression is always $>0$ for $x\geq1.$\\\\ So The Given Expression i.e $x^6-x^5+x^4-x^3+x^2-x+\frac{3}{4}>0\forall x\in R.$