$The no. of Real roots of the equation $x^6-x^5+x^4-x^3+x^2-x+\frac{3}{4}=0$
1NO 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
62rewrite the eqtn as
x^6-x^5+x^4...-x+1=1/4
Multiply both sides by x+1
x^7+1=1/4(x+1)
Now the graph shows clearly (one that I have not drawn :P) that there is only 1 real root? which is -ve and from the argument of souradeep, we can say that that root is -1 which has been created because of our factor (x+1)
Hence the given equation has no real roots..
341Its obvious that for x≤0, the expression is positive.
For x≥1, since xn+1≥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
1708$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 $x^6-x^5+x^4-x^3+x^2-x+\frac{3}{4}>0\forall x\in R.$
