<s>[INST] Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$
Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$
and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$
$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$
Now I did not understand how can i solve it
Help Required
Thanks
 [/INST] Observe that 
$(x - 1)f'(x) = (x - 1)(\sum_0^6 x^i) = x^7 - 1, \tag{1}$
and that the polynomial $x^7 - 1$ has exactly one real zero, $x = 1$.  Thus the zeroes of $f'(x)$ must be the remaining zeroes of $x^7 - 1$, which are the six complex $7$-th roots of unity $e^{2 \pi i / 7}$ for $1 \le i \le 6$.  This shows that $f'(x)$ has no real zeroes; and since $f'(1) = 7 > 0$, $f'(x) > 0$ for all $x \in \Bbb R$.  Thus
$f(x) = \sum_0^7 (x^r /.r) \tag{2}$
is monotonically increasing everywhere.  Being of odd degree, $f(x)$ has at least one real zero; being monotonically increasing, it can have no more.
Hope this helps.  Good Yule to One and All,
and as always,
Fiat Lux!!!
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