Solve for x when the ten powers of (1 + x) with exponents 27 through 36 are added together and the sum equals 20.

Step-by-step solution

1. 1

   Substitute $y=1+x$ and factor out the lowest power. The ten exponents are consecutive, so

   $y^{27}\bigl(1+y+y^2+\cdots+y^{9}\bigr)=20$

   This is the key move: without it you are staring at a degree-36 polynomial; with it you are looking at a geometric series.

2. 2

   Collapse the bracket with the geometric-series formula. For $y\ne 1$,

   $1+y+\cdots+y^{9}=\frac{y^{10}-1}{y-1}\quad\Longrightarrow\quad y^{27}\cdot\frac{y^{10}-1}{y-1}=20$

   No rational root survives this — $y=1$ gives $10$, $y=2$ gives more than $10^{10}$ — so the roots have to be located numerically.

3. 3

   Bracket the positive root by a sign change. Write $S(y)=y^{27}+\cdots+y^{36}$. Then $S(1)=10<20$ and

   $S(1.05)=1.05^{27}\cdot\frac{1.05^{10}-1}{0.05}=46.959>20$

   so a root lies in $(1,1.05)$. Bisecting to seven decimals gives $y=1.0221841$, i.e. $x=0.0221841$, and $S(1.0221841)=19.99997$.

4. 4

   Find the second real root by making the sign pattern explicit. For $y=-u$ with $u>0$ the alternating signs regroup exactly as

   $S(-u)=u^{27}(u-1)\bigl(1+u^2+u^4+u^6+u^8\bigr)$

   This is negative for $0<u<1$, zero at $u=1$ (matching $S(-1)=0$), and strictly increasing for $u>1$, so it meets $20$ exactly once. Bisecting gives $u=1.1184421$, where the factored expression evaluates to $19.99998$; hence $y=-1.1184421$ and $x=-2.1184421$.

5. 5

   Argue that there are no other real roots. On $y>0$ every term $y^k$ is increasing, so $S$ rises strictly from $0$ to $\infty$ and crosses $20$ once. On $y<0$ the factored form above is monotone past $u=1$ and never reaches $20$ for $u\le 1$. Two real roots in total; the remaining $34$ roots of the degree-36 polynomial are complex.

6. 6

   State the answer in terms of $x$. Undoing the substitution $x=y-1$:

   $x\approx 0.0221841\qquad\text{or}\qquad x\approx -2.1184421$