Solve (1 + x) to the 12th power minus 11.0528x = 1.

Step-by-step solution

1. 1

   Rewrite as a root-finding problem. Move the $1$ to the left and define $f(x)=(1+x)^{12}-11.0528x-1,$ so we need $f(x)=0$. This is the shape of an annuity/IRR equation: an accumulation factor over $12$ periods set against $11.0528$ times the rate.

2. 2

   Spot the trivial root. $f(0)=1^{12}-0-1=0$, so $x=0$ is an exact solution. It is the degenerate 'zero interest' case, so a second, economically meaningful root is expected.

3. 3

   Locate the turning point before hunting for the second root. $f'(x)=12(1+x)^{11}-11.0528$, which vanishes when $(1+x)^{11}=\tfrac{11.0528}{12}=0.921067$, i.e. $x=0.921067^{1/11}-1\approx-0.0074469$. Since $f'$ increases in $x$, this is the single minimum, with $f(-0.0074469)\approx-0.0034829<0$.

4. 4

   Conclude there are exactly two real roots and bracket the other one. $f$ falls from $f(-1)=10.0528$ to the minimum, then rises back to $f(0)=0$, so one root sits to the left of the minimum. Testing: $f(-0.02)=+0.0057727$ and $f(-0.01)=-0.0030871$, so the root lies in $(-0.02,-0.01)$.

5. 5

   Bisect to full precision. Halving the bracket repeatedly gives $x \approx -0.01508489,$ where $f(x)$ is zero to within $10^{-16}$. Beware of stopping near $x\approx-0.00745$: that is the minimum, where $f=-0.0035\ne0$.

6. 6

   Verify and interpret. $(1-0.01508489)^{12}=0.8332697$ and $-11.0528(-0.01508489)=0.1667303$; their sum is $1.000000$, matching the right-hand side. The two real solutions are $x=0$ and $x\approx-1.5085\%$ per period; there is no positive-rate solution, because $f>0$ for all $x>0$.