Find the square root of 2706.225166081663.

Step-by-step solution

1. 1

   Bracket the answer with nearby perfect squares. $52^2 = 2704$ and $53^2 = 2809$. Since $2704 < 2706.225 < 2809$, the root lies between $52$ and $53$, and much closer to $52$ because the target is only $2.225$ above $2704$ out of a $105$-wide gap.

2. 2

   Get a first-order estimate. Near $x_0 = 52$ the square function has slope $2x_0 = 104$, so a rise of $2.225166$ in the square corresponds to about

   $\frac{2.225166}{104} \approx 0.021396$

   in the root, giving the estimate $52.021396$. This is exactly one step of Newton's method.

3. 3

   Refine with a second Newton iteration. The iteration is $x_{n+1} = \tfrac{1}{2}\left(x_n + \tfrac{N}{x_n}\right)$ with $N = 2706.225166081663$. From $x_1 = 52.021396$:

   $\frac{N}{x_1} = 52.021387, \qquad x_2 = \frac{52.021396 + 52.021387}{2} = 52.021391$

   Newton's method roughly doubles the number of correct digits per step, which is why one more pass already stabilises six decimals.

4. 4

   State the result.

   $\sqrt{2706.225166081663} \approx 52.021391$

5. 5

   Verify by squaring. $52.021391^2 = 2706.225122$, which falls short of the target by $4.45 \times 10^{-5}$ — consistent with the root being truncated at the $10^{-6}$ level, since the derivative $2x \approx 104$ magnifies that error by about a hundred. Carrying more digits, $52.0213914278^2 = 2706.2251660844$, matching the given number to eleven significant figures.