Estimating Square Roots

What's the square root of 55? You know it's between 7 (49) and 8 (64), but "somewhere between 7 and 8" isn't a usable answer in an interview. The fix is a one-step linear correction: start from the nearest perfect square, then nudge your guess using how far off the square of that guess is, scaled by roughly $\tfrac{1}{2 \times \text{guess}}$.

The correction step

If $g$ is a nearby guess with $g^2$ close to $x$, a better estimate is:

$$\sqrt{x} \approx g + \frac{x - g^2}{2g} .$$

In plain English: measure how far $g^2$ is from the target $x$, then convert that gap into a correction to $g$ by dividing it by roughly twice your guess. This comes from the same idea as a first-order Taylor approximation — treating the square-root function as locally linear near a point you already know — and one application is almost always accurate enough for interview purposes.

Worked example 1: √55

Nearest perfect square below 55 is $49 = 7^2$, so start with $g = 7$. The gap is $55 - 49 = 6$. Correction: $\frac{6}{2 \times 7} = \frac{6}{14} \approx 0.429$. So:

$$\sqrt{55} \approx 7 + 0.429 = 7.429 .$$

The true value is $\sqrt{55} \approx 7.416$ — the estimate is accurate to about 0.01, reached in two small mental steps with no calculator. That level of accuracy is far more than most interview questions actually require, but it comes essentially free once the correction formula is automatic.

Worked example 2: √170

Nearest perfect square is $169 = 13^2$, remarkably close, so $g = 13$. Gap: $170 - 169 = 1$. Correction: $\frac{1}{2\times13} = \frac{1}{26} \approx 0.0385$. So:

$$\sqrt{170} \approx 13 + 0.0385 = 13.038 .$$

The true value is $\sqrt{170} \approx 13.038$ — essentially exact, because the starting guess was already so close that the linear approximation barely needed to correct anything. The lesson generalizes: the closer your starting perfect square, the more accurate one correction step is.

What this means in practice

Fast square-root estimation is the backbone of quick standard-deviation and volatility calculations — variance often needs to be converted to a standard deviation on the spot, and this trick makes that instantaneous instead of a stalling point. It's also a direct, hands-on instance of a linear (first-order) approximation, the same underlying idea that shows up later in far more advanced settings like local approximations to option Greeks. Once it's automatic, it removes one of the last places a candidate visibly stalls out mid-calculation.