Estimating Irrational Roots

Trap the root between perfect squares

List the perfect squares you know: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$. To estimate $\sqrt{n}$, find the two neighbors on that list that surround $n$.

For $\sqrt{40}$: the perfect squares around $40$ are $36$ and $49$, so $36 < 40 < 49$. Taking square roots across the whole inequality gives $6 < \sqrt{40} < 7$. The root is between $6$ and $7$ — guaranteed. On the number line below, $\sqrt{40}$ sits inside the shaded band between $6$ and $7$.

The classic mistake is dividing by $2$: half of $40$ is $20$, and $20$ is nowhere near $\sqrt{40}$. Square roots don't come from halving. Trap between perfect squares, always.

Which whole number is it closer to?

Once the root is trapped, compare your number to the two perfect squares. $40$ is $4$ away from $36$ but $9$ away from $49$, so $\sqrt{40}$ sits closer to $6$ than to $7$. Indeed $\sqrt{40} = 6.32\ldots$

For a one-decimal estimate, test a candidate by squaring it: $6.3^2 = 39.69$ and $6.4^2 = 40.96$, so $\sqrt{40}$ is between $6.3$ and $6.4$. You can keep tightening the trap as far as the problem demands.

Cube roots work the same way

To estimate a cube root, trap the number between perfect cubes instead: $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000$.

For $\sqrt[3]{100}$: since $64 < 100 < 125$, taking cube roots gives $4 < \sqrt[3]{100} < 5$. Mixing up the lists is the trap here — using perfect squares would wrongly put the answer near $10$.

Worked examples

Try one yourself

Common questions