A question pops up in class IX or X: between which two integers does \sqrt{50} lie? You don't need a calculator. You don't even need decimals. What you need is the squares table up to \sim 20^2 burned into your memory, and one small trick: find consecutive perfect squares k^2 and (k+1)^2 that sandwich N, and \sqrt{N} will sandwich between k and k+1.

That is the entire technique. Let's make it automatic.

The rule

If N is a positive integer that is not a perfect square, and k^2 \leq N < (k+1)^2 for some non-negative integer k, then

k < \sqrt{N} < k + 1.

Why: the square-root function is strictly increasing on non-negative numbers. So a < b \Rightarrow \sqrt{a} < \sqrt{b}. From k^2 < N < (k+1)^2 you take square roots of all three to get k < \sqrt{N} < k + 1. The monotonicity is doing all the work.

The squares you need to know cold

Perfect squares up to twentyA table of integers one through twenty with their squares beneath each. One squared is one. Two squared is four. Three squared is nine. And so on up to twenty squared is four hundred. The integers are written in a row and their squares in the row below, so the reader can scan and find the two consecutive squares that bracket a given number. k k² 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 memorise this table to instantly sandwich √N for any N up to 400
The squares $1^2$ through $20^2$. For any $N$ up to $400$, scan this table for the two consecutive squares that bracket $N$. The integer parts of those squares bracket $\sqrt{N}$.

If this table is not in your head, stop and memorise it before continuing. It is a five-minute investment that pays off in every exam question involving roots.

Walking through it: √50

Between which two integers does $\sqrt{50}$ lie?

Step 1. Scan the squares for the two that sandwich 50.

7^2 = 49, \qquad 8^2 = 64.

50 sits between 49 and 64.

Step 2. Take square roots of all three.

7 < \sqrt{50} < 8.

Answer: \sqrt{50} lies between 7 and 8. (Much closer to 7, actually, since 50 is very close to 49. \sqrt{50} \approx 7.07.)

Three lines. No calculator. Ten seconds if the squares are in your head.

More examples

Three more in quick succession

(a) \sqrt{200}. Sandwich: 14^2 = 196 and 15^2 = 225. Since 196 < 200 < 225: \quad 14 < \sqrt{200} < 15.

(b) \sqrt{300}. Sandwich: 17^2 = 289 and 18^2 = 324. Since 289 < 300 < 324: \quad 17 < \sqrt{300} < 18.

(c) \sqrt{2}. Sandwich: 1^2 = 1 and 2^2 = 4. Since 1 < 2 < 4: \quad 1 < \sqrt{2} < 2.

Each took one line once the squares are memorised.

Why no calculator is needed

The question "between which two integers does \sqrt{N} lie?" is asking for the integer part (also called the floor) of \sqrt{N}:

\lfloor \sqrt{N} \rfloor = k \quad \Leftrightarrow \quad k^2 \leq N < (k+1)^2.

That is exactly the sandwich condition. The answer is an integer pair, not a decimal. Calculators give you decimals; you want the pair of integers around them. The squares table skips the decimal detour.

When N is bigger than 400

You either push your squares table further (memorise up to 25^2 = 625 and 30^2 = 900 for JEE), or you decompose.

Decomposition trick for large N. If N is a multiple of a perfect square, pull the square out. For N = 800: 800 = 400 \cdot 2, so \sqrt{800} = 20\sqrt{2}. You know \sqrt{2} \approx 1.414, so \sqrt{800} \approx 20 \cdot 1.414 \approx 28.28. Sandwich: between 28 and 29. (Or with integers only: 28^2 = 784 and 29^2 = 841, and 784 < 800 < 841. ✓)

For N = 2500: that's 50^2 exactly, so \sqrt{2500} = 50.

For N = 10000: 100^2, so \sqrt{10000} = 100.

A sharper version: which integer is √N closer to?

Sometimes the exam asks not just "between which integers" but "closest integer to \sqrt{N}." Quick refinement: once you have k^2 \leq N < (k+1)^2, check where N sits relative to the midpoint.

The midpoint is k^2 + k + \tfrac{1}{2}, because (k+1)^2 = k^2 + 2k + 1, so the midpoint of [k^2, (k+1)^2] is at k^2 + k + 0.5.

For \sqrt{50}: k = 7, k^2 = 49, midpoint is 49 + 7 + 0.5 = 56.5. Since 50 < 56.5, \sqrt{50} is closer to 7 than to 8. (Indeed \sqrt{50} \approx 7.07.)

This works because \sqrt{N} is closer to k than to k+1 when N is on the lower half of [k^2, (k+1)^2]. It's not exact (the square root is slightly concave so it bends the midpoint a little) but it's right for the vast majority of N and close enough for "closest integer."

One warning about negatives

The rule gives k < \sqrt{N} < k+1 only for N \geq 0. There is no real \sqrt{N} for negative N — and trying to sandwich it with integers is meaningless. If a problem gives you \sqrt{-7}, that's a complex-numbers question, not a sandwich-with-squares question.

Also: the rule gives the principal (non-negative) square root. Some problems ask "the solutions to x^2 = 50," which are \pm\sqrt{50}. The sandwich gives you that \sqrt{50} is between 7 and 8, so the solutions are near \pm 7.07. Just remember the \pm if the problem is an equation.

The reflex

Read "between which two integers does \sqrt{N} lie" and your first mental motion should be:

"Scan squares table for the two consecutive squares that bracket N."

The scanning takes one second if the table is memorised. If you find yourself reaching for a calculator, stop — you are solving the wrong kind of problem. This is a recognition skill, and it is trained by drilling the squares table, not by computing.

Related: Number Systems · Constructing √2 on the Number Line · Root of a Non-Perfect-Square Integer: Default to Irrational · Is √4 Irrational Because It's Under a Root Sign?