When a question ends with \sqrt{72} on the answer sheet, the fast student does not pause. They write 6\sqrt{2} and move on. The slow student stares, maybe reaches for the calculator, maybe writes "approx 8.49" and loses a mark. The difference is not intelligence — it is a single well-drilled habit.
The habit is: see any radical, mentally split the radicand into perfect square × remainder, pull the square root of the perfect square out, and leave the remainder inside. Done.
The pattern you want in your muscle memory
For every \sqrt{n} that is not already simplified, rewrite it as
where k^2 is the largest perfect-square factor of n, and m is what is left. The three steps — find k^2, split, pull out — take less than five seconds once the pattern is in place.
Examples to warm up on:
- \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}
- \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}
- \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}
- \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}
- \sqrt{500} = \sqrt{100 \cdot 5} = 10\sqrt{5}
- \sqrt{162} = \sqrt{81 \cdot 2} = 9\sqrt{2}
Notice how the same few "remainders" (\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}) keep appearing. Most textbook answers leak one of these small irrationals at the end.
Why this always works: the product rule \sqrt{ab} = \sqrt{a}\cdot\sqrt{b} lets you split any radical across a product. If one factor is a perfect square k^2, then \sqrt{k^2} = k (an integer), which can be lifted out of the radical sign. Anything that is not a perfect-square factor must stay inside.
The list to memorise (if you must memorise one thing)
Keep the first few perfect squares in your head, because you will be scanning for them:
When you meet a radicand, scan this list from largest downward and ask: "does this divide my number cleanly?" The first hit is the largest perfect-square factor.
Three ways to find the perfect-square factor
The same answer, reached by different methods — use whichever fits the radicand's size.
Method 1: scan perfect squares top-down. Works for small radicands. For \sqrt{72}: is 64 a factor? 72/64 is not an integer — no. Is 49? No. Is 36? Yes, 72 = 36 \cdot 2. Stop. Answer: 6\sqrt{2}.
Method 2: prime factorisation, pair up primes. Works always. For \sqrt{72}: prime factorise, 72 = 2^3 \cdot 3^2 = 2\cdot 2\cdot 2\cdot 3\cdot 3. Pair the primes: one pair of 2s, one pair of 3s, one leftover 2. Each pair gives one copy of the prime outside the root: 2 \cdot 3 = 6 outside, leftover 2 inside. Answer: 6\sqrt{2}.
Method 3: pull out squares progressively. Divide out 4s until you cannot, then 9s, then 25s. For \sqrt{72}: 72 / 4 = 18, so \sqrt{72} = 2\sqrt{18}. Now simplify \sqrt{18}: 18 / 9 = 2, so \sqrt{18} = 3\sqrt{2}. Combine: \sqrt{72} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}.
All three methods give the same 6\sqrt{2}. Method 1 is fastest for quick recognition, method 2 never fails, method 3 is useful when you do not spot a big square right away.
Slide the radicand and practise. The habit strengthens with repetition — after twenty reps, the perfect-square factor jumps off the page.
The factoring habit in action
Here is what "automatic" looks like in a quadratic-formula answer.
Problem: solve x^2 - 4x - 5 = 0.
\sqrt{36} = 6 (this is a perfect square on its own, no splitting). So x = (4 \pm 6)/2, giving x = 5 or x = -1.
Problem: solve x^2 - 6x + 3 = 0.
Simplify \sqrt{24}: perfect-square factor is 4, since 24 = 4 \cdot 6. So \sqrt{24} = 2\sqrt{6}, and
The final answer is clean. Without the split, \sqrt{24} would just sit there, the fraction would not reduce, and the answer would look messy. The simplification habit is what produces the neat final line every Board examiner expects.
Simplify $\sqrt{288}$.
Scan for a perfect-square factor. 288 is large; try 144: 288 / 144 = 2. Yes. So 288 = 144 \cdot 2.
Cross-check by prime factorisation. 288 = 2^5 \cdot 3^2 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3. Pairs: two pairs of 2s, one pair of 3s, one leftover 2. Outside: 2 \cdot 2 \cdot 3 = 12. Inside: 2. Answer: 12\sqrt{2}. Matches.
Why the largest perfect square matters: if you split with a smaller perfect square — say 288 = 4 \cdot 72 — you get \sqrt{288} = 2\sqrt{72}, which is not yet simplified because \sqrt{72} simplifies further to 6\sqrt{2}. You have to keep going. Going for the largest perfect square finishes the job in one step.
The common mistake: splitting into a non-square
Do not split into two arbitrary factors and think you have simplified.
- \sqrt{72} = \sqrt{8 \cdot 9} = \sqrt{8} \cdot \sqrt{9} = 3\sqrt{8}. Correct so far, but \sqrt{8} = 2\sqrt{2} is not simplified yet. So the split was not enough.
- \sqrt{72} = \sqrt{12 \cdot 6}? Technically correct but useless — neither 12 nor 6 is a perfect square.
- \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}. This is the right move: 36 is a perfect square, 2 has no perfect-square factor other than 1.
Always aim for the largest perfect-square factor — it finishes the simplification in one step.
When the radicand has variables
The same split works with algebraic expressions. For \sqrt{50 x^3}, treat 50 = 25 \cdot 2 and x^3 = x^2 \cdot x:
assuming x \geq 0. Powers of variables get split into pairs the same way primes do — x^3 = x^2 \cdot x, with the x^2 coming out as |x| (or just x under the non-negativity assumption). For even powers, the whole thing comes out: \sqrt{x^4} = x^2. For odd powers, pair all you can: \sqrt{x^{11}} = \sqrt{x^{10} \cdot x} = x^5 \sqrt{x}.
The one-sentence reflex
See a radical → scan the radicand for the largest perfect-square factor → write \sqrt{k^2 m} = k\sqrt{m} and you are done. Keep the first fifteen perfect squares in your head, and every surd in Class 9 through JEE simplifies itself in under ten seconds.
Related: Roots and Radicals · How Do I Simplify Root-50 Without a Calculator? The Systematic Method · Surd Simplifier: Watch Root-72 Split Into 6 Root-2 Step by Step · Why √a · √b = √(ab) But √a + √b ≠ √(a+b)