You meet \sqrt{50} in a quadratic answer and your instinct is to reach for the calculator. But every exam board — Board, JEE, NEET — wants the simplified surd form, not a seven-digit decimal. And the calculator will not give you 5\sqrt{2}; it will give you 7.0710678\ldots, which is usually the wrong answer on the answer key.
The good news: simplifying any surd like \sqrt{50} is a three-step routine that takes maybe ten seconds once you have done it a few times. Here is the systematic method.
The idea behind the method
The one identity powering the whole technique is the product rule for radicals:
valid for any non-negative a and b. This says: a root can be split across a product.
The trick is to split the radicand into a perfect square times something else, because \sqrt{\text{perfect square}} simplifies to an integer, which you can then pull out of the radical. Everything left under the root is what cannot be pulled out.
Why: \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5 \cdot \sqrt{2} = 5\sqrt{2}. The 25 is a perfect square, so its root comes out as the integer 5. The 2 has no perfect-square factor other than 1, so it stays inside. The 50 has been simplified without any calculator.
The three-step method
Step 1: Find the largest perfect-square factor of the radicand.
Look at the number under the root and ask: what is the biggest perfect square (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \ldots) that divides it cleanly?
For 50: is 100 a factor? No, 50 < 100. Is 49? No, 50/49 is not an integer. Is 25? Yes — 50 = 25 \cdot 2. Stop there; 25 is the largest.
Step 2: Split the radical using the product rule.
Step 3: Pull the perfect square out as its root.
Done. \sqrt{50} = 5\sqrt{2}.
Five worked reps
Drill the routine on a range of numbers until the perfect-square factor jumps out at you without effort.
- \sqrt{72}: largest perfect-square factor is 36, since 72 = 36 \cdot 2. So \sqrt{72} = \sqrt{36}\sqrt{2} = 6\sqrt{2}.
- \sqrt{48}: largest perfect-square factor is 16, since 48 = 16 \cdot 3. So \sqrt{48} = 4\sqrt{3}.
- \sqrt{98}: 98 = 49 \cdot 2. So \sqrt{98} = 7\sqrt{2}.
- \sqrt{200}: 200 = 100 \cdot 2. So \sqrt{200} = 10\sqrt{2}.
- \sqrt{75}: 75 = 25 \cdot 3. So \sqrt{75} = 5\sqrt{3}.
Notice that \sqrt{2} and \sqrt{3} keep appearing, and \sqrt{5} often, because these are the small primes whose multiples by perfect squares come up all the time in textbook answers.
Slide the radicand and mentally find the largest perfect-square factor each time. That is the whole skill.
What if the radicand has no perfect-square factor (other than 1)?
Then the surd is already in simplest form. \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{11}, \sqrt{13}, \sqrt{15}, \sqrt{17} — these are all already as simple as they get, because no perfect square divides them (check: 15 = 3 \cdot 5, neither 4 nor 9 divides 15).
A quick test: if the radicand's prime factorisation has every prime appearing exactly once, the surd is already simplified. If any prime appears twice or more, you can pull out a copy.
For 50 = 2 \cdot 5 \cdot 5, the 5 appears twice — so a 5^2 = 25 factor can be pulled out. That is exactly the 5 that came out of the root.
The slow-motion method using prime factorisation
If the "find the largest square factor" step feels guessy, here is the guaranteed method. Write the prime factorisation of the radicand, then pair up primes:
- 50 = 2 \cdot 5 \cdot 5. Pairs: one pair of 5s. Unpaired: 2.
- Each pair comes out as a single copy of that prime: the pair of 5s comes out as a 5. The unpaired 2 stays inside the root.
- Result: 5\sqrt{2}.
Use prime factorisation to simplify $\sqrt{720}$.
Prime factorise: 720 = 2^4 \cdot 3^2 \cdot 5 = 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5.
Pair up: two pairs of 2s, one pair of 3s, one lonely 5.
Each pair comes out as one copy. The two pairs of 2s give 2 \cdot 2 = 4. The pair of 3s gives 3. The unpaired 5 stays inside.
Why pairing works: \sqrt{p \cdot p} = \sqrt{p^2} = p. Every pair of the same prime under the root cancels to that prime, outside the root. Anything unpaired stays trapped inside because you need two copies to form a square.
The prime-factorisation method never fails, even for ugly radicands like \sqrt{2520} where the largest perfect-square factor is not obvious at a glance. Use it as a fallback whenever the direct method stalls.
The mistake you must not make
Do not split the radicand into just any product — only into a product that contains a perfect square.
- \sqrt{50} = \sqrt{10 \cdot 5} = \sqrt{10} \cdot \sqrt{5}? Technically correct, but useless: neither \sqrt{10} nor \sqrt{5} simplifies, so you have just rewritten a single surd as a product of two surds.
- \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}? Correct and useful: \sqrt{25} simplifies to 5, actually reducing the expression.
The point of splitting is to extract a perfect square. If the split does not contain one, it has not simplified anything.
The one-line reflex
See a surd → find its largest perfect-square factor → split, root, pull out. That is the whole routine, and it handles every surd you will meet in Class 9 through JEE. No calculator needed.
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