You have seen your teacher scribble \sqrt{72} = 6\sqrt{2} in one line and move on. That one line is actually three separate moves stitched together, and every one of them follows from a property you already know. This page lets you slow the simplification down, choose which perfect square to pull out, and watch the integer coefficient walk outside the radical sign.

What "simplified" means for a surd

A surd is "simplified" when the radicand — the number under the \sqrt{\,} — has no perfect-square factor other than 1. So \sqrt{2} is simplified, \sqrt{50} is not (it contains a factor of 25), and \sqrt{72} is not (it contains a factor of 36).

The whole trick is the product property from Roots and Radicals:

\sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \qquad \text{for } a, b \geq 0.

Why: this is the power-of-a-product law (ab)^{1/2} = a^{1/2} b^{1/2} written in radical notation. Nothing new to memorise — it is an exponent law in disguise.

Once you know the property, simplification is a search problem: find the largest perfect square k^2 that divides the radicand N, write N = k^2 \cdot m, split the radical, and pull \sqrt{k^2} = k outside.

Pick a factorisation and watch it simplify

The visualiser below lets you choose how to factor 72. Three decompositions are available: 72 = 4 \cdot 18, 72 = 9 \cdot 8, and 72 = 36 \cdot 2. Only the last one finishes the job in a single step, because 36 is the largest perfect-square factor of 72. The other two need a second pass.

Interactive surd simplifier for the square root of seventy-two A three-stop slider where each stop corresponds to a factorisation of seventy-two. The slider positions one, two, and three show seventy-two equals four times eighteen, seventy-two equals nine times eight, and seventy-two equals thirty-six times two. Below the slider, three boxed expressions show the current factored form, the split radical, and the partially simplified result with any leftover perfect-square factor still inside the radical highlighted. 4 · 18 9 · 8 36 · 2 √72 = √(k² · m) = √(k²) · √m = k · √m ↔ drag to pick a factorisation of 72
Three factorisations of $72$. Only $72 = 36 \cdot 2$ uses the *largest* perfect-square factor, which is why it finishes in one step: $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$. The other two factorisations are correct but leave a perfect square still inside the radical, so they need a second round.

The three-move simplification of √72

Here is the fastest route, in detail.

Move 1. Factor the radicand.

72 = 36 \cdot 2

Why: you are looking for the largest perfect-square divisor of 72. Start from the top of the perfect-square list — 64, 49, 36, 25, 16, 9, 4, 1 — and take the first one that divides 72. That is 36, with quotient 2.

Move 2. Split the radical using the product property.

\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2}

Why: the property \sqrt{ab} = \sqrt{a} \sqrt{b} lets you break one radical into a product of two radicals, whenever both factors are non-negative.

Move 3. Evaluate the perfect-square part.

\sqrt{36} = 6, \quad \text{so} \quad \sqrt{72} = 6\sqrt{2}.

That is the finished surd. 6 is an integer, \sqrt{2} cannot be simplified further, and there is no perfect square hiding inside the radical any more.

What the other factorisations do

If you slid to 72 = 4 \cdot 18, the product property gives

\sqrt{72} = \sqrt{4 \cdot 18} = 2\sqrt{18}.

That is correct, but not simplified — 18 = 9 \cdot 2 still has a perfect-square factor. One more pass:

2\sqrt{18} = 2 \cdot \sqrt{9 \cdot 2} = 2 \cdot 3 \sqrt{2} = 6\sqrt{2}.

The answer is the same. The difference is that a smaller choice of perfect square just means more rounds of the same work. Mathematically you cannot go wrong; practically, finding the largest perfect square first saves you from chasing tails.

A quick recognition drill

Simplify $\sqrt{200}$

Find the largest perfect square dividing 200. Work down the perfect-square list: 196 does not divide 200; 144 does not; 100 does, with quotient 2. So

\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}.

Why: 100 is the largest perfect square dividing 200, so \sqrt{100} = 10 walks outside and \sqrt{2} stays put. One pass, done.

The general recipe

For any positive integer N that is not itself a perfect square:

  1. Find the largest perfect-square factor k^2 of N; write N = k^2 \cdot m.
  2. Use \sqrt{ab} = \sqrt{a}\sqrt{b} to split: \sqrt{N} = \sqrt{k^2} \cdot \sqrt{m}.
  3. Replace \sqrt{k^2} by k, leaving k\sqrt{m} as the simplified surd.

If you are ever unsure whether you found the largest perfect square, it does not matter — as the 4 \cdot 18 and 9 \cdot 8 branches show, you can always take whatever perfect square you spotted first, split the radical, and repeat on whatever is left inside. The final simplified form is unique.

Related: Roots and Radicals · Rationalise the Denominator: Watch the Conjugate Flush the Irrational Upstairs · Spot a Conjugate Surd in a JEE Problem and Rationalise on Sight · Spiral of Theodorus: Root-2, Root-3, Root-4, Unfolding Into a Curve · A Square of Side √2 Has Area Exactly 2 — Watch the Tiles Rearrange to Prove It