# How to simplify square roots and radicals

Canonical: https://duckyhelper.com/learn/algebra-1/simplifying-radicals/
Updated: 2026-10-01

To simplify a square root, find the largest perfect square that divides the number under the root, split it off, and take its square root. For example, \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\). You can only add or subtract radicals with the same number under the root, like \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\). A simplified answer has no square root left in a denominator.

## The key idea

A square root of a product splits into a product of square roots. If one of the pieces is a perfect square, its root is a whole number and can leave the radical sign:

$$
\sqrt{ab} = \sqrt{a}\cdot\sqrt{b} \qquad\text{so}\qquad \sqrt{72} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}
$$

It helps to know the perfect squares by sight: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. A radical is fully simplified when no perfect square (other than 1) divides the number under the root, and there is no root in a denominator.

## Worked examples

**Example 1: pull out the largest perfect square**

Problem: Simplify \(\sqrt{72}\).

1. The largest perfect square that divides 72 is 36.

   $$
   \sqrt{72} = \sqrt{36 \cdot 2}
   $$
2. Split the root and take \(\sqrt{36} = 6\).

   $$
   \sqrt{36 \cdot 2} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}
   $$

Answer: \(6\sqrt{2}\)

**Example 2: a radical with a variable**

Problem: Simplify \(\sqrt{50x^3}\), where \(x \ge 0\).

1. Split into the perfect-square part and the rest. \(50 = 25 \cdot 2\) and \(x^3 = x^2 \cdot x\).

   $$
   \sqrt{50x^3} = \sqrt{25x^2 \cdot 2x}
   $$
2. \(\sqrt{25x^2} = 5x\), because \((5x)^2 = 25x^2\). The \(2x\) stays inside.

   $$
   \sqrt{25x^2 \cdot 2x} = 5x\,\sqrt{2x}
   $$

Answer: \(5x\,\sqrt{2x}\)

**Example 3: add and subtract radicals**

Problem: Simplify \(3\sqrt{12} + \sqrt{27} - \sqrt{75}\).

1. These look unlike, so simplify each one first. \(12 = 4 \cdot 3\), \(27 = 9 \cdot 3\), \(75 = 25 \cdot 3\).

   $$
   3\sqrt{12} + \sqrt{27} - \sqrt{75} = 3 \cdot 2\sqrt{3} + 3\sqrt{3} - 5\sqrt{3}
   $$
2. Now all three are multiples of \(\sqrt{3}\). Combine them like \(6y + 3y - 5y\).

   $$
   6\sqrt{3} + 3\sqrt{3} - 5\sqrt{3} = 4\sqrt{3}
   $$

Answer: \(4\sqrt{3}\)

**Example 4 (test-hard): rationalize, then combine**

Problem: Simplify \(\frac{10}{\sqrt{5}} + \sqrt{20}\).

1. Rationalize: multiply the top and the bottom by \(\sqrt{5}\). That is multiplying by 1, so the value does not change, and \(\sqrt{5} \cdot \sqrt{5} = 5\).

   $$
   \frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}
   $$
2. Simplify the second radical.

   $$
   \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}
   $$
3. Now they match, so add.

   $$
   2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5}
   $$

Answer: \(4\sqrt{5}\)

## Common mistakes

- **Splitting a root over a sum.** \(\sqrt{9 + 16} = \sqrt{25} = 5\), not \(3 + 4 = 7\). Splitting only works for multiplication.
- **Stopping too early.** \(\sqrt{72} = 2\sqrt{18}\) is true but not finished, since 18 still contains 9. Fix: use the largest perfect square, or keep going.
- **Adding unlike radicals.** \(\sqrt{2} + \sqrt{3}\) is not \(\sqrt{5}\). Only radicals with the same number inside combine.
- **Losing the outside number.** In \(3\sqrt{12}\), the 2 that comes out multiplies the 3: \(3 \cdot 2\sqrt{3} = 6\sqrt{3}\).
- **Leaving a root in the denominator.** Most teachers want \(\frac{10}{\sqrt{5}}\) written as \(2\sqrt{5}\).

## Quick methods

> **Tip: Use a factor tree when you can't spot the square**
>
> Break the number into primes and circle pairs. Each pair leaves the root as one number. For 72: \(2 \cdot 2 \cdot 2 \cdot 3 \cdot 3\). The pairs \(2 \cdot 2\) and \(3 \cdot 3\) come out as 2 and 3, which multiply to 6, and one 2 stays inside. So \(\sqrt{72} = 6\sqrt{2}\).

> **Note: Check with a calculator**
>
> Type both the original and your answer as decimals. \(\sqrt{72} \approx 8.485\) and \(6\sqrt{2} \approx 8.485\). If they match, your simplification is right. Your teacher still wants the exact form, not the decimal.

## Practice

**5 practice questions**

1. Simplify \(\sqrt{48}\).
   A. \(4\sqrt{3}\)
   B. \(3\sqrt{4}\)
   C. \(6\sqrt{2}\)
   D. \(16\sqrt{3}\)

   Answer: \(4\sqrt{3}\). \(48 = 16 \cdot 3\), and \(\sqrt{16} = 4\). The choice \(16\sqrt{3}\) forgets to take the square root of 16. \(6\sqrt{2}\) is \(\sqrt{72}\), and \(3\sqrt{4}\) is just 6.

2. Simplify \(2\sqrt{18} + \sqrt{8}\).
   A. \(8\sqrt{2}\)
   B. \(2\sqrt{26}\)
   C. \(3\sqrt{26}\)
   D. \(10\sqrt{2}\)

   Answer: \(8\sqrt{2}\). \(\sqrt{18} = 3\sqrt{2}\), so \(2\sqrt{18} = 6\sqrt{2}\). \(\sqrt{8} = 2\sqrt{2}\). Together that is \(8\sqrt{2}\). The choices with 26 add the numbers under the roots, which is not allowed.

3. Simplify \((3\sqrt{2})(4\sqrt{6})\).
   A. \(24\sqrt{3}\)
   B. \(12\sqrt{3}\)
   C. \(7\sqrt{12}\)
   D. \(12\sqrt{8}\)

   Answer: \(24\sqrt{3}\). Multiply outside by outside and inside by inside: \(12\sqrt{12}\). Then \(\sqrt{12} = 2\sqrt{3}\), so the result is \(24\sqrt{3}\). The choice \(12\sqrt{3}\) forgets the 2 that came out of \(\sqrt{12}\).

4. Rationalize the denominator: \(\frac{8}{\sqrt{2}}\).
   A. \(4\sqrt{2}\)
   B. \(8\sqrt{2}\)
   C. \(4\)
   D. \(2\sqrt{2}\)

   Answer: \(4\sqrt{2}\). Multiply top and bottom by \(\sqrt{2}\): \(\frac{8\sqrt{2}}{2} = 4\sqrt{2}\). The choice \(8\sqrt{2}\) multiplies the top but forgets that the bottom became 2.

5. What is \(\sqrt{81 + 144}\)?
   A. \(15\)
   B. \(21\)
   C. \(12.5\)
   D. \(225\)

   Answer: \(15\). Add first: \(81 + 144 = 225\), and \(\sqrt{225} = 15\). The trap 21 is \(9 + 12\), which splits the root over a sum. 225 forgets to take the root.

## Frequently asked questions

### What makes a square root fully simplified?

Three things. No perfect square other than 1 divides the number under the root. There is no fraction under the root. And there is no root in the denominator. So \(6\sqrt{2}\) is simplified, but \(2\sqrt{18}\) and \(\frac{10}{\sqrt{5}}\) are not.

### Can you add square roots?

Only like ones, with the same number under the root. Treat \(\sqrt{3}\) like a variable: \(6\sqrt{3} + 3\sqrt{3} = 9\sqrt{3}\). If they look different, simplify each first, because \(\sqrt{12}\) and \(\sqrt{27}\) both turn into multiples of \(\sqrt{3}\).

### Why do we rationalize the denominator?

It is a convention that makes answers easy to compare and combine. \(\frac{10}{\sqrt{5}}\) and \(2\sqrt{5}\) are the same number, but the second form can be added to other multiples of \(\sqrt{5}\) right away, as in Example 4.

### How do I simplify a square root with variables?

Split each variable power into an even part and what is left. Even powers come out at half the exponent: \(\sqrt{x^2} = x\) and \(\sqrt{x^6} = x^3\) when x is not negative. So \(\sqrt{x^5} = \sqrt{x^4 \cdot x} = x^2\sqrt{x}\).

## Related

- [Exponent rules and how to use them](https://duckyhelper.com/learn/algebra-1/exponent-rules/)
- [How to solve quadratic equations](https://duckyhelper.com/learn/algebra-1/solving-quadratics/)
- [The Pythagorean theorem](https://duckyhelper.com/learn/geometry/pythagorean-theorem/)
- [Algebra 1 study guides](https://duckyhelper.com/learn/algebra-1/)

## Try asking Ducky

- "What's the biggest perfect square inside 180?"
- "Why can't I just add the numbers under the square roots?"
- "Walk me through rationalizing this denominator."

## Get DuckyHelper

Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. [Try it free in your browser](https://app.duckyhelper.com/?utm_source=duckyhelper.com&utm_medium=learn) or [Get the Mac app](https://duckyhelper.com/download/)