Simplifying Radicals

A radical like $\sqrt{72}$ can often be written in a simpler, exact form, here $6\sqrt{2}$. Exact radical answers come up in quadratic formulas, trigonometry, and distances, so you’ll need to simplify them and combine them confidently.

Key ideas

Square roots and perfect squares

$\sqrt{a}$ means the positive number that squares to $a$, so $\sqrt{49} = 7$. Numbers like $4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$ are perfect squares, and it helps to know them on sight.

The multiplication property

For $a \ge 0$ and $b \ge 0$:

$$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$

Check with numbers: $\sqrt{4} \times \sqrt{9} = 2 \times 3 = 6$, and $\sqrt{36} = 6$. ✓

Simplest radical form

To simplify $\sqrt{n}$, split $n$ into a perfect square times another number, using the largest perfect square factor you can find:

$$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$$

A radical is in simplest form when the number under the root has no perfect square factor other than $1$.

• $6\sqrt{2}$ is a mixed radical (a number times a root).
• $\sqrt{72}$ is an entire radical (everything under the root).

Adding and subtracting

You can only combine like radicals, which have the same number under the root, just like combining like terms:

$$2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$$

$\sqrt{2} + \sqrt{3}$ can’t be combined. Simplify each radical first, because some like radicals are hiding: $\sqrt{12} = 2\sqrt{3}$.

Multiplying

Multiply the numbers outside together and the numbers inside together, then simplify:

$$2\sqrt{6} \times 5\sqrt{3} = 10\sqrt{18} = 10 \times 3\sqrt{2} = 30\sqrt{2}$$

Remember that $\sqrt{a} \times \sqrt{a} = a$, so $(\sqrt{5})^2 = 5$.

On the SAT

Desmos gives decimals, not simplest radical form, but that’s still useful on the SAT: type sqrt(72) and 6sqrt(2) and both show $8.485\ldots$, so they’re equal. To pick an answer choice, compare its decimal with the decimal of the original expression. Simplifying by hand with the largest perfect-square factor is usually quicker for small numbers, and you’ll need it when the question asks for a form like $a\sqrt{b}$. See using Desmos on the SAT.

Worked examples

Example 1: Simplest radical form

Simplify $\sqrt{48}$ and $\sqrt{50}$.

Solution.

$$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$ $$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$

Example 2: Mixed to entire

Write $3\sqrt{5}$ as an entire radical.

Solution. Write the $3$ as $\sqrt{9}$ and multiply:

$$3\sqrt{5} = \sqrt{9} \times \sqrt{5} = \sqrt{45}$$

Example 3: Adding and subtracting

Simplify $\sqrt{12} + \sqrt{27} - \sqrt{75}$.

Solution. Simplify each radical first:

$$\sqrt{12} = 2\sqrt{3}, \qquad \sqrt{27} = 3\sqrt{3}, \qquad \sqrt{75} = 5\sqrt{3}$$

Now they’re like radicals:

$$2\sqrt{3} + 3\sqrt{3} - 5\sqrt{3} = 0$$

Example 4: Expanding with radicals

Expand and simplify $(3 + \sqrt{2})(4 - \sqrt{2})$ and $(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})$.

Solution. Multiply every term by every term, as with polynomials:

$$\begin{aligned} (3 + \sqrt{2})(4 - \sqrt{2}) &= 12 - 3\sqrt{2} + 4\sqrt{2} - (\sqrt{2})^2 \\ &= 12 + \sqrt{2} - 2 \\ &= 10 + \sqrt{2} \end{aligned}$$

The second is a difference of squares, $(a + b)(a - b) = a^2 - b^2$:

$$(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = 5 - 3 = 2$$

Common mistakes

Splitting a root over addition. $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$. For example, $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 7$. The multiplication property only works for multiplication.

Not simplifying all the way. $\sqrt{72} = 2\sqrt{18}$ is true but not finished, because $18$ still has the factor $9$. Use the largest perfect square factor, or keep going until no perfect square is left.

Adding unlike radicals. $\sqrt{2} + \sqrt{3} \ne \sqrt{5}$. Only like radicals combine.

Mixing up outside and inside numbers. In $2\sqrt{3} \times 4\sqrt{5}$, multiply $2 \times 4$ outside and $3 \times 5$ inside: $8\sqrt{15}$.

Leaving $(\sqrt{2})^2$ unsimplified. $(\sqrt{2})^2 = 2$. It’s a whole number.

Practice

1. (Warm-up) Simplify $\sqrt{20}$, $\sqrt{98}$, and $\sqrt{200}$.

Solution

$$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}, \qquad \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}, \qquad \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}$$

2. (Warm-up) Write $4\sqrt{3}$ as an entire radical.

Solution

$$4\sqrt{3} = \sqrt{16} \times \sqrt{3} = \sqrt{48}$$

3. (Warm-up) Simplify $5\sqrt{7} - 2\sqrt{7} + \sqrt{7}$.

Solution

$$(5 - 2 + 1)\sqrt{7} = 4\sqrt{7}$$

4. (Core) Simplify $\sqrt{18} + \sqrt{50} - \sqrt{8}$.

Solution

$$3\sqrt{2} + 5\sqrt{2} - 2\sqrt{2} = 6\sqrt{2}$$

5. (Core) Simplify $3\sqrt{10} \times 2\sqrt{15}$.

Solution

$$3\sqrt{10} \times 2\sqrt{15} = 6\sqrt{150} = 6\sqrt{25 \times 6} = 6 \times 5\sqrt{6} = 30\sqrt{6}$$

6. (Core) Expand and simplify $\sqrt{3}\,(2\sqrt{6} - \sqrt{3})$.

Solution

$$2\sqrt{18} - (\sqrt{3})^2 = 2 \times 3\sqrt{2} - 3 = 6\sqrt{2} - 3$$

7. (Core) Expand and simplify $(2 + \sqrt{5})(3 - \sqrt{5})$.

Solution

$$\begin{aligned} (2 + \sqrt{5})(3 - \sqrt{5}) &= 6 - 2\sqrt{5} + 3\sqrt{5} - 5 \\ &= 1 + \sqrt{5} \end{aligned}$$

8. (Core) Expand and simplify $(\sqrt{7} - 2)^2$.

Solution

Use $(a - b)^2 = a^2 - 2ab + b^2$:

$$(\sqrt{7})^2 - 2(2)\sqrt{7} + 4 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}$$

9. (Challenge) A rectangle has sides of $\sqrt{12}$ cm and $\sqrt{27}$ cm. Find its exact area, perimeter, and diagonal length.

Solution

Area: $\sqrt{12} \times \sqrt{27} = \sqrt{324} = 18$ cm².

Perimeter: $2(\sqrt{12} + \sqrt{27}) = 2(2\sqrt{3} + 3\sqrt{3}) = 10\sqrt{3}$ cm.

Diagonal, by the Pythagorean theorem: $\sqrt{(\sqrt{12})^2 + (\sqrt{27})^2} = \sqrt{12 + 27} = \sqrt{39}$ cm. ($39 = 3 \times 13$ has no perfect square factor, so this is already simplest form.)

10. (Challenge) A classmate says $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$. Use $a = 9$ and $b = 16$ to show they’re wrong. Can you find any positive values where it is true?

Solution

$\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$.

Squaring both sides of $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ gives $a + b = a + 2\sqrt{ab} + b$, so $\sqrt{ab} = 0$. That only happens when $a = 0$ or $b = 0$. So for two positive numbers, it’s never true.