Radicals and Square Roots

Understanding Radical Notation

In $\sqrt[n]{x}$:

• $n$ is the index (when omitted, it's 2 for square root)
• $x$ is the radicand (the number under the radical)

Examples:

• $\sqrt{16} = 4$ because $4^2 = 16$
• $\sqrt[3]{27} = 3$ because $3^3 = 27$
• $\sqrt[4]{81} = 3$ because $3^4 = 81$

The Product Property

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

Example 1: Simplify $\sqrt{50}$

Step 1: Find perfect square factors

$\sqrt{50} = \sqrt{25 \cdot 2}$

Step 2: Separate and simplify

$$\sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$$

Example 2: Simplify $\sqrt{72}$

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

Like Radicals

Just like combining like terms in algebra!

$$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$$

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

Unlike Radicals

$\sqrt{2} + \sqrt{3}$ cannot be simplified (different radicands)

Example: Simplify First!

$\sqrt{12} + \sqrt{27} = \sqrt{4 \cdot 3} + \sqrt{9 \cdot 3}$

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

Example 1: Simple Multiplication

$$\sqrt{3} \cdot \sqrt{12} = \sqrt{36} = 6$$

Example 2: With Coefficients

$$2\sqrt{5} \cdot 3\sqrt{10} = 6\sqrt{50} = 6 \cdot 5\sqrt{2} = 30\sqrt{2}$$

Example 3: FOIL with Radicals

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

$= \sqrt{3} \cdot \sqrt{3} - 2\sqrt{3} + 2\sqrt{3} - 4$

$$= 3 - 4 = -1$$

The Quotient Property

$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$

Example

$$\sqrt{\frac{50}{2}} = \sqrt{25} = 5$$

Or: $$\frac{\sqrt{50}}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2}} = 5$$

Method 1: One Term

$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

Method 2: Conjugate

For denominators like $a + \sqrt{b}$, multiply by the conjugate!

$$\frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$$

$$= \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$$