Operations with Radical Expressions

Operations with Radicals

By the end of this lesson, you will be able to add and subtract like radicals, multiply and divide them, rationalize a denominator, and use a conjugate when the denominator has two terms.

Adding: They Have to Match

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

Like radicals combine, unlike ones do not. The radical part has to be identical, and then you add the numbers in front and leave the radical alone.

$$3\sqrt{2} + 5\sqrt{3} \quad \text{does not combine}$$

This is combining like terms, exactly as it was in the first unit. $\sqrt{2}$ is behaving like $x$ and $\sqrt{3}$ like $y$: same thing adds, different things stay apart.

Simplify FIRST, because unlike radicals often become like ones.

$$\sqrt{8} + \sqrt{18}$$

Those look unrelated. Simplify each:

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

They matched all along. Never conclude "these do not combine" before simplifying both, and that single habit accounts for most of the errors in addition.

Sort It: Do They Combine

Simplify each one in your head first.

Example 1: Adding After Simplifying

Here is every step on one expression, worked all the way through. $$\sqrt{12} + \sqrt{27}$$

• Simplify the square root of 12 2\sqrt{3}
• Simplify the square root of 27 3\sqrt{3}
• Both are root 3 now. What do the outside numbers add to? 5
• Write the whole answer 5\sqrt{3}

Multiplying: Straight Across

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

Multiplication has no matching requirement at all, which makes it easier than addition rather than harder.

$$\sqrt{3} \cdot \sqrt{5} = \sqrt{15}$$

$$2\sqrt{3} \cdot 4\sqrt{5} = 8\sqrt{15}$$

Numbers with numbers, radicals with radicals.

Simplify afterwards, because the product often contains a square that neither factor had:

$$\sqrt{6} \cdot \sqrt{8} = \sqrt{48} = 4\sqrt{3}$$

And a radical times itself removes it entirely.

$$\sqrt{7} \cdot \sqrt{7} = \sqrt{49} = 7$$

That fact is the engine of everything in the rest of this lesson: it is how a radical gets out of a denominator.

Worked Example 1: Multiplying and Simplifying

$$3\sqrt{6} \cdot 2\sqrt{10}$$

Rationalizing a Denominator

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

Conventionally, a radical does not stay in a denominator. Multiply the top and bottom by that radical:

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

<!-- SLATE_IF_STRONG: math.advanced-math.radical-expressions --> <!-- END_SLATE_IF_STRONG -->