Operations with Radicals Lesson

Now that we understand how to simplify radicals, we are ready to learn how to add and subtract radicals. First and foremost, we must understand the concept of "like radicals". Like radicals have the same index and the same radicand. The numbers multiplying the radicals can be different.
Like Radicals: $$-12\sqrt{3}, 7\sqrt{3}$$ $$4\sqrt[3]{17}, 5\sqrt[3]{17}$$ $$13\sqrt[5]{22}, -8\sqrt[5]{22}$$ Not Like Radicals: $$5\sqrt{7}, 2\sqrt{3}$$ $$2\sqrt[4]{19}, 9\sqrt[4]{13}$$ $$-7\sqrt[5]{21}, -3\sqrt[3]{15}$$ We can combine "like radicals" using the distributive property. $$5\sqrt{2x}+ 7\sqrt{2x}=(5 + 7)\sqrt{2x}=12\sqrt{2x}$$ Let's look at a few examples.
Example 1: Simplify each. $$4\sqrt{2}+ 3\sqrt{2}$$ Since we have like radicals, we can perform operations with the numbers that are multiplying our radicals. $$4\sqrt{2}+ 3\sqrt{2}=(4 + 3)\sqrt{2}=7\sqrt{2}$$ Example 2: Simplify each. $$2\sqrt{27}- \sqrt{3}$$ In this case, it appears that we do not have like radicals. When this scenario occurs, try to simplify each radical. $$2\sqrt{27}=2 \cdot \sqrt{9}\cdot \sqrt{3}=6\sqrt{3}$$ Now that we have simplified the first radical, we can see that we have like radicals. We will rewrite our problem as: $$6\sqrt{3}- \sqrt{3}=(6 - 1)\sqrt{3}=5\sqrt{3}$$ Example 3: Simplify each. Assume all variables represent positive real numbers.$$-\sqrt[4]{486xy}- 4\sqrt[4]{96xy}$$ Let's first simplify each radical: $$-\sqrt[4]{486xy}=-\sqrt[4]{81}\cdot \sqrt[4]{6xy}=-3\sqrt[4]{6xy}$$ $$4\sqrt[4]{96xy}=4\sqrt[4]{16}\cdot \sqrt[4]{6xy}=8\sqrt[4]{6xy}$$ Now we can rewrite our problem as: $$-3\sqrt[4]{6xy}- 8\sqrt[4]{6xy}$$ $$-3\sqrt[4]{6xy}- 8\sqrt[4]{6xy}=(-3 - 8)\sqrt[4]{6xy}=-11\sqrt[4]{6xy}$$ Example 4: Simplify each. Assume all variables represent positive real numbers.$$3\sqrt{72x^3}- 5x\sqrt{32x}- 3\sqrt{18x^3}$$ Let's first simplify each radical: $$3\sqrt{72x^3}=3 \cdot \sqrt{36x^2}\cdot \sqrt{2x}=18x\sqrt{2x}$$ $$5x\sqrt{32x}=5x \cdot \sqrt{16}\cdot \sqrt{2x}=20x\sqrt{2x}$$ $$3\sqrt{18x^3}=3 \cdot \sqrt{9x^2}\cdot \sqrt{2x}=9x\sqrt{2x}$$ Now we can rewrite our problem as: $$18x\sqrt{2x}- 20x\sqrt{2x}- 9x\sqrt{2x}$$ $$18x\sqrt{2x}- 20x\sqrt{2x}- 9x\sqrt{2x}=(18x - 20x - 9x)\sqrt{2x}=-11x\sqrt{2x}$$

Multiplying Radical Expressions

In some cases, we will have more complex problems that deal with multiplying radicals. Let's look at a few examples.
Example 5: Simplify each. $$2\sqrt{10}\left(5 + 4\sqrt{5}\right)$$ To perform our multiplication, we will use the distributive property: $$2\sqrt{10}\left(5 + 4\sqrt{5}\right)=$$ $$2\sqrt{10}\cdot 5 + 2\sqrt{10}\cdot 4\sqrt{5}=$$ $$(2 \cdot 5)\sqrt{10}+ (2 \cdot 4)\sqrt{10 \cdot 5}=$$ $$10\sqrt{10}+ 8 \cdot \sqrt{25}\cdot \sqrt{2}=$$ $$10\sqrt{10}+ (8 \cdot 5)\sqrt{2}=$$ $$10\sqrt{10}+ 40\sqrt{2}$$ Example 6: Simplify each. $$\left(-\sqrt{2}- 6\right)\left(3\sqrt{2}+ 6\right)$$ Since we have two terms multiplied by two terms, we can use FOIL
First Terms: $$-\sqrt{2}\cdot 3\sqrt{2}=$$ $$(-1 \cdot 3) \sqrt{2 \cdot 2}=$$ $$-3 \cdot 2=-6$$ Outer Terms: $$-\sqrt{2}\cdot 6=$$ $$(-1 \cdot 6)\sqrt{2}=$$ $$-6\sqrt{2}$$ Inner Terms: $$-6 \cdot 3\sqrt{2}=$$ $$(-6 \cdot 3)\sqrt{2}=$$ $$-18\sqrt{2}$$ Last Terms: $$-6 \cdot 6=-36$$ Combine Like Terms: $$-6 - 6\sqrt{2}- 18\sqrt{2}- 36=$$ $$(-6 - 36) + (-6\sqrt{2}- 18\sqrt{2})=$$ $$-42 + (-6 - 18)\sqrt{2}=$$ $$-42 - 24\sqrt{2}$$

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