Lesson Objectives
• Demonstrate an understanding of how to simplify a radical
• Learn how to multiply radical expressions
• Learn how to rationalize a binomial denominator
• Learn how to write radical expressions with quotients in lowest terms

How to Perform Operations with Radicals

Now that we have a good understanding of radicals, we can look at some more challenging scenarios. Let's begin by restating the rules for the simplified form of a radical.

• The radicand contains no factor (except 1) that is a:
• Perfect Square » Square Root
• Perfect Cube » Cube Root
• Perfect Fourth » Fourth Root
• So on and so forth...
• The radicand cannot contain fractions
• There is no radical present in any denominator

In some cases, we will have more complex problems that deal with multiplying radicals. Let's look at a few examples.
Example 1: 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 2: 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: $$-6 - 6\sqrt{2}- 18\sqrt{2}- 36=$$ $$(-6 - 36) + (-6\sqrt{2}- 18\sqrt{2})=$$ $$-42 + (-6 - 18)\sqrt{2}=$$ $$-42 - 24\sqrt{2}$$

Rationalizing a Binomial Denominator

In our special products lesson, we learned about conjugates. Conjugates occur when we have the sum and difference of the same two terms. If we multiply conjugates together using FOIL, the O and I step cancel: $$(5x + 1)(5x - 1)$$ First Terms: $$5x \cdot 5x=25x^2$$ Outer Terms: $$5x \cdot -1=-5x$$ Inner Terms: $$1 \cdot 5x=5x$$ Last Terms: $$1 \cdot -1=-1$$ If we combine like terms, the O and I steps will cancel: $$\require{cancel}25x^2 + (-5x) + 5x - 1=$$ $$\require{cancel}25x^2 + \cancel{(-5x)}+ \cancel{5x}- 1=$$ $$25x^2 - 1$$ When we simplify a radical expression with two terms in the denominator and at least one is a radical, we multiply the numerator and denominator by the conjugate of the denominator. Let's look at an example.
Example 3: Simplify each. $$\frac{8}{6 + 2\sqrt{7}}$$ To rationalize the denominator, we will multiply the numerator and denominator by the conjugate of the denominator: $$\frac{8}{6 + 2\sqrt{7}}\cdot \frac{6 - 2\sqrt{7}}{6 - 2\sqrt{7}}$$ Our numerator: $$8(6 - 2\sqrt{7})=$$ $$48 - 16\sqrt{7}$$ Our denominator: $$(6 + 2\sqrt{7})(6 - 2\sqrt{7})$$ If we use FOIL, the O and I steps will cancel. We just need to perform the F and L steps:
First Terms: $$6 \cdot 6=36$$ Last Terms: $$2\sqrt{7}\cdot -2\sqrt{7}=$$ $$(2 \cdot -2) \cdot (\sqrt{7 \cdot 7})$$ $$-4 \cdot 7=-28$$ Combine Like Terms: $$36 - 28=8$$ Our Fraction becomes: $$\frac{48 - 16\sqrt{7}}{8}$$ We can simplify this fraction by factoring an 8 out of the numerator. This will cancel with the 8 in the denominator: $$\frac{8(6 - 2\sqrt{7})}{8}=$$ $$\frac{\cancel{8}(6 - 2\sqrt{7})}{\cancel{8}}=$$ $$6 - 2\sqrt{7}$$ Example 4: Simplify each. $$\frac{4y}{9x^3 - 6y\sqrt{7x}}$$ To rationalize the denominator, we will multiply the numerator and denominator by the conjugate of the denominator: $$\frac{4y}{9x^3 - 6y\sqrt{7x}}\cdot \frac{9x^3 + 6y\sqrt{7x}}{9x^3 + 6y\sqrt{7x}}$$ Our numerator: $$4y(9x^3 + 6y\sqrt{7x})=$$ $$4y \cdot 9x^3 + 4y \cdot 6y\sqrt{7x}=$$ $$36x^3y + 24y^2\sqrt{7x}$$ Our denominator: $$(9x^3 - 6y\sqrt{7x})(9x^3 + 6y\sqrt{7x})$$ If we use FOIL, the O and I steps will cancel. We just need to perform the F and L steps:
First Terms: $$9x^3 \cdot 9x^3=81x^6$$ Last Terms: $$-6y\sqrt{7x}\cdot 6y\sqrt{7x}=$$ $$(-6y \cdot 6y) \cdot \sqrt{7x \cdot 7x}=$$ $$-36y^2 \cdot 7x=-252xy^2$$ Combine Like Terms: $$81x^6 - 252xy^2$$ Our Fraction becomes: $$\frac{36x^3y + 24y^2\sqrt{7x}}{81x^6 - 252xy^2}$$ We can simplify this fraction by factoring a 3 out of the numerator and the denominator. We can then cancel this 3 between the numerator and denominator: $$\frac{3(12x^3y + 8y^2\sqrt{7x})}{3(27x^6 - 84xy^2)}=$$ $$\frac{\cancel{3}(12x^3y + 8y^2\sqrt{7x})}{\cancel{3}(27x^6 - 84xy^2)}=$$ $$\frac{12x^3y + 8y^2\sqrt{7x}}{27x^6 - 84xy^2}$$

Skills Check:

Example #1

Simplify each. $$(-\sqrt{2}+ 3\sqrt{3})(4\sqrt{2}- 2\sqrt{3})$$

A
$$2\sqrt{30}+ 5$$
B
$$-26$$
C
$$-14 + 26\sqrt{6}$$
D
$$\frac{1 + 14\sqrt{6}}{26}$$
E
$$-26 + 14\sqrt{6}$$

Example #2

Simplify each. $$\frac{2}{2x^{3}- 4\sqrt{x^{2}}}$$

A
$$\frac{-3\sqrt{2}}{2x}$$
B
$$-3\sqrt{2}$$
C
$$\frac{1}{2}$$
D
$$\frac{-16 + 20\sqrt{3}}{59}$$
E
$$\frac{1}{x^{3}- 2x}$$

Example #3

Simplify each. $$\frac{4}{-2 - 4\sqrt{5x^{4}}}$$

A
$$\frac{\sqrt{3}- \sqrt{15}}{4}$$
B
$$\frac{5 + 5\sqrt{2}}{4}$$
C
$$\frac{2x\sqrt{3}+ x}{2}$$
D
$$\frac{-2 + 4x^{2}\sqrt{5}}{1 - 20x^{4}}$$
E
$$4\sqrt{3}- 2\sqrt{15}$$

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