Radical Expressions

What is a radical expression?

Before we can begin to learn trigonometry, it’s important that we have a firm understanding of radical expressions. This is a review from algebra.

A radical expression denotes the square root of a given number or term, and is expressed by a symbol that looks similar to a check mark with an extended tail at the top.

Anytime you see a radical sign, you are dealing with a radical expression.

For example, √4 = 2. You can prove this by squaring the number. 2 x 2 = 4. Therefore, the expression √4 = 2 is true.

How can radical expressions be manipulated?

There are several different ways you will see radical expressions used in trigonometry.

Binomials

A binomial is simply two terms being added or subtracted. Ex: (4+2) or (1000-3452).

Binomials come up frequently within trigonometry, so it’s important to get comfortable with them.

(\sqrt{45}-\sqrt{28})(\sqrt{45}+\sqrt{28})

To simplify this expression, you must FOIL (First, Outer, Inner, Last) the binomials.

1. Multiply the first terms in both expressions together: √45 x √45 = 45
2. Multiply the first term in the first expression with the outer term in the second expression: √45 x √28 = √1260
3. Multiply the two inner terms (pay attention to the signs): -√28 x √45 = -√1260
4. Multiply the two last terms: -√28 x √28 = -28

When you put all of these terms together, you get:

45+\sqrt{1260}-\sqrt{1260}-28

The two middle terms cancel out, and you are left with:

45-28=17

17 is the answer to this expression.

Fractions

Another way you will see radical expressions is used within fractions.

\sqrt{\frac{3}{16}}

Another way to write the expression above is:

\frac{\sqrt{3}}{\sqrt{16}}

Now we may not know what the square root of 3 is, but we do know that 16 has a perfect square of 4, and thus we can now write the radical expression in it’s final form as:

\frac{\sqrt{3}}{4}

Pulling a Perfect Square out of a Radical

11\sqrt{2} + \sqrt{18}
\\ = 11\sqrt{2}+\sqrt{9}\cdot\sqrt{2}
\\=11\sqrt{2}+3\sqrt{2}
\\=14\sqrt{2}

Here we can see that we can break apart numbers under the radical. Since two square roots multiplied together is just the product of those two numbers under a single radical sign, we can also reverse this process to pull a perfect square out.

Because the two terms have the same number under the radical, we can then add the 11 and 3 to end up with 14 square root 2.

Squared Binomial Radical Expressions

(\sqrt{3}-\sqrt{7})^2 = (\sqrt{3}-\sqrt{7})(\sqrt{3}-\sqrt{7})

When you see that a binomial is squared, you simply expand the expression by multiplying it by itself.

Make sure you distribute each term in the parentheses to each term in the next.

(\sqrt{3}-\sqrt{7})^2 = 3 - \sqrt{21}-\sqrt{21}+7
\\=3-2\sqrt{21}+7
\\=10-2\sqrt{21}

Rationalizing Denominators

\frac{12}{\sqrt{18}}=\frac{12}{\sqrt{9}\cdot\sqrt{2}}=\frac{12}{3\sqrt{2}}

Now wait, this is not your final answer. You must rationalize the denominator – do not leave any radical signs in your denominator! Rationalize by multiplying the numerator and the denominator by the radical term.

\frac{12}{3\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{12\sqrt{2}}{3\cdot2}=\frac{12\sqrt{2}}{6}=2\sqrt{2}

Rationalizing a Denominator with a Binomial

\frac{9}{(6+\sqrt{2})}

Multiply the numerator and the denominator by the binomial. Distribute the 9 to the binomial, and FOIL the denominator with the binomial.

\frac{9}{(6+\sqrt{2})}\cdot\frac{(6-\sqrt{2})}{(6-\sqrt{2})}=\frac{54-9\sqrt{2}}{36-2}=\frac{54-9\sqrt{2}}{34}

Boom. Rationalized.

More Complex Radical Expressions

\sqrt{\frac{5-\frac{\sqrt{7}}{2}}{2}}

Let’s break it down.

\sqrt{\frac{5-\frac{\sqrt{7}}{2}}{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}

\\=\sqrt{\frac{2(5-\frac{\sqrt{7}}{2})}{2\cdot2}}

Remember that the 2 is under a radical sign, so to rationalize, we will multiply the numerator and the denominator by square root 2. Because both the numerator and denominator are still under the radical sign, it reads like “the square root of 2 times the binomial 5 minus the square root of 7 divided by 2 — divided by the square root of 2 times the square root of 2”.

=\sqrt{\frac{10-\sqrt{7}}{4}}=\frac{\sqrt{10-\sqrt{7}}}{\sqrt{4}}=\frac{\sqrt{10-\sqrt{7}}}{2}

Distribute the two to the binomial.

*Note: The 2’s cancel with the fraction leaving you with just square root 7.

You end up with the square root of 10 minus square root 7 — divided by square root 4 — you can then simplify the denominator and leave the radicals in the numerator.

Conclusion

If you understand how these different examples work, then you are ready to move on to some real trig! The last example is more extreme than most problems you will see in your trig class, but it is good to be able to understand the processes happening to prove a greater understanding of the operations taking place.

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