Difference between revisions of "Simplifying Radicals"

Latest revision as of 11:18, 21 October 2021

Radicals

A radical is a special kind of number that is the root of a polynomial equation. First, let us look at one specific kind of radical, the "square root". It is a special kind of number related to squaring.

When we have a number, say 2, what positive number will give us, when we square it, the number 2?

1 × 1 = 1, so the number must be higher than 1.
2 × 2 = 4, so the number must be lower than 2.
1.5 × 1.5 = 2.25, so the number must be lower than 1.5.

We could continue like this forever, and with each step get closer and closer to the answers (this kind of process for refining the answer is called iteration). The number we are looking for is approximately 1.41421...

Obviously this is very difficult for us to work with, so we have a special notation. We write for a number a, ${\sqrt {a}}$ to represent the number when squared will give us a back.

Since this is the inverse operation from squaring, it can also be denoted as a^1/2 and
$(a^{1/2})^{2}=a^{1}=a$ .

We can extend this idea to other kinds of radicals. ${\sqrt[{3}]{a}}$ indicates the number $x$ such that
$x^{3}=a$ . For example, ${\sqrt[{3}]{7}}=1.91293...$

Note that it is not possible to find any real number whose square would be negative. Multiplying with a negative number changes the sign of the number being multiplied and two negative signs thus eliminate each other. For example, -7 × -7 = 49 and also 7 × 7 = 49 . Therefore the square root of a negative number is an undefined operation unless imaginary numbers are allowed as answers.

Simplifying Radicals

To simplify a radical, you look for the largest perfect square factor of the number under the radical.

{\sqrt {162}}

Once you find a perfect square factor, you can express the number under the radical as the product of two factors.

{\sqrt {81*2}}

Then you can take the square root of the perfect square factor out of the radical and place it outside.

9{\sqrt {2}}

Once you have factored out all perfect square factors, you have simplified the radical.

Rationalizing the Denominator

"Rationalizing the denominator" is simply taking the roots out of the denominator. This is needed because it is not proper to leave roots in the denominator. To rationalize the denominator you simply multiply the numerator and the denominator by the root that is in the denominator. For example:

${4 \over {\sqrt {5}}}={4{\sqrt {5}} \over ({\sqrt {5}})^{2}}={4{\sqrt {5}} \over 5}$

If you can simplify, do so.

{10 \over {\sqrt {5}}}={10{\sqrt {5}} \over ({\sqrt {5}})^{2}}={10{\sqrt {5}} \over 5}={2{\sqrt {5}}}

{4 \over {\sqrt {8}}}={{4} \over {\sqrt {(4*2)}}}={{4} \over {2}{\sqrt {2}}}={{4{\sqrt {2}}} \over {2{{\sqrt {2}}^{2}}}}={{4{\sqrt {2}}} \over {4}}={\sqrt {2}}

Applying Operations to Radicals

Addition and Subtraction

Adding and subtracting radicals can only be done when the numbers inside the radicals are the same. For instance, consider the following expression:

{{\sqrt {7}}+{\sqrt {28}}}

You can not add those two terms as they are. However, the following equation simplifies the second term, and you get this:

{{\sqrt {7}}+{\sqrt {4*7}}}={{\sqrt {7}}+2{\sqrt {7}}}

These terms can be added (or subtracted, for that matter). Also, coefficients of 1 are usually not written, so:

{{\sqrt {7}}+2{\sqrt {7}}}={1{\sqrt {7}}+2{\sqrt {7}}}={3{\sqrt {7}}}

You can subtract radicals using the same procedure as adding, just instead of adding the coefficients, you subtract them.

The reason we can do this is the distributive property. Proving it is quite simple, and would look like this:

{{\sqrt {7}}+2{\sqrt {7}}}

Remember that coefficients of 1 are not written, so this equation could also be written as:

{1{\sqrt {7}}+2{\sqrt {7}}}

We then extract the term ${\sqrt {7}}$ from the expression:

{{\sqrt {7}}*(1+2)}

We all know that ${1+2}={3}$ , so:

{{\sqrt {7}}*(3)}={3{\sqrt {7}}}

Multiplication and Division

Multiplying and dividing radicals is quite simple. Multiplication requires very little work. When you are given two radicals to multiply, all you do is multiply the numbers inside the radicals and put the product under a radical. Consider the following equation:

{{\sqrt {11}}*{\sqrt {5}}}={\sqrt {55}}

Division is a bit different, however. One way to divide radicals that uses a concept mentioned earlier on this page is to set up the division problem as a fraction. The following equation illustrates this concept using constants (regular numbers):

{{1}\div {2}}={{1} \over {2}}={0.5}

The same concept applies to the following equation:

{{\sqrt {10}}\div {\sqrt {3}}}={{\sqrt {10}} \over {\sqrt {3}}}

You can then rationalize the expression ${{\sqrt {10}} \over {\sqrt {3}}}$ .

Resources

Introduction to Radicals, Lumen Learning: Boundless Algebra
Simplifying Square Roots, Khan Academy
Useful Radical/Root Rules for Simplification, Mathwords.com
Example Problems of Simplifying Radicals, LibreTexts

Licensing

Content obtained from:

Radicals, Wikibooks: Arithmetic under a CC BY-SA license

@@ Line 1: / Line 1: @@
-We will use the following conventions for simplifying expressions involving radicals:
+== Radicals ==
-# Given the expression <math>a^\frac{b}{c}</math>, write this as <math>\sqrt[c]{a^b}</math>
-# No fractions under the radical sign
-# No radicals in the denominator
-# The radicand has no exponentiated factors with exponent greater than or equal to the index of the radical
-Example: Simplify the expression <math>\left(\frac{1}{8}\right)^\frac{1}{2}</math>
+<!--Maybe we should bring part of the Exponents section over here before we get into radicals? Alternately, maybe this isn't the place for discussion of radicals?-->
-Using convention 1, we rewrite the given expression as
+A '''radical''' is a special kind of number that is the root of a polynomial equation. First, let us look at one specific kind of radical, the "square root". It is a special kind of number related to squaring.
-: <math>\left(\frac{1}{8}\right)^\frac{1}{2} = \sqrt[2]{\left(\frac{1}{8}\right)^1} = \sqrt{\frac{1}{8}}</math>
-The expression now violates convention 2. To get rid of the fraction in the radical, apply the rule
-<math>\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}</math> and simplify the result:
-: <math>\sqrt{\frac{1}{8}} = \frac{\sqrt{1}}{\sqrt{8}} = \frac{1}{\sqrt{8}}</math>
-The resulting expression violates convention 3. To get rid of the radical in the denominator, multiply by <math>\frac{\sqrt{8}}{\sqrt{8}}</math>:
-: <math>\frac{1}{\sqrt{8}} = \frac{1}{\sqrt{8}}\cdot\frac{\sqrt{8}}{\sqrt{8}} = \frac{\sqrt{8}}{8}</math>
-Notice that <math>8=2^3</math>. Since the index of the radical is 2, our expression violates convention 4. We can reduce the exponent of the expression under the radical as follows:
-: <math>\frac{\sqrt{8}}{8} = \frac{\sqrt{2^3}}{8} = \frac{\sqrt{2^2\cdot2}}{8} = \frac{2\cdot\sqrt{2}}{8} = \frac{\sqrt{2}}{4}</math>
-==The Conjugate of a Radical Expression==
+When we have a number, say 2, what positive number will give us, when we square it, the number 2?
-The conjugate of the two term expression <math> a + b </math> is <math> a - b </math> (likewise, the conjugate of <math> a - b </math> is <math> a + b </math>). The product of a two term expression and its conjugate, <math> a + b </math> and <math> a - b </math>, is <math> (a+b)(a-b) = a^2 + ba - ba - b^2 = a^2 - b^2 </math>. This property is useful for getting rid of square roots in two term expressions.
+* 1 &times; 1 = 1, so the number must be higher than 1.
+* 2 &times; 2 = 4, so the number must be lower than 2.
+* 1.5 &times; 1.5 = 2.25, so the number must be lower than 1.5.
-For example, consider the limit
+We could continue like this forever, and with each step get closer and closer to the answers (this kind of process for refining the answer is called iteration). The number we are looking for is approximately 1.41421...
-: <math> \lim_{h\to 0} \frac{\sqrt{3+h} - \sqrt{3}}{h}</math>
+Obviously this is very difficult for us to work with, so we have a special notation. We write for a number ''a'', <math>\sqrt{a}</math> to represent the number when squared will give us ''a'' back.
-If we plug in 0 for <math> h </math>, we get the indeterminate form <math> 0/0 </math>. However, using the conjugate of the numerator (<math> \sqrt{3+h} + \sqrt{3} </math>) and multiplying by a clever form of 1 (the conjugate divided by itself), we can rewrite this limit in a way that allows us to evaluate it.
+Since this is the inverse operation from squaring, it can also be denoted as a<sup>1/2</sup> and <br><math>(a^{1/2})^2=a^1=a</math>.
-: <math> \lim_{h\to 0} \frac{\sqrt{3+h} - \sqrt{3}}{h} = \lim_{h\to 0}\frac{(\sqrt{3+h} - \sqrt{3})}{h}\cdot\frac{(\sqrt{3+h} + \sqrt{3})}{(\sqrt{3+h} + \sqrt{3})} </math>
+We can extend this idea to other kinds of radicals. <math>\sqrt[3]{a}</math> indicates the number <math>x</math> such that <br><math>x^3=a</math>. For example, <math>\sqrt[3]{7}=1.91293...</math>
-:::::::<math>  = \lim_{h\to 0} \frac{(\sqrt{3+h})^2 - (\sqrt{3})^2}{h(\sqrt{3+h} + \sqrt{3})} </math>
-:::::::<math>  = \lim_{h\to 0} \frac{((3 + h) - 3}{h(\sqrt{3+h} + \sqrt{3})} </math>
+Note that it is not possible to find any real number whose square would be negative. Multiplying with a negative number changes the sign of the number being multiplied and two negative signs thus eliminate each other. For example, -7 &times; -7 = 49 and also 7 &times; 7 = 49 . Therefore the square root of a negative number is an undefined operation unless imaginary numbers are allowed as answers.
-:::::::<math>  = \lim_{h\to 0} \frac{h}{h(\sqrt{3 + h} + \sqrt{3})} </math>
-:::::::<math>  = \lim_{h\to 0} \frac{1}{\sqrt{3 + h} + \sqrt{3}} </math>
+== Simplifying Radicals ==
-Now, setting <math> h = 0 </math> will not result in an indeterminate form, so we can evaluate the limit.
+To simplify a radical, you look for the largest perfect square factor of the number under the radical.
-:::::::<math>  = \frac{1}{\sqrt{3+0} + \sqrt{3}} = \frac{1}{2\sqrt{3}}</math>
+:<math>{\sqrt{162}}</math>
+Once you find a perfect square factor, you can express the number under the radical as the product of two factors.
+:<math>{\sqrt{81*2}}</math>
+Then you can take the square root of the perfect square factor out of the radical and place it outside.
+:<math>9{\sqrt{2}}</math>
+Once you have factored out all perfect square factors, you have simplified the radical.
+== Rationalizing the Denominator ==
+"Rationalizing the denominator" is simply taking the roots out of the denominator. This is needed because it is not proper to leave roots in the denominator. To rationalize the denominator you simply multiply the numerator and the denominator by the root that is in the denominator. For example:
+<math>{4\over\sqrt{5}}={4\sqrt{5}\over(\sqrt{5})^2}={4\sqrt{5}\over5}</math>
+If you can simplify, do so.
+:<math>{10\over\sqrt{5}}={10\sqrt{5}\over(\sqrt{5})^2}={10\sqrt{5}\over5}={2\sqrt{5}}</math>
+:<math>{4\over\sqrt{8}}={{4}\over{\sqrt{(4*2)}}}={{4}\over{2}{\sqrt{2}}}={{4{\sqrt{2}}}\over{2{{{\sqrt{2}}}^2}}}={{4{\sqrt{2}}}\over{4}}=\sqrt{2}</math>
+==Applying Operations to Radicals==
+===Addition and Subtraction===
+Adding and subtracting radicals can only be done when the numbers inside the radicals are the same. For instance, consider the following expression:
+:<math>{\sqrt{7}+\sqrt{28}}</math>
+You can not add those two terms as they are. However, the following equation simplifies the second term, and you get this:
+:<math>{\sqrt{7}+\sqrt{4*7}}={\sqrt{7}+2\sqrt{7}}</math>
+These terms can be added (or subtracted, for that matter). Also, coefficients of 1 are usually not written, so:
+:<math>{\sqrt{7}+2\sqrt{7}}={1\sqrt{7}+2\sqrt{7}}={3\sqrt{7}}</math>
+You can subtract radicals using the same procedure as adding, just instead of adding the coefficients, you subtract them.
+The reason we can do this is the distributive property. Proving it is quite simple, and would look like this:
+:<math>{\sqrt{7}+2\sqrt{7}}</math>
+Remember that coefficients of 1 are not written, so this equation could also be written as:
+:<math>{1\sqrt{7}+2\sqrt{7}}</math>
+We then extract the term <math>{\sqrt{7}}</math> from the expression:
+:<math>{\sqrt{7}*(1+2)}</math>
+We all know that <math>{1+2}={3}</math>, so:
+:<math>{\sqrt{7}*(3)}={3\sqrt{7}}</math>
+===Multiplication and Division===
+Multiplying and dividing radicals is quite simple. Multiplication requires very little work. When you are given two radicals to multiply, all you do is multiply the numbers inside the radicals and put the product under a radical. Consider the following equation:
+:<math>{\sqrt{11}*\sqrt{5}}={\sqrt{55}}</math>
+Division is a bit different, however. One way to divide radicals that uses a concept mentioned earlier on this page is to set up the division problem as a fraction. The following equation illustrates this concept using constants (regular numbers):
+:<math>{{1}\div{2}}={{1}\over{2}}={0.5}</math>
+The same concept applies to the following equation:
+:<math>{{\sqrt{10}}\div{\sqrt{3}}}={{\sqrt{10}}\over{\sqrt{3}}}</math>
+You can then rationalize the expression <math>{{\sqrt{10}}\over{\sqrt{3}}}</math>.
@@ Line 40: / Line 82: @@
 * [https://www.mathwords.com/r/radical_rules.htm Useful Radical/Root Rules for Simplification], Mathwords.com
 * [https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Advanced_Algebra_(Redden)/05%3A_Radical_Functions_and_Equations/5.02%3A_Simplifying_Radical_Expressions Example Problems of Simplifying Radicals], LibreTexts
+== Licensing ==
+Content obtained from:
+* [https://en.wikibooks.org/wiki/Arithmetic/Radicals Radicals, Wikibooks: Arithmetic] under a CC BY-SA license

Difference between revisions of "Simplifying Radicals"

Latest revision as of 11:18, 21 October 2021

Contents

Radicals

Simplifying Radicals

Rationalizing the Denominator

Applying Operations to Radicals

Addition and Subtraction

Multiplication and Division

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