3.1: Radical Expressions Encountered in Calculus
- Refresher Topics
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Simplifying Expressions
- Simplifying Radical Expression
- Rationalizing Denominators and Numerators
- Adding, Subtracting, and Multiplying Radical Expressions
- Dividing Radical Expressions
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Simplifying Expressions
In this section, you will:
- Simplify expressions of the form \( \sqrt[n]{a^n} \).
- Rationalize denominators and numerators.
- Compute the Difference Quotient of radical functions.
We dealt with inverses of polynomial functions (over restricted domains) at the end of the previous chapter. At that time, we discovered that such inverses are often radical functions . Other than the square root and cubed root functions, however, radical functions (as a class of functions) are not very common in Calculus. On the other hand, manipulating functions and expressions involving radicals (and, alternatively, expressions involving rational exponents) occurs so frequently in Calculus that we would benefit from improving on our base skills.
Simplification of radical expressions is a skill thoroughly covered in Elementary and Intermediate Algebra courses. Moreover, the content in this course is too dense to include a full review of those skills. As such, we must assume you have mastered the following topics:
- Understand the definitions of a square root (\( \sqrt{ \, \, } \)) and an \( n^{\text{th}} \) root (\( \sqrt[n]{\, \,} \)).
- Compute the square root and the \( n^{\text{th}} \) root of a number.
- Simplify the square root and the \( n^{\text{th}} \) root of an algebraic expression.
- Know when a square root and an \( n^{\text{th}} \) root are not real numbers.
- Approximate the value of the square root and the \( n^{\text{th}} \) root of a number (to the nearest integer) without technology.
- Understand what it means for a radical expression to be simplified.
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Understand and use the Product and Quotient Properties of Radicals.
If \(\sqrt[n]{a}\) and \(\sqrt[n]{b}\) are real numbers, and \(n\geq 2\) is a natural number, then\[\sqrt[n]{a b}=\sqrt[n]{a} \cdot \sqrt[n]{b} \quad \text { and } \quad \sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a b}. \nonumber \]Moreover, if \( b \neq 0 \), then\[\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} \text { and } \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}. \nonumber \] - Understand the concept of "like" radical expressions.
- Add and subtract radical expressions.
- Multiply radical expressions.
- Divide radical expressions.
If you require a review of any of these topics, it's best to refer to the links in the Algebra Refresher section above.
Simplify Variable Expressions of the Form \( \sqrt[n]{a^n} \)
With all the topics we must assume have been mastered, it seems like there is nothing left to cover; however, this cannot be further from the truth. There are many subtleties in Mathematics when dealing with radical expressions and, unfortunately, almost all of these occur in Calculus. The first such subtlety involves the \( n^{\text{th}} \) root of an expression raised to the \( n^{\text{th}} \) power. The following theorem reminds us of the complication.
For any natural numer \(n \geq 2\),\[ \sqrt[n]{a^n} = \begin{cases}
a, & \text{ if }n\text{ is odd} \\[6pt]
|a|, & \text{ if }n\text{ is even} \\[6pt]
\end{cases} \nonumber \]
An easy way to remember this theorem is to know that even-indexed radicals only return non-negative values. Therefore, \( \sqrt[10]{a^10} \), for example, must always return \( |a| \) and \( \sqrt[10]{a^20} = \sqrt[10]{(a^2)^10} \) always returns \( \left| a^2 \right| \). Of course, after forcing the absolute values on the result, we need to use some logic to simplify. In the latter case, we know that \( a^2 \geq 0 \). Therefore, \( \left| a^2 \right| = a^2 \).
Let's try some examples to solidify this concept.
Simplify each expression.
- \(\sqrt{x^{2}}\)
- \(\sqrt[3]{n^{3}}\)
- \(\sqrt[4]{p^{4}}\)
- \(\sqrt[5]{y^{5}}\)
- Solutions
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- Since the index, \(2\), is even, \(\sqrt{x^2} = \sqrt[2]{x^{2}}=|x|\).
- This is an odd indexed root so there is no need for an absolute value sign.\[\sqrt[3]{m^{3}} = m.\nonumber \]
- Since the index is even \(\sqrt[4]{p^{4}}=|p|\).
- Since the index is odd, \(\sqrt[5]{y^{5}}= y\).
Simplify:
- \(\sqrt{b^{2}}\)
- \(\sqrt[3]{w^{3}}\)
- \(\sqrt[4]{m^{4}}\)
- \(\sqrt[5]{q^{5}}\)
- Answers
-
- \(|b|\)
- \(w\)
- \(|m|\)
- \(q\)
What about square roots of higher powers of variables? The Power Property of the Laws of Exponents says \(\left(a^{m}\right)^{n}=a^{m \cdot n}\). So if we square \(a^{m}\), the exponent will become \(2m\).\[\left(a^{m}\right)^{2}=a^{2 m}. \nonumber \]Looking now at the square root, we get\[\sqrt{a^{2 m}}. \nonumber \]Since \(\left(a^{m}\right)^{2}=a^{2 m}\),\[\sqrt{\left(a^{m}\right)^{2}}. \nonumber \]Since \(n\) is even, \(\sqrt[n]{a^{n}}=|a|\). Therefore,\[\left|a^{m}\right| \implies \sqrt{a^{2 m}}=\left|a^{m}\right|.\nonumber \]We apply this concept in the next example.
Simplify:
- \(\sqrt{x^{6}}\)
- \(\sqrt{y^{16}}\)
- \(\sqrt[3]{y^{18}}\)
- \(\sqrt[4]{z^{8}}\)
- \(\sqrt{16 n^{2}}\)
- \(-\sqrt{81 c^{2}}\)
- \(\sqrt[3]{64 p^{6}}\)
- \(\sqrt[4]{16 q^{12}}\)
- Solutions
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- Since \(\left(x^{3}\right)^{2}=x^{6}\),\[\sqrt{\left(x^{3}\right)^{2}}.\nonumber \]Since the index is even, this becomes \(\left|x^{3}\right|\). Hence,\[ \sqrt{x^{6}} = \left|x^{3}\right|. \nonumber \]
- Since \(\left(y^{8}\right)^{2}=y^{16}\),\[\sqrt{\left(y^{8}\right)^{2}}.\nonumber \]Since the index is even, this becomes \( |y^8| \). In this case the absolute value sign is not needed as \(y^{8}\) is not negative. Therefore,\[ \sqrt{y^{16}} = y^8 \nonumber \]
- Since \(\left(y^{6}\right)^{3}=y^{18}\),\[\sqrt[3]{\left(y^{6}\right)^{3}}.\nonumber \]Since \(3\) is odd, this becomes \(y^{6}\).
- Since \(\left(z^{2}\right)^{4}=z^{8}\),\[\sqrt[4]{\left(z^{2}\right)^{4}}.\nonumber \]Since \(z^{2}\) is positive, we do not need an absolute value sign. Therefore, this simplifies to \(z^{2}\)
- Since \((4 n)^{2}=16 n^{2}\),\[\sqrt{(4 n)^{2}}.\nonumber \]Since the index is even, this reduces to \(4|n|\).
- Since \((9 c)^{2}=81 c^{2}\),\[-\sqrt{(9 c)^{2}}.\nonumber \]Since the index is even, this becomes \(-9|c|\).
- Rewrite \(64p^{6}\) as \(\left(4 p^{2}\right)^{3}\). Therefore, we have\[\sqrt[3]{\left(4 p^{2}\right)^{3}}.\nonumber \]Taking the cube root, we get\[4p^{2}.\nonumber \]
- Rewrite the radicand as a fourth power.\[\sqrt[4]{\left(2 q^{3}\right)^{4}}.\nonumber \]Take the fourth root to get\[2|q^{3}|.\nonumber \]
Simplify:
- \(\sqrt{y^{18}}\)
- \(\sqrt{z^{12}}\)
- \(\sqrt[4]{u^{12}}\)
- \(\sqrt[3]{v^{15}}\)
- \(\sqrt{64 x^{2}}\)
- \(-\sqrt{100 p^{2}}\)
- \(\sqrt[3]{27 x^{27}}\)
- \(\sqrt[4]{81 q^{28}}\)
- Answers
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- \(|y^{9}|\)
- \(z^{6}\)
- \(|u^{3}|\)
- \(v^{5}\)
- \(8|x|\)
- \(-10|p|\)
- \(3x^{9}\)
- \(3|q^{7}|\)
Rationalizing Numerators and Denominators
By now, you have probably figured out that finding the Difference Quotient of a function is a big deal in Calculus. After all, we have covered the Difference Quotient in multiple sections up to this point. It turns out that in order to find the Difference Quotient of a radical function, we need to be able to rationalize denominators (and, sometimes, numerators) of fractions involving radicals.
Recall, when the denominator of a fraction is a sum or difference involving at least one square root, we use the
conjugate multiplication
to
rationalize the denominator
.\[\begin{array}{c|c}
(a-b)(a+b) & (2-\sqrt{5})(2+\sqrt{5}) \\[6pt]
= a^{2}-b^{2} & = 2^{2} - (\sqrt{5})^{2} \\[6pt]
& = 4-5 \\[6pt]
& = -1 \\[6pt]
\end{array} \nonumber \]As seen above, when we multiply a binomial that includes a square root by its conjugate, the product has no square roots. This process is so important and common in Calculus, we should codify it with a theorem.
Given algebraic expressions \( a \) and \( b \), the product of \( a + b \) with its conjugate, \( a - b \), is always \( a^2 - b^2 \).
Of course, this is simply the Difference of Squares factorization theorem stated in reverse, so it should not be a surprising result.
Simplify: \(\frac{5}{2-\sqrt{3}}\)
- Solution
-
\[ \begin{array}{rclcl}
\dfrac{5}{2-\sqrt{3}} & = & \dfrac{5}{\left(2-\sqrt{3}\right)} \cdot \dfrac{\left(2 + \sqrt{3}\right)}{\left(2+\sqrt{3}\right)} & \quad & \left( \text{multiplying numerator and denominator by the conjugate of the denominator} \right) \\[6pt]
& = & \dfrac{5\left(2 + \sqrt{3}\right)}{2^2 - \left(\sqrt{3}\right)^2} & \quad & \left( \text{distributing (denominator only)} \right) \\[6pt]
& = & \dfrac{5\left(2 + \sqrt{3}\right)}{4 - 3} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{5\left(2 + \sqrt{3}\right)}{1} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 5\left(2 + \sqrt{3}\right) & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Notice we did not distribute the \(5\) in the answer of Example \( \PageIndex{ 3 } \). By leaving the result factored we can see if there are any factors that may be common to both the numerator and denominator.
Simplify: \(\frac{3}{1-\sqrt{5}}\).
- Solution
-
\(-\frac{3(1+\sqrt{5})}{4}\)
Simplify: \(\frac{\sqrt{3}}{\sqrt{u}-\sqrt{6}}\).
- Solution
-
\[ \begin{array}{rclcl}
\dfrac{\sqrt{3}}{\sqrt{u}-\sqrt{6}} & = & \dfrac{\sqrt{3}}{\left(\sqrt{u}-\sqrt{6}\right)} \cdot \dfrac{\left(\sqrt{u} + \sqrt{6}\right)}{\left(\sqrt{u}+\sqrt{6}\right)} & \quad & \left( \text{multiplying numerator and denominator by the conjugate of the denominator} \right) \\[6pt]
& = & \dfrac{\sqrt{3}\left(\sqrt{u} + \sqrt{6}\right)}{\left( \sqrt{u} \right)^2 - \left(\sqrt{6}\right)^2} & \quad & \left( \text{distributing (denominator only)} \right) \\[6pt]
& = & \dfrac{\sqrt{3}\left(\sqrt{u} + \sqrt{6}\right)}{u - 6} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Simplify: \(\frac{\sqrt{5}}{\sqrt{x}+\sqrt{2}}\).
- Solution
-
\(\frac{\sqrt{5}(\sqrt{x}-\sqrt{2})}{x-2}\)
Focusing on Calculus - Revisiting the Difference Quotient
As with simplifying compound fractions, you must instinctively know when to rationalize a numerator or denominator throughout Calculus (without explicit instruction from your instructor).
Find the Difference Quotient of the following function.\[ f(x) = \dfrac{1}{\sqrt{x}} \nonumber \]Simplify this expression as much as possible, and make sure your final simplification does not have a factor of \( h \) in the denominator.
- Solution
-
\[ \begin{array}{rclcl}
\dfrac{f(x + h) - f(x)}{h} & = & \dfrac{ \frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}} }{h} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{ \left(\frac{1}{\sqrt{x + h}} - \dfrac{1}{\sqrt{x}} \right) }{h} \cdot \dfrac{\sqrt{x} \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}} & \quad & \left( \text{multiplying numerator and denominator by the LCD} \right) \\[6pt]
& = & \dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{distributing (numerator only)} \right) \\[6pt]
\end{array} \nonumber \]Now that the compound fraction has been simplified, we are faced with an expression that still has the \( h \) in the denominator. Remember, a hidden, underlying reason will compel us to somehow remove that through creative uses of Algebra. One such use is conjugate multiplication!\[ \begin{array}{rclcl}
\dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & = & \dfrac{ \left( \sqrt{x} - \sqrt{x + h} \right)}{h \sqrt{x} \sqrt{x + h}} \cdot \dfrac{ \left( \sqrt{x} + \sqrt{x + h} \right) }{ \left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{multiplying numerator and denominator by the conjugate of the numerator} \right) \\[6pt]
& = & \dfrac{ \left( \sqrt{x} \right)^2 - \left(\sqrt{x + h}\right)^2}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{distributing (numerator only)} \right) \\[6pt]
& = & \dfrac{x - (x + h)}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{x - x - h)}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{-h}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{-\cancelto{1}{h}}{\cancelto{1}{h} \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{canceling like factors} \right) \\[6pt]
& = & \dfrac{-1}{\sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Our final expression is no longer a compound fraction, nor does it have that pesky \( h \) in the denominator!
In the solution to Example \( \PageIndex{ 5 } \), there were a couple of times when we did not (and, technically, could not) distribute the denominator. We were forced to multiply by the LCD because of the fractions in the numerator . Remember: only use distribution on the offending piece (the numerator in both cases).
Find and simplify the difference quotient for the following function.\[r(x)=\sqrt{x}\nonumber \]
- Solution
-
\( \frac{1}{\sqrt{x+h}+\sqrt{x}} \)
With difference quotients, you will commonly need to multiply the numerator and denominator of an expression by either
- the LCD of all fractions within a compound fraction (a.k.a. simplifying compound fractions), or
- the conjugate of either the numerator or the denominator (a.k.a. rationalizing).
When doing so, knowing what to distribute and what not to distribute is imperative.
Simplifying Compound Fractions
The entire point of multiplying the numerator and denominator of your compound fraction by the LCD of all the minor fractions is to "get rid of" denominators in those minor fractions; however, with difference quotients, it is common that the denominator of the entire (major) fraction is only \( h \). In this case, do not distribute out the denominator! Doing so will complicate the mathematics, and you will lose visibility of factors that cancel.
Rationalizing
In a difference quotient, the entire point of multiplying the numerator and denominator by the conjugate of either the numerator or the denominator is to clear radicals. This will only happen with the conjugate pairs. So, definitely distribute the conjugate pairs, but do not distribute the non-conjugates.
For example, in Example \( \PageIndex{5} \), we multiplied the numerator and denominator by the conjugate of the numerator. We then distributed the numerator but did not bother with the distribution in the denominator. This is because distribution in the numerator cleared the radicals, but distribution in the denominator would do nothing other than make a mess.