The six exponent properties restated for rational exponents, a mammal surface-area model, the product and quotient properties of radicals, the two conditions for simplest form, combining like radicals, and simplifying variable expressions including the absolute-value rule.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 6 — Rational Exponents and Radical Functions
Apply Properties of Rational Exponents
Objectives
Five outcomes. The first is the payoff for Lesson 6.1's definitions.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 420-427 — the lesson these objectives are drawn from
Warm-up
Lesson 6.1 defined a to the m over n so that the power of a power rule would survive. This lesson collects the interest.
Discussion prompt
Simplify 5 to the one half times 5 to the three halves using the product of powers rule. Does the rule still work when the exponents are fractions?
Hint: Add the exponents as usual.
Answer:
\[ 5^{1/2}\cdot 5^{3/2} = 5^{(1/2 + 3/2)} = 5^2 = 25 \]
It works, and so do the other five properties. That is not luck: Lesson 6.1's definition was chosen to make them work, so this lesson is a list of consequences rather than a list of new rules.
Concept
Every exponent property from Lesson 5.1 holds unchanged when the exponents are rational. Two of them can be rewritten with radical signs, giving the product and quotient properties of radicals, and everything else in this lesson follows from those.
simplest form of a radical — A radical with index n is in simplest form when its radicand has no perfect nth power as a factor and no radical remains in any denominator.
\[ a^m a^n = a^{m+n}, \quad (ab)^m = a^mb^m, \quad \sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b} \]
Working in exponent form is usually faster, and converting to radical form at the end is usually clearer. Fluency means being able to move between them without thinking about it.
Figure (svg): The six exponent properties written with rational exponents, each with a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 420-421
Section
Section 1
Concept
For real a and b and rational m and n, powers of the same base multiply by adding exponents and divide by subtracting them, a power of a power multiplies them, and a power of a product or quotient distributes.
\[ a^ma^n = a^{m+n}, \; (a^m)^n = a^{mn}, \; (ab)^m = a^mb^m, \; a^{-m} = \tfrac{1}{a^m} \]
The arithmetic on the exponents is now fraction arithmetic, so a common denominator is often needed before the exponents can be added or subtracted.
Figure (svg): The six exponent properties written with rational exponents, each with a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 420-420 — Properties of Rational Exponents
Picture it
Every one is an instance of a rule you already know.
Figure (svg): The six exponent properties written with rational exponents, each with a worked instance
Only the exponents changed. The textbook prints this table with the same numbering as the integer version on page 330, which is the point being made.
Worked example
Example 1, all five parts.
\[ \text{Simplify } 7^{1/4}7^{1/2}, \; (6^{1/2}4^{1/3})^2, \; (4^5 3^5)^{-1/5}, \; \tfrac{5}{5^{1/3}}, \; \left(\tfrac{42^{1/3}}{6^{1/3}}\right)^2. \]
First: add the exponents
Why: One quarter plus one half is three quarters.
\[ 7 ^{\frac{3}{4}} \]
Second: distribute the outer power
Why: Each exponent is multiplied by 2, giving 6 to the first and 4 to the two thirds.
\[ 6 x 4 ^{\frac{2}{3}} \]
Third: combine the bases first
Why: Four to the fifth times 3 to the fifth is 12 to the fifth, and raising that to negative one fifth gives 12 to the negative 1.
\[ \frac{1}{12} \]
Fourth: subtract the exponents
Why: The numerator is 5 to the first, so the exponent is 1 minus one third.
\[ 5 ^{\frac{2}{3}} \]
Fifth: combine inside the bracket
Why: Forty-two over 6 is 7, so the bracket is 7 to the one third, and squaring gives 7 to the two thirds.
\[ 7 ^{\frac{2}{3}} \]
Figure (svg): The solution to Worked example simplify five expressions shown as a ladder of expressions, one row per algebraic move
\[ 7^{3/4}, \; 6\cdot 4^{2/3}, \; \tfrac{1}{12}, \; 5^{2/3}, \; 7^{2/3} \]
Verify: check the third numerically
Why: Twelve to the fifth is 248,832, and its negative fifth root is one over 12. Combining the bases before applying the exponent turned an enormous number into a one-line answer — which is the same labour-saving instinct as taking the root before the power in Lesson 6.1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 420-420
Matching
Add, subtract or multiply the exponents as the rule says.
Match the pairs
Why: The last two are the interesting ones: both combine their bases before touching the exponent, turning 20 over 5 into 4 and 4 times 3 into 12. Looking for that simplification first often removes the fractional exponent entirely.
Worked example
Example 2 and Guided Practice 5. Scientific notation makes the two-thirds power manageable.
\[ \text{With } S = km^{2/3}, \text{ find the surface area of a } 3.4 \text{ kg rabbit } (k = 9.75) \text{ and a } 95 \text{ kg sheep } (k = 8.4). \]
Write the rabbit's mass in scientific notation
Why: Three point four kilograms is 3.4 times 10 cubed grams.
\[ m = 3.4 x 10 ^{3} \]
Distribute the exponent over the product
Why: The two-thirds power applies to 3.4 and to 10 cubed separately.
\[ 9.75(3.4) ^{\frac{2}{3}}(10 ^{3}) ^{\frac{2}{3}} \]
Apply the power of a power to the ten
Why: Three times two thirds is 2, so the power of ten is 10 squared.
\[ 9.75(2.26) (100) \]
Evaluate
Why: The product is about 2200 square centimetres.
\[ \text{about } 2200 \]
Repeat for the sheep
Why: Ninety-five kilograms is 9.5 times 10 to the fourth grams, and 8.4 times 9.5 to the two thirds times 10 to the eight thirds is about 17,000.
\[ \text{about } 17, 000 \]
Figure (svg): The solution to Worked example a surface-area model shown as a ladder of expressions, one row per algebraic move
\[ 2200 \text{ cm}^2; \qquad 17{,}000 \text{ cm}^2 \]
Verify: check the scaling between the two
Why: The sheep is about 28 times the rabbit's mass, and surface area scales with the two-thirds power, so it should be about 28 to the two thirds, or about 9.2, times larger before the different k values are considered. The ratio 17,000 over 2200 is about 7.9, and the smaller k for sheep accounts for the rest — consistent, as it should be.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 421-421
Trap
\[ 2^{3/4}\cdot 2^{1/2} \]
Add numerators and add denominators
Why: The two fractions are combined as if adding across.
\[ 2^{4/6} = 2^{2/3} \quad \text{(wrong)} \]
Three quarters plus one half is five quarters, not two thirds. Two to the five quarters is about 2.38, while 2 to the two thirds is about 1.59.
\[ \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{3}{4} + \tfrac{2}{4} = \tfrac{5}{4} \]
Add the exponents as ordinary fractions, with a common denominator
Why: The exponent rule says add; how to add fractions has not changed.
\[ 2^{3/4}\cdot 2^{1/2} = 2^{5/4} \]
Most errors in this lesson are fraction-arithmetic errors rather than exponent errors. Writing the fraction addition on its own line is worth the space.
Fill the middle
Guided Practice 2.
Fill in the blanks
2^5/4\cdot 2^___ = 2^___ = 2^___}
Why: Rewriting one half as two quarters makes the addition immediate: three quarters plus two quarters is five quarters. The exponent rule was never in doubt; the fraction arithmetic is where the care goes.
Sorting
Read the structure before touching the exponents.
Sort into buckets
Sort each expression by the first property to use.
Five expressions, five different first moves. Reading the structure is what decides which rule is even relevant.
Prediction
Commit before reasoning.
Predict first
Why do the exponent properties still hold when the exponents are fractions?
Correct: Because a to the one over n was defined precisely so that they would.
\[ (a^{1/n})^n = a \;\Longrightarrow\; \text{every other rule follows} \]
Why: Lesson 6.1 fixed a to the one over n by insisting that the power of a power rule keep working, and a to the m over n followed from that. So the properties are not being rediscovered here; they are being inherited. That is why the textbook prints this table with the same numbering as the integer table — it really is the same list.
Section
Section 2
Concept
The product and quotient properties of exponents, written with an exponent of one over n, become rules about radicals: the nth root of a product is the product of the nth roots, and the nth root of a quotient is the quotient of the nth roots.
\[ \sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b}, \qquad \sqrt[n]{\tfrac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \]
Both indices must match. There is no rule at all for combining a square root with a cube root while they are written as radicals, though converting both to exponent form makes it routine.
Figure (svg): The product and quotient properties of radicals, each with a worked instance
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 421-421 — Properties of Radicals
Picture it
Example 3: one of each.
Figure (svg): The product and quotient properties of radicals, each with a worked instance
Both were used right to left, combining two radicals into one so that the resulting radicand turned out to be a perfect power. That is the usual reason for reaching for them.
Worked example
Example 3, both parts.
\[ \text{Simplify } \sqrt[3]{12}\cdot\sqrt[3]{18} \text{ and } \frac{\sqrt[4]{80}}{\sqrt[4]{5}}. \]
First: check the indices match
Why: Both are cube roots, so the product property applies.
First: combine and evaluate
Why: Twelve times 18 is 216, and 6 cubed is 216.
Second: check the indices
Why: Both are fourth roots, so the quotient property applies.
Second: combine and evaluate
Why: Eighty over 5 is 16, and 2 to the fourth is 16.
Figure (svg): The solution to Worked example combine radicals shown as a ladder of expressions, one row per algebraic move
\[ 6 \qquad \text{and} \qquad 2 \]
Verify: check by approximating separately
Why: The cube root of 12 is about 2.289 and the cube root of 18 about 2.621; their product is about 6.00. Combining first turned two irrational numbers into one whole number, which no amount of separate evaluation would have revealed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 421-421
Sorting
The indices must match.
Sort into buckets
Sort each product or quotient.
The last one becomes 5 to the one half times 5 to the one fifth, which is 5 to the seven tenths — routine in exponent form and impossible in radical form.
Worked example
Guided Practice 6 and 7.
\[ \text{Simplify } \sqrt[4]{27}\cdot\sqrt[4]{3} \text{ and } \frac{\sqrt[3]{250}}{\sqrt[3]{2}}. \]
First: combine under one fourth root
Why: Twenty-seven times 3 is 81.
First: evaluate
Why: Three to the fourth is 81.
\[ 3 \]
Second: combine under one cube root
Why: Two hundred fifty over 2 is 125.
Second: evaluate
Why: Five cubed is 125.
\[ 5 \]
Figure (svg): The solution to Worked example two more shown as a ladder of expressions, one row per algebraic move
\[ 3 \qquad \text{and} \qquad 5 \]
Verify: notice why the numbers were chosen
Why: In both, the combined radicand is a perfect power of the index — 81 is 3 to the fourth and 125 is 5 cubed. That is not an accident of the exercise; recognising when a combination will produce a perfect power is exactly the judgement these problems are training.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422
Error analysis
A student multiplies a square root by a cube root.
Annotate
On: \( \sqrt{4}\cdot\sqrt[3]{2} = \sqrt{8} = 2\sqrt{2} \)
Check the indices before combining. When they differ, convert both to exponent form, where a common denominator makes the combination routine.
Fill the middle
Example 3a.
Fill in the blanks
\sqrt[3]216\cdot\sqrt[3]___ = \sqrt[3]___ = \sqrt[3]___} = 6
Why: Twelve times 18 is 216, which is 6 cubed. Neither original radicand is a perfect cube, so combining first was essential — checking each separately would have suggested that nothing simplifies.
Prediction
Commit before reasoning.
Predict first
Which exponent properties become the product and quotient properties of radicals?
Correct: The third and sixth, with m equal to one over n.
\[ (ab)^{1/n} = a^{1/n}b^{1/n} \;\Longleftrightarrow\; \sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b} \]
Why: The power of a product says a b, all to the m, is a to the m times b to the m; setting m to one over n turns that into the nth root of a product. The power of a quotient does the same for division. So the radical rules are not new — they are two of the six, rewritten in a notation that makes the index visible.
Comparison
Fill the blanks. Each notation is better at something.
Comparison matrix
| Question | Exponent form | Radical form |
|---|---|---|
| How many rules apply? | all six | two |
| Different indices | handled by a common denominator | no rule at all |
| Spotting a perfect power | harder | easier |
| Which to use | for the work | for the final answer |
The convention in this chapter is to work in whichever form is convenient and present the answer in radical form, since that is what simplest form is defined for.
Section
Section 3
Concept
A radical of index n is in simplest form when its radicand contains no perfect nth power as a factor and no radical remains in any denominator. The first condition is fixed by factoring, the second by rationalising.
\[ \sqrt[3]{135} = \sqrt[3]{27\cdot 5} = 3\sqrt[3]{5} \]
Rationalising with a general index needs care: to clear a fifth root of 8 from a denominator, multiply by the fifth root of 4, because 8 times 4 is 32, which is 2 to the fifth.
Figure (svg): The two conditions for a radical to be in simplest form, each with an expression failing and then passing
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422 — Write radicals in simplest form
Picture it
Example 4, both parts.
Figure (svg): The two conditions for a radical to be in simplest form, each with an expression failing and then passing
The second condition is Lesson 4.5's rationalising raised to a general index. What has to be supplied is whatever completes the denominator to a perfect nth power.
Worked example
Example 4, both parts.
\[ \text{Simplify } \sqrt[3]{135} \text{ and } \frac{\sqrt[5]{7}}{\sqrt[5]{8}}. \]
First: factor out a perfect cube
Why: One hundred thirty-five is 27 times 5, and 27 is 3 cubed.
First: split and evaluate
Why: The cube root of 27 is 3, and 5 has no cube factor.
Second: decide what to multiply by
Why: The denominator is the fifth root of 8, which is 2 cubed; multiplying by the fifth root of 4 gives 2 to the fifth.
Second: carry it out
Why: The numerator becomes the fifth root of 28 and the denominator becomes the fifth root of 32, which is 2.
Figure (svg): The solution to Worked example write in simplest form shown as a ladder of expressions, one row per algebraic move
\[ 3\sqrt[3]{5} \qquad \text{and} \qquad \frac{\sqrt[5]{28}}{2} \]
Verify: check the second numerically
Why: The fifth root of 7 is about 1.4758 and the fifth root of 8 about 1.5157, giving about 0.9737. The fifth root of 28 is about 1.9474, halved is 0.9737. The two forms agree, as multiplying by a fraction equal to 1 requires.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422
Sorting
Test both conditions.
Sort into buckets
Sort each expression.
The definition has two clauses, exactly as in Lesson 4.5, and an expression must pass both. Value is never the issue; form is.
Worked example
Guided Practice 8. Decide the multiplier before doing anything.
\[ \text{Simplify } \sqrt[5]{\tfrac{3}{4}}. \]
Split with the quotient property
Why: The expression becomes the fifth root of 3 over the fifth root of 4.
Decide what completes a fifth power
Why: Four is 2 squared, and 2 to the fifth is needed, so three more factors of 2 are required — that is 8.
Multiply top and bottom
Why: The numerator becomes the fifth root of 24 and the denominator the fifth root of 32.
Evaluate the denominator
Why: The fifth root of 32 is 2.
Figure (svg): The solution to Worked example a fifth root of a fraction shown as a ladder of expressions, one row per algebraic move
\[ \frac{\sqrt[5]{24}}{2} \]
Verify: check that the numerator cannot simplify further
Why: Twenty-four is 8 times 3, and 8 is 2 cubed rather than a fifth power, so nothing comes out of the radical. With index 5 only factors that are fifth powers can escape, which is why a factor of 8 is useless here and essential in the denominator.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422
Trap
\[ \frac{\sqrt[5]{7}}{\sqrt[5]{8}} \]
Multiply top and bottom by the fifth root of 8
Why: The habit from square roots is carried across unchanged.
\[ \frac{\sqrt[5]{56}}{\sqrt[5]{64}} \quad \text{(still a radical below)} \]
Sixty-four is 2 to the sixth, not a fifth power, so the denominator is still irrational and the expression is no closer to simplest form.
\[ 8 = 2^3, \text{ and } 2^5 \text{ is wanted, so multiply by } 2^2 = 4 \]
Multiply by whatever completes the denominator to a perfect nth power
Why: For a square root that happens to be the radical itself; for other indices it usually is not.
\[ \frac{\sqrt[5]{7}}{\sqrt[5]{8}}\cdot\frac{\sqrt[5]{4}}{\sqrt[5]{4}} = \frac{\sqrt[5]{28}}{2} \]
Write the denominator as a power of a prime first. Then the question is only how many more factors are needed to reach the index, which is arithmetic rather than guesswork.
Fill the middle
Example 4b.
Fill in the blanks
\sqrt[5]4: \; 8 = 2^3, \text___ 2^5, \text___ \sqrt[5]___}
Why: Eight is 2 cubed and a fifth power needs 2 to the fifth, so two more factors of 2 are required — that is 4. Writing the denominator as a power of a prime makes the multiplier a subtraction rather than a guess: 5 minus 3 is 2, so multiply by 2 squared.
Matching
Complete it to a perfect nth power.
Match the pairs
Why: In every case the denominator is a power of 2, and the multiplier supplies exactly the missing factors to reach the index. Only the first needs the radical itself, which is why the square-root habit misleads at other indices.
Prediction
Commit before reasoning.
Predict first
From the fifth root of 24, does the factor 8 come out?
Correct: No — with index 5 only perfect fifth powers escape.
\[ \sqrt[5]{2^5} = 2, \text{ but } \sqrt[5]{2^3} \text{ stays inside} \]
Why: The index and the power must match: a cube root releases cubes, a fifth root releases fifth powers. Since 24 is 8 times 3 and neither factor is a fifth power, nothing comes out and the fifth root of 24 is already in simplest form. Matching the index to the power is the single idea behind the first condition.
Section
Section 4
Concept
Two radicals are like radicals when they have the same index and the same radicand. Like radicals are added and subtracted by the distributive property, exactly as like terms are combined in Lesson 1.2.
like radicals — Radicals with the same index and the same radicand. Only like radicals can be added or subtracted by combining their coefficients.
\[ \sqrt[4]{10} + 7\sqrt[4]{10} = 8\sqrt[4]{10} \]
Radicals that look unlike may become like after simplifying. The cube root of 54 and the cube root of 2 are unlike as written, but the first is 3 times the cube root of 2.
Figure (svg): Like radicals defined, and three sums combined with the distributive property
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422 — Add and subtract like radicals and roots
Picture it
Example 5: one in radical form, one in exponent form, one needing simplification first.
Figure (svg): Like radicals defined, and three sums combined with the distributive property
The third is the instructive one: nothing could be combined until the cube root of 54 was rewritten as 3 times the cube root of 2.
Worked example
Example 5, all three parts.
\[ \text{Simplify } \sqrt[4]{10} + 7\sqrt[4]{10}, \; 2(8^{1/5}) + 10(8^{1/5}), \; \sqrt[3]{54} - \sqrt[3]{2}. \]
First: the radicals already match
Why: Both are fourth roots of 10, so the coefficients 1 and 7 add.
Second: the same in exponent form
Why: Both terms are multiples of 8 to the one fifth, so 2 and 10 add.
\[ 12(8 ^{\frac{1}{5}}) \]
Third: simplify before comparing
Why: Fifty-four is 27 times 2, so the cube root of 54 is 3 times the cube root of 2.
Third: now combine
Why: Three minus 1 is 2.
Figure (svg): The solution to Worked example combine like radicals shown as a ladder of expressions, one row per algebraic move
\[ 8\sqrt[4]{10}, \; 12\cdot 8^{1/5}, \; 2\sqrt[3]{2} \]
Verify: check the third numerically
Why: The cube root of 54 is about 3.780 and the cube root of 2 about 1.260; their difference is about 2.520, which is twice 1.260. Simplifying revealed that the two terms were multiples of the same quantity all along.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-422
Sorting
Compare the index and the radicand, after simplifying.
Sort into buckets
Sort each pair.
Always simplify before deciding. Three of these four like pairs look unlike until one term is rewritten.
Worked example
Guided Practice 9, plus a pair that cannot be combined.
\[ \text{Simplify } \sqrt[3]{5} + \sqrt[3]{40}, \text{ and try } 2\sqrt{5} + 2\sqrt[3]{5}. \]
First: simplify the second term
Why: Forty is 8 times 5, and the cube root of 8 is 2.
First: combine
Why: One plus 2 is 3.
Second: compare the two radicals
Why: Both have radicand 5, but one has index 2 and the other index 3.
Second: conclude
Why: They are not like radicals, so the expression cannot be simplified further.
Figure (svg): The solution to Worked example one more, and a near miss shown as a ladder of expressions, one row per algebraic move
\[ 3\sqrt[3]{5}; \qquad 2\sqrt{5} + 2\sqrt[3]{5} \]
Verify: check the second numerically
Why: Two root 5 is about 4.472 and 2 times the cube root of 5 is about 3.420; their sum is about 7.892, which is not a whole multiple of either. Exercise 1 of the lesson asks about exactly this pair, and the answer is that the indices differ.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-424
Error analysis
A student combines two square roots.
Annotate
On: \( \sqrt{9} + \sqrt{16} = \sqrt{25} = 5 \)
The same warning as in Lesson 4.5, one lesson later and one index higher. Only like radicals combine, and they combine by adding coefficients, never radicands.
Fill the middle
Guided Practice 9.
Fill in the blanks
\sqrt[3]2 + \sqrt[3]___ = \sqrt[3]___ + ___\sqrt[3]___ = 3\sqrt[3]___
Why: Forty is 8 times 5, and the cube root of 8 is 2, so the second term is 2 times the cube root of 5. Only after that rewriting do the two terms share a radicand and become combinable.
Ranking
Adding or subtracting radicals.
Put in order
Why: Step two before step three is the step people skip, and it is where most combinable pairs are discovered. Step five is a real part of the answer: an expression with two unlike radicals is finished, not unfinished.
Prediction
Commit before reasoning.
Predict first
What plays the part of the variable when radicals are added?
Correct: The radical itself, index and radicand together.
\[ r + 7r = (1+7)r = 8r \quad \text{with } r = \sqrt[4]{10} \]
Why: Writing r for the fourth root of 10 turns the first example into r plus 7r, which is 8r by the distributive property — the same move as combining x and 7x in Lesson 1.2. Two radicals are like exactly when they represent the same r, which needs both the index and the radicand to match. Nothing about the rule is new.
Section
Section 5
Concept
The properties apply unchanged to expressions with variables, with one caution: the nth root of x to the n is x when n is odd, but the absolute value of x when n is even, because an even root is never negative.
\[ \sqrt[n]{x^n} = x \; (n \text{ odd}), \qquad \sqrt[n]{x^n} = \lvert x \rvert \; (n \text{ even}) \]
When every variable is assumed positive, the absolute value bars are unnecessary and are usually dropped. Most textbook problems say so explicitly, and this one does.
Figure (svg): The rule for the nth root of x to the n, split by the parity of the index
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 422-423 — Simplify expressions involving variables
Picture it
Two rules with two instances each.
Figure (svg): The rule for the nth root of x to the n, split by the parity of the index
The seventh root of negative 5 to the seventh is negative 5, but the fourth root of negative 3 to the fourth is positive 3. The bars record that loss of sign.
Worked example
Example 6, all four parts, with variables assumed positive.
\[ \text{Simplify } \sqrt[3]{64y^6}, \; (27p^3q^{12})^{1/3}, \; \sqrt[4]{\tfrac{m^4}{n^8}}, \; \frac{14xy^{1/3}}{2x^{3/4}z^{-6}}. \]
First: write each factor as a cube
Why: Sixty-four is 4 cubed and y to the sixth is y squared, cubed.
\[ 4 y ^{2} \]
Second: distribute the one-third power
Why: Twenty-seven to the one third is 3, p cubed to the one third is p, and q to the twelfth to the one third is q to the fourth.
\[ 3 p q ^{4} \]
Third: split with the quotient property
Why: M to the fourth is a perfect fourth power and n to the eighth is n squared, to the fourth.
\[ m / n ^{2} \]
Fourth: divide coefficients and subtract exponents
Why: Fourteen over 2 is 7; x is 1 minus three quarters; z to the negative negative 6 is z to the sixth.
\[ 7 x ^{\frac{1}{4}} y ^{\frac{1}{3}} z ^{6} \]
Figure (svg): The solution to Worked example simplify with variables shown as a ladder of expressions, one row per algebraic move
\[ 4y^2, \; 3pq^4, \; \tfrac{m}{n^2}, \; 7x^{1/4}y^{1/3}z^6 \]
Verify: check the exponents divide evenly
Why: In the first, 6 divided by 3 is 2, so y to the sixth is a perfect cube. In the third, 8 divided by 4 is 2, so n to the eighth is a perfect fourth power. A variable's exponent escapes a radical exactly when the index divides it, which is the same rule as for numbers.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 423-423
Sorting
Even index and unknown sign.
Sort into buckets
Sort each expression by whether absolute value is required.
The fourth item shows how much the standard assumption does. Nearly every exercise in the chapter states it, which is why the bars are so rarely seen.
Worked example
Examples 7 and 8, plus Guided Practice 13.
\[ \text{Simplify } \sqrt[5]{4a^8b^{14}c^5}, \; 12\sqrt[3]{2z^5} - z\sqrt[3]{54z^2}, \; \sqrt{9w^5} - w\sqrt{w^3}. \]
First: split each exponent at a multiple of 5
Why: A to the eighth is a to the fifth times a cubed; b to the fourteenth is b to the tenth times b to the fourth.
\[ a ^{5} b ^{10} c ^{5}\text{ outside} \]
First: take the fifth root of the perfect powers
Why: That gives a, b squared and c outside the radical.
Second: simplify each term
Why: Two z to the fifth is 2z squared times z cubed, and 54z squared is 27 times 2z squared.
Second: combine like radicals
Why: Twelve z minus 3z is 9z.
Third: simplify and combine
Why: Nine w to the fifth gives 3w squared root w, and w times w root w gives w squared root w.
\[ 2 w ^{2} \sqrt{w} \]
Figure (svg): The solution to Worked example simplest form and like radicals with variables shown as a ladder of expressions, one row per algebraic move
\[ ab^2c\sqrt[5]{4a^3b^4}, \; 9z\sqrt[3]{2z^2}, \; 2w^2\sqrt{w} \]
Verify: check what stayed inside the first radical
Why: The exponents left inside are 3 for a and 4 for b, both less than the index 5 — as they must be, or more could come out. Checking that every remaining exponent is below the index is the quickest test that a variable radical is fully simplified.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 423-423
Trap
\[ \sqrt[4]{x^4} \]
Cancel the root against the power
Why: The index and the exponent match, so both are removed.
\[ = x \quad \text{(wrong when } x < 0) \]
At x equal to negative 3 the radicand is 81 and its fourth root is positive 3, not negative 3.
\[ \sqrt[4]{x^4} = \lvert x \rvert \]
Use absolute value whenever the index is even
Why: An even root is never negative, so the answer cannot be either.
\[ \sqrt[3]{x^3} = x \quad \text{(odd index: no bars needed)} \]
Most exercises say assume all variables are positive, which makes the bars unnecessary. The assumption is doing real work, and it is worth noticing when a problem does not make it.
Fill the middle
Example 7a.
Fill in the blanks
\sqrt[5]3 = \sqrt[5]___}}} = a\sqrt[5]___
Why: Eight is 5 plus 3, so a to the eighth is a to the fifth times a cubed, and only the fifth power escapes. The remaining exponent must be less than the index, which is why 3 is the right split rather than, say, a to the fourth times a to the fourth.
Matching
Variables assumed positive.
Match the pairs
Why: In the first two every exponent is a multiple of the index, so everything escapes the radical. The third is pure exponent arithmetic with fractions, and the fourth needed both terms simplified before the like radicals appeared.
Prediction
Commit before reasoning.
Predict first
From the fifth root of b to the fourteenth, what comes out?
Correct: b squared, since 5 goes into 14 twice.
\[ \sqrt[5]{b^{14}} = \sqrt[5]{b^{10}\cdot b^4} = b^2\sqrt[5]{b^4} \]
Why: Fourteen is 5 times 2 plus 4, so b to the fourteenth is b to the tenth times b to the fourth, and only the b to the tenth is a perfect fifth power. It escapes as b squared, leaving b to the fourth inside. The rule is division with remainder: the quotient comes out and the remainder stays in.
Comparison
Fill the blanks. The same grouping as Lesson 5.1.
Comparison matrix
| Group | Properties | New difficulty |
|---|---|---|
| Combining like bases | product and quotient of powers | adding and subtracting fractions |
| Nested powers | power of a power | multiplying fractions |
| Distributing | power of a product and of a quotient | none: the exponent is copied |
| Rewriting | negative exponent | none: the rule is unchanged |
The only genuinely new work is fraction arithmetic, which is why most errors in this lesson are arithmetic rather than conceptual.
Pattern
One routine for simplifying anything with rational exponents or radicals.
An expression is finished when no exponent is negative or zero, no radicand contains a perfect nth power, and no radical sits in a denominator.
OpenStax Algebra and Trigonometry 2e, §1.3 Radicals and Rational Exponents §1.3
Check
Rational exponents. Watch the fractions.
Check your understanding
Simplify 2^(3/4) x 2^(1/2).
Answer: A
Why: The exponents add: 3/4 + 1/2 = 3/4 + 2/4 = 5/4.
Check
Simplest form. Complete the denominator.
Check your understanding
Write (fifth root of 7)/(fifth root of 8) in simplest form.
Answer: A
Why: Multiplying top and bottom by the fifth root of 4 makes the denominator the fifth root of 32, which is 2.
Check
Variables. Split the exponent by the index.
Check your understanding
Simplify the fifth root of b^14, with b positive.
Answer: A
Why: Fourteen is 10 plus 4, so b^10 escapes as b^2 and b^4 stays inside.
Real world
The period of a simple pendulum, in seconds, is about 2 pi times the square root of L over g, where L is its length in metres and g is 9.8 metres per second squared.
Discussion prompt
Write the period as a power of L, find how the period changes when the length is quadrupled, and rationalise the expression for a 2-metre pendulum.
Hint: Split the radical with the quotient property and use a rational exponent.
Answer:
\[ T = 2\pi\sqrt{\tfrac{L}{g}} = \frac{2\pi}{\sqrt{g}}L^{1/2} \]
The period is proportional to L to the one half, so quadrupling the length multiplies the period by the square root of 4, which is 2 — not by 4.
\[ L = 2: \; T = 2\pi\sqrt{\tfrac{2}{9.8}} = 2\pi\frac{\sqrt{2}}{\sqrt{9.8}} = 2\pi\frac{\sqrt{19.6}}{9.8} \approx 2.84 \text{ s} \]
Two things are worth noticing. Writing the model with a rational exponent makes the scaling immediate: an exponent of one half means the output changes by the square root of whatever factor the input changes by, which is the same reasoning as the coral cod's cube root in Lesson 6.1. And rationalising the denominator turned a division by an irrational number into a division by 9.8, which is what makes the arithmetic doable at all by hand.
Commit first
Answer, then rate your confidence honestly.
Predict first
Are 2 root 5 and 2 times the cube root of 5 like radicals?
Correct: No — their indices differ.
\[ 2\sqrt{5} = 2\cdot 5^{1/2}, \quad 2\sqrt[3]{5} = 2\cdot 5^{1/3}: \; \text{different exponents} \]
Why: Like radicals need the same index and the same radicand, and here only the radicand matches. No simplification can change an index, so the two terms stay separate permanently. Exercise 1 of the lesson asks exactly this, and the distinction matters because a same-looking radicand is the most tempting reason to combine terms that cannot be combined. Numerically, 2 root 5 is about 4.47 and 2 times the cube root of 5 about 3.42.
Explain it
They can simplify square roots and are unsure what changes at higher indices.
Discussion prompt
In four sentences or fewer, explain what comes out of a radical and how to rationalise a denominator when the index is not 2.
Hint: Match the power to the index.
Answer:
Only a factor that is a perfect nth power can escape an nth root, so a cube root releases cubes and a fifth root releases fifth powers. For a variable, divide its exponent by the index: the quotient comes out and the remainder stays in.
To rationalise, write the denominator as a power of a prime and multiply by whatever brings that power up to the index. For a square root that happens to be the radical itself, but for a fifth root of 8 you multiply by the fifth root of 4, because 8 is 2 cubed and you need 2 to the fifth.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For fractions, write the common-denominator step on its own line. For rationalising, write the denominator as a power of a prime and subtract exponents to find what is missing. For like radicals, simplify every term first and only then compare. For splitting, divide the exponent by the index and keep the remainder inside. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference. Top left: write the six exponent properties with rational exponents and one instance of each that you invent and verify. Top right: write the two conditions for simplest form and give an expression failing each, with its corrected version beside it. Middle: take the fifth root of 4a to the eighth b to the fourteenth c to the fifth and simplify it completely, showing how each exponent is split into a multiple of 5 plus a remainder. Bottom left: rationalise three denominators — a square root, a cube root and a fifth root — writing each denominator as a power of a prime first and showing how you chose the multiplier. Bottom right: write two pairs of radicals, one like and one unlike, and say in a sentence what makes the difference. In a margin, write the two cases of the nth root of x to the n rule.
If any exponent left inside a radical is as large as the index, go back: more can still come out.
Recap
Five things, all inherited from Lesson 5.1 by way of Lesson 6.1's definition.
| If you see | Then |
|---|---|
| Same base, powers multiplied | Add the exponents, with a common denominator |
| Different indices | Convert to exponent form |
| A perfect nth power inside | Take it out |
| A radical in a denominator | Complete it to a perfect nth power |
| Radicals with matching index and radicand | Combine the coefficients |
| An even index on a variable | Use absolute value unless positivity is assumed |
Lesson 6.3 changes subject: instead of simplifying one expression, it combines two whole functions by adding, multiplying and composing them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.2 Apply Properties of Rational Exponents §6.2, pp. 420-427 — everything on these slides traces back here
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