nth roots and how many real ones a number has, why a to the power one over n is the nth root, evaluating expressions with rational exponents in both forms, solving equations by taking nth roots, and a cube-root model for a fish's length.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 6 — Rational Exponents and Radical Functions
Evaluate nth Roots and Use Rational Exponents
Objectives
Five outcomes. The second is a definition chosen to keep Lesson 5.1's rules alive.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 414-419 — the lesson these objectives are drawn from
Warm-up
Lesson 5.1 gave the exponent rules for whole-number exponents. Lesson 4.5 gave square roots. This lesson joins them.
Discussion prompt
Assume the power of a power rule still works when the exponent is a fraction. What is the quantity a to the one half, raised to the second power, and what does that tell you about a to the one half?
Hint: Multiply the two exponents.
Answer:
\[ (a^{1/2})^2 = a^{(1/2)\cdot 2} = a^1 = a \]
So a to the one half is a number whose square is a — which is exactly what a square root is. The same argument gives a to the one third as a cube root and a to the one fourth as a fourth root, and it works because the exponent rule was assumed to survive.
Concept
For an integer n greater than 1, an nth root of a is a number whose nth power is a. Writing it as a to the power one over n makes the exponent rules of Lesson 5.1 apply to roots without change, which is why the notation was invented.
nth root of a — A number b with b to the n equal to a. It is written as the nth root of a, where n is the index of the radical, or equivalently as a to the power one over n.
\[ \sqrt[n]{a} = a^{1/n}, \quad n > 1 \]
How many real nth roots a number has depends on two things: whether the index is even or odd, and whether the number is positive, zero or negative.
Figure (svg): The power of a power property used to show that a to the one over n is the nth root of a
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 414-414
Section
Section 1
Concept
An odd index gives exactly one real nth root, whatever the sign of the number. An even index gives two real roots for a positive number, one for zero, and none at all for a negative number.
index of a radical — The number n in the nth root symbol, saying which power is being undone. A square root has index 2, and the index is usually written above the radical sign.
\[ \sqrt[3]{-216} = -6, \qquad \pm\sqrt[4]{81} = \pm 3 \]
The reason is Lesson 5.1's sign rule: an even power of any real number is non-negative, so nothing raised to an even power can give a negative result.
Figure (svg): A table of how many real nth roots a number has, split by the parity of the index and the sign of the number
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 414-414 — Real nth Roots of a
Picture it
Two parities across, three signs down.
Figure (svg): A table of how many real nth roots a number has, split by the parity of the index and the sign of the number
Only one cell has no real root at all: an even index applied to a negative number. That cell is what Lesson 4.6's imaginary numbers were built to fill for the square-root case.
Worked example
Example 1, both parts.
\[ \text{Find the real } n\text{th roots when } n = 3, a = -216 \text{ and when } n = 4, a = 81. \]
First: read the parity and the sign
Why: The index 3 is odd and negative 216 is negative, so there is exactly one real root.
First: find it
Why: Negative 6 cubed is negative 216.
\[ -6 \]
Second: read the parity and the sign
Why: The index 4 is even and 81 is positive, so there are two real roots.
Second: find them
Why: Three to the fourth is 81, and so is negative 3 to the fourth.
\[ +- 3 \]
Figure (svg): The solution to Worked example find the real nth roots shown as a ladder of expressions, one row per algebraic move
\[ \sqrt[3]{-216} = -6; \qquad \pm\sqrt[4]{81} = \pm 3 \]
Verify: check by raising each back
Why: Negative 6 cubed is negative 216, since three negative factors leave one unpaired. Three to the fourth and negative 3 to the fourth are both 81, since four negative factors pair off completely. The parity of the index is doing exactly the same work as the parity of the exponent did in Lesson 5.1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 414-414
Sorting
Read the parity, then the sign.
Sort into buckets
Sort each case by its number of real nth roots.
The last bucket is where Lesson 4.6 came from: the square root of negative 9 has no real value but does have two imaginary ones.
Worked example
Guided Practice 1 to 4.
\[ \text{Find the real roots for } n=4, a=625; \; n=6, a=64; \; n=3, a=-64; \; n=5, a=243. \]
First: even index, positive number
Why: Five to the fourth is 625, so both 5 and negative 5 work.
\[ +- 5 \]
Second: even index, positive number
Why: Two to the sixth is 64, and so is negative 2 to the sixth.
\[ +- 2 \]
Third: odd index, negative number
Why: Negative 4 cubed is negative 64, and that is the only real root.
\[ -4 \]
Fourth: odd index, positive number
Why: Three to the fifth is 243.
\[ 3 \]
Figure (svg): The solution to Worked example four more shown as a ladder of expressions, one row per algebraic move
\[ \pm 5, \; \pm 2, \; -4, \; 3 \]
Verify: notice the pattern in the answers
Why: The two even-index problems produced pairs and the two odd-index ones produced single values, regardless of the sign of the number. That is the table in action: parity decides the count, and the sign only decides whether the count for an even index is two or zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415
Trap
\[ \sqrt[3]{-216} \]
Write both signs, as for a square root
Why: The plus-or-minus habit from Lesson 4.5 is carried across.
\[ \pm 6 \quad \text{(wrong)} \]
Positive 6 cubed is positive 216, not negative 216, so 6 is not a cube root of negative 216 at all.
\[ \sqrt[3]{-216} = -6 \text{ only} \]
Check the parity of the index before writing any sign
Why: An odd index gives exactly one real root, whose sign matches the radicand's.
\[ (-6)^3 = -216 \quad \checkmark, \qquad 6^3 = 216 \neq -216 \]
The book flags the opposite mistake too: with an even index and a positive number, forgetting the negative root loses half the answer. Both errors come from not reading the parity first.
Fill the middle
Guided Practice 3.
Fill in the blanks
\sqrt[3]-4 = ___ \quad \text___ (-4)^3 = -64
Why: Negative 4 cubed is negative 64, because three negative factors leave the product negative. With an odd index there is no second answer to look for, so the plus-or-minus that a square root would need is absent here.
Matching
Parity first, sign second.
Match the pairs
Why: Zero is the only value with exactly one root regardless of parity, because zero to any power is zero and nothing else is. The four rows together are the whole table, and the parity is always the first thing to check.
Prediction
Commit before reasoning.
Predict first
Why does a negative number have no real even-index root?
Correct: Because an even power of any real number is non-negative.
\[ b^4 \ge 0 \text{ for every real } b \;\Longrightarrow\; \sqrt[4]{-81} \text{ is not real} \]
Why: Raising any real number to an even power pairs its factors, so every negative sign cancels and the result is at least zero. There is therefore nothing real to raise to the fourth power and get negative 81. This is exactly why the square-root case needed the imaginary unit of Lesson 4.6, and the same construction extends to every even index.
Section
Section 2
Concept
For a rational exponent m over n, the denominator gives the root and the numerator gives the power. A negative rational exponent means the reciprocal, exactly as it did for integer exponents in Lesson 5.1.
\[ a^{m/n} = (a^{1/n})^m = (\sqrt[n]{a})^m, \qquad a^{-m/n} = \frac{1}{a^{m/n}} \]
Taking the root before the power keeps the numbers small. Sixteen to the three halves is 4 cubed, which is 64, and computing 16 cubed first would mean taking the square root of 4096.
Figure (svg): Two expressions with rational exponents evaluated twice, once in exponent form and once in radical form
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415 — Rational Exponents
Picture it
Example 2: a positive and a negative rational exponent.
Figure (svg): Two expressions with rational exponents evaluated twice, once in exponent form and once in radical form
The exponent form and the radical form are the same calculation written two ways. Which to use is a matter of taste, and the exponent form is usually shorter to write.
Worked example
Example 2, in both forms.
\[ \text{Evaluate } 16^{3/2} \text{ and } 32^{-3/5}. \]
First: split the exponent
Why: Three halves means the square root, then the cube.
\[ (16 ^{\frac{1}{2}}) ^{3} \]
First: evaluate
Why: The square root of 16 is 4, and 4 cubed is 64.
\[ 64 \]
Second: handle the negative sign
Why: A negative exponent means the reciprocal of the positive-exponent value.
\[ 1 / 32 ^{\frac{3}{5}} \]
Second: split and evaluate
Why: The fifth root of 32 is 2, and 2 cubed is 8.
\[ \frac{1}{8} \]
Figure (svg): The solution to Worked example evaluate two expressions shown as a ladder of expressions, one row per algebraic move
\[ 16^{3/2} = 64; \qquad 32^{-3/5} = \tfrac{1}{8} \]
Verify: check by raising the answer back
Why: Sixty-four to the two thirds should return 16: the cube root of 64 is 4, and 4 squared is 16. And one eighth raised to negative five thirds should return 32. Both check, which confirms that a rational exponent behaves like an ordinary one under the power of a power rule.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415
Matching
The denominator is the index.
Match the pairs
Why: In every case the denominator became the index and the numerator the power. The two negative exponents also became reciprocals, which is the same rule as for integer exponents in Lesson 5.1 — a negative exponent has always meant a reciprocal.
Worked example
Guided Practice 5 to 8.
\[ \text{Evaluate } 4^{5/2}, \; 9^{-1/2}, \; 81^{3/4}, \; 1^{7/8}. \]
First: root then power
Why: The square root of 4 is 2, and 2 to the fifth is 32.
\[ 32 \]
Second: negative exponent
Why: The reciprocal of the square root of 9, which is one third.
\[ \frac{1}{3} \]
Third: fourth root then cube
Why: The fourth root of 81 is 3, and 3 cubed is 27.
\[ 27 \]
Fourth: any root of 1
Why: One raised to any power is 1, and every root of 1 is 1.
\[ 1 \]
Figure (svg): The solution to Worked example four more without a calculator shown as a ladder of expressions, one row per algebraic move
\[ 32, \; \tfrac{1}{3}, \; 27, \; 1 \]
Verify: compare the size of the two routes
Why: For 81 to the three quarters, taking the root first gives 3 cubed, or 27. Taking the power first would mean 81 cubed, which is 531,441, and then its fourth root. Both give 27, and only one is doable in your head — which is the whole reason for taking the root first.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415
Error analysis
A student evaluates an expression with a rational exponent.
Annotate
On: \( 16^{3/2} = \sqrt[3]{16^2} = \sqrt[3]{256} \approx 6.35 \)
Denominator down below, index down below. Checking with a familiar case such as a to the one half settles the convention in a second.
Fill the middle
Example 2a.
Fill in the blanks
16^4 = (16^___)^3 = ___^3 = 64
Why: The square root of 16 is 4, and 4 cubed is 64. Taking the root first is legal because the power of a power rule lets the exponents be multiplied in either order, and it is preferable because it keeps the intermediate number small.
Sorting
Root first, or power first.
Sort into buckets
For each expression, sort by which order keeps the arithmetic manageable.
Both orders are legal and give identical answers. Taking the root first is a habit worth forming for no reason other than the size of the numbers.
Prediction
Commit before reasoning.
Predict first
Why does a to the m over n have to equal the nth root of a, all raised to the m?
Correct: Because the power of a power rule forces it once a to the one over n is fixed.
\[ a^{m/n} = a^{m \cdot (1/n)} = (a^{1/n})^m \]
Why: Writing m over n as m times one over n and applying the power of a power rule gives a to the one over n, raised to the m — there is no freedom left. Every definition in this lesson is forced by insisting that Lesson 5.1's rules keep holding, which is the same principle that fixed a to the zero as 1 and a negative exponent as a reciprocal.
Section
Section 3
Concept
Most rational exponents give irrational values that must be approximated. On a calculator the whole exponent must be enclosed in brackets, or the machine computes a power and then divides.
\[ 9^{1/5} \approx 1.5518 \]
Without brackets, 9 to the power 1 divided by 5 is read as 9 to the first, divided by 5, which is 1.8 — a plausible-looking wrong answer rather than an obvious error.
Figure (svg): Two expressions with rational exponents evaluated twice, once in exponent form and once in radical form
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415 — Approximate roots with a calculator
Picture it
Exact where possible, approximate where not.
Figure (svg): Two expressions with rational exponents evaluated twice, once in exponent form and once in radical form
Sixteen to the three halves came out exactly as 64 because 16 is a perfect square. Nine to the one fifth does not, because 9 is not a fifth power, so an approximation is the best available.
Worked example
Example 3. Brackets around every exponent.
\[ \text{Approximate } 9^{1/5}, \; 12^{3/8}, \; (\sqrt[4]{7})^3. \]
First: enter with brackets
Why: Nine to the power in brackets 1 over 5.
\[ \text{about } 1.5518 \]
Second: the same
Why: Twelve to the power in brackets 3 over 8.
\[ \text{about } 2.5392 \]
Third: convert to exponent form first
Why: The fourth root of 7, cubed, is 7 to the three quarters.
\[ 7 ^{\frac{3}{4}} \]
Third: evaluate
Why: Seven to the power in brackets 3 over 4.
\[ \text{about } 4.3035 \]
Figure (svg): The solution to Worked example approximate three values shown as a ladder of expressions, one row per algebraic move
\[ 1.5518, \; 2.5392, \; 4.3035 \]
Verify: sanity-check the sizes
Why: Nine to the one fifth should be a little above 1, since 1 to the fifth is 1 and 2 to the fifth is 32 — and 1.55 sits comfortably there. Seven to the three quarters should be a bit less than 7, since the exponent is less than 1, and 4.30 is. Estimating the size before reading the display catches a mistyped exponent.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415
Fill the middle
Example 3c.
Fill in the blanks
(\sqrt[4]4)^3 = 7^___}}
Why: The index 4 becomes the denominator and the outer power 3 becomes the numerator, so the expression is 7 to the three quarters. Converting to exponent form before typing is worth doing because most calculators have no root key beyond the square root.
Worked example
Guided Practice 9 to 12.
\[ \text{Evaluate } 4^{2/5}, \; 64^{-2/3}, \; (\sqrt[4]{16})^5, \; (\sqrt[3]{-30})^2. \]
First: approximate
Why: Four to the two fifths is about 1.7411.
\[ \text{about } 1.74 \]
Second: exact
Why: The cube root of 64 is 4, 4 squared is 16, and the negative exponent gives one sixteenth.
\[ \frac{1}{16},\text{ or } 0.06 \]
Third: exact
Why: The fourth root of 16 is 2, and 2 to the fifth is 32.
\[ 32 \]
Fourth: approximate
Why: The cube root of negative 30 is about negative 3.107, and squaring makes it positive.
\[ \text{about } 9.65 \]
Figure (svg): The solution to Worked example four with mixed forms shown as a ladder of expressions, one row per algebraic move
\[ 1.74, \; \tfrac{1}{16}, \; 32, \; 9.65 \]
Verify: notice which came out exact
Why: The second and third were exact because 64 is a perfect cube and 16 a perfect fourth power. The other two were not, so an approximation was the honest answer. Looking for a perfect power before reaching for the calculator is worth a moment on every problem of this kind.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 415-415
Error analysis
A student evaluates a rational-exponent expression.
Annotate
On: \( 64^{-2/3} \approx 0.06 \)
Approximate only when the value is irrational. Checking whether the base is a perfect power takes a second and often keeps the answer exact.
Prediction
Commit before reasoning.
Predict first
Typing nine, caret, one, divide, five gives 1.8 rather than 1.55. Why?
Correct: It computes 9 to the first power and then divides by 5.
\[ \text{no brackets: } \tfrac{9^1}{5} = 1.8 \qquad \text{brackets: } 9^{1/5} \approx 1.5518 \]
Why: Exponentiation binds tighter than division, so the machine reads it as nine to the power one, all divided by 5 — which is 9 over 5, or 1.8. The result looks like a plausible number rather than an error, which is what makes the mistake dangerous. Brackets around the whole exponent are the fix, and the textbook flags it in an Avoid Errors note.
Sorting
Look for a perfect power.
Sort into buckets
Sort each value by whether it can be written exactly without a radical.
Checking for a perfect power costs nothing and often turns a calculator problem into a mental one.
Comparison
Fill the blanks. Two notations, one meaning.
Comparison matrix
| Question | Radical form | Exponent form |
|---|---|---|
| Written as | the nth root of a, to the m | a to the m over n |
| Exponent rules apply? | not directly | yes, unchanged from Lesson 5.1 |
| Easier to type | no | yes |
| Same value? | yes | yes |
The second row is why the exponent form was invented: it lets one set of rules cover roots and powers together, which is what Lesson 6.2 exploits.
Section
Section 4
Concept
To solve an equation of the form a x to the n equals b, divide by a and take the nth root of both sides. If the index is even, both signs must be written; if it is odd, there is exactly one root.
\[ 4x^5 = 128 \;\Longrightarrow\; x^5 = 32 \;\Longrightarrow\; x = 2 \]
This is Lesson 4.5's method generalised from index 2 to any index. The plus-or-minus decision is now made by the parity of the index rather than being automatic.
Figure (svg): Two equations solved by taking nth roots, with the plus-or-minus appearing only for an even index
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416 — Solve equations using nth roots
Picture it
Example 4: one equation of each kind.
Figure (svg): Two equations solved by taking nth roots, with the plus-or-minus appearing only for an even index
The fifth-root equation has one solution and the fourth-root equation has two. Nothing else about the two procedures differs.
Worked example
Example 4, both parts.
\[ \text{Solve } 4x^5 = 128 \text{ and } (x-3)^4 = 21. \]
First: isolate the power
Why: Dividing both sides by 4 gives x to the fifth equal to 32.
\[ x ^{5} = 32 \]
First: take the fifth root
Why: The index is odd, so there is exactly one real root, and 2 to the fifth is 32.
\[ x = 2 \]
Second: take the fourth root of both sides
Why: The index is even and 21 is positive, so both signs are written.
\[ x - 3 = +- 21 ^{\frac{1}{4}} \]
Second: isolate x and evaluate
Why: Adding 3 gives 3 plus or minus the fourth root of 21, which is about 2.14.
\[ x\text{ about } 5.14\text{ or } 0.86 \]
Figure (svg): The solution to Worked example solve two equations shown as a ladder of expressions, one row per algebraic move
\[ x = 2; \qquad x = 3 \pm \sqrt[4]{21} \]
Verify: substitute back into each
Why: Two to the fifth is 32, and 4 times 32 is 128. For the second, 5.14 minus 3 is 2.14, whose fourth power is about 21. Both check, and note that the second's two solutions are symmetric about 3 — the value that makes the bracket vanish, exactly as in Lesson 4.5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416
Sorting
Read the index's parity.
Sort into buckets
Sort each equation by its number of real solutions.
Parity decides the count and the sign of the right side decides whether an even index gives two or none. Two readings, no arithmetic.
Worked example
Guided Practice 13 to 18.
\[ \text{Solve } x^3=64, \; \tfrac{1}{2}x^5=512, \; 3x^2=108, \; \tfrac{1}{4}x^3=2, \; (x-2)^3=-14, \; (x+5)^4=16. \]
The three odd-index ones without brackets
Why: X cubed equal to 64 gives 4; x to the fifth equal to 1024 gives 4; x cubed equal to 8 gives 2.
\[ 4, 4, 2 \]
The even-index one without brackets
Why: Three x squared equal to 108 gives x squared equal to 36, so both signs.
\[ +- 6 \]
The odd-index one with a bracket
Why: The bracket cubed is negative 14, so the bracket is the cube root of negative 14, about negative 2.41.
\[ x\text{ about } -0.41 \]
The even-index one with a bracket
Why: The bracket to the fourth is 16, so the bracket is plus or minus 2.
\[ x = -3\text{ or } -7 \]
Figure (svg): The solution to Worked example six more equations shown as a ladder of expressions, one row per algebraic move
\[ 4, \; 4, \; \pm 6, \; 2, \; \approx -0.41, \; -3 \text{ or } -7 \]
Verify: check the two bracketed answers
Why: For the fifth, negative 0.41 minus 2 is negative 2.41, whose cube is about negative 14. For the sixth, negative 3 plus 5 is 2 and negative 7 plus 5 is negative 2, and both raise to 16 at the fourth power. The two solutions of the last are symmetric about negative 5, which is where the bracket vanishes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416
Trap
\[ 4x^5 = 128 \]
Take the fifth root of both sides first
Why: The root is applied before the coefficient is removed.
\[ \sqrt[5]{4}\cdot x = \sqrt[5]{128} \;\Longrightarrow\; x = \sqrt[5]{32} = 2 \quad \text{(right answer, wrong reason)} \]
Here it happens to work because 128 over 4 is 32 and roots distribute over products. But writing the fifth root of 4x to the fifth as the fifth root of 4 times x is exactly the step that fails when a is negative or the index is even.
\[ 4x^5 = 128 \;\Longrightarrow\; x^5 = 32 \;\Longrightarrow\; x = 2 \]
Divide by the coefficient first, then take the root
Why: Isolating the power keeps every later step unambiguous.
\[ \text{same order as Lesson 4.5: isolate, then undo} \]
The habit matters more than this example does. With an even index and a negative coefficient, distributing the root over the product produces a root of a negative number that need not exist.
Fill the middle
Example 4b.
Fill in the blanks
(x-3)^4 = 21 \;\Longrightarrow\; x - 3 = +-\sqrt[4]___
Why: An even index applied to a positive number gives two real roots, so both signs must be written. Omitting the plus-or-minus would lose the solution near 0.86 and leave only 5.14, exactly as dropping it lost half the answers in Lesson 4.5.
Ranking
Solving an equation with a power.
Put in order
Why: This is Lesson 4.5's routine with one change: the sign decision at step three now depends on the parity of the index rather than being automatic. The bracket step is an ordinary linear move and comes after the root, never before it.
Prediction
Commit before reasoning.
Predict first
Lesson 4.5 solved x squared equals s. What has changed?
Correct: Only the index, and with it the number of solutions.
\[ x^3 = -8 \;\Longrightarrow\; x = -2, \quad \text{but } x^2 = -8 \text{ has no real solution} \]
Why: Lesson 4.5 was the case n equal to 2, where the index is always even and the plus-or-minus is always needed. Allowing any index makes the sign decision conditional, and it also makes a negative right side solvable when the index is odd — x cubed equals negative 8 has the perfectly ordinary solution negative 2, where x squared equals negative 8 has none.
Section
Section 5
Concept
A model of the form y equals a times x to the n gives an output from an input. To go the other way, substitute the output, divide by the coefficient, and take the nth root.
\[ w = 0.0167 l^3 \;\Longrightarrow\; l = \sqrt[3]{w/0.0167} \]
In a physical model the index is usually 2 or 3 and the quantities are positive, so the root is unique and no solution has to be rejected.
Figure (svg): A cube model relating a fish's length to its weight, with the length recovered from a weight of two hundred grams
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416 — Use nth roots in problem solving
Picture it
Example 5: the coral cod model, run backwards.
Figure (svg): A cube model relating a fish's length to its weight, with the length recovered from a weight of two hundred grams
A weight of 200 grams gives a length of about 23 centimetres. The cube in the model came from volume scaling with the cube of a length, exactly as in Lesson 5.1.
Worked example
Example 5. Substitute, divide, take the root.
\[ \text{With } w = 0.0167l^3, \text{ find the length of a } 200 \text{ gram coral cod.} \]
Substitute the weight
Why: The model becomes 200 equals 0.0167 times l cubed.
\[ 200 = 0.0167 l ^{3} \]
Divide by the coefficient
Why: Two hundred over 0.0167 is about 11,976.
\[ l ^{3}\text{ about } 11, 976 \]
Take the cube root
Why: The index is 3, which is odd, so there is exactly one real root.
Evaluate
Why: The cube root of 11,976 is about 22.9.
\[ \text{about } 22.9 \text{cm} \]
Figure (svg): The solution to Worked example find the length from the weight shown as a ladder of expressions, one row per algebraic move
\[ l \approx 22.9 \text{ cm} \]
Verify: substitute the length back
Why: At l equal to 22.9, l cubed is about 12,009, and 0.0167 times that is about 200.6 grams — close to 200, with the small gap coming from rounding the length to one decimal place. Using the exact cube root would give exactly 200.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416
Fill the middle
Example 5, at the division.
Fill in the blanks
200 = 0.0167l^3 \;\Longrightarrow\; l^3 \approx 11976
Why: Two hundred divided by 0.0167 is about 11,976. Dividing by a small decimal produces a large number, which is worth expecting: a coefficient of 0.0167 means the cube of the length is roughly sixty times the weight.
Worked example
Guided Practice 19. Notice how slowly the length grows.
\[ \text{Find the length of a coral cod weighing } 275, \; 340 \text{ and } 450 \text{ grams.} \]
Divide each weight by the coefficient
Why: The three quotients are about 16,467, 20,359 and 26,946.
Take the cube root of each
Why: The roots are about 25.4, 27.3 and 30.0.
State the answers
Why: About 25, 27 and 30 centimetres.
\[ 25, 27, 30 \text{cm} \]
Compare the growth
Why: The weight rose by more than half from 275 to 450, while the length rose by less than a fifth.
Figure (svg): The solution to Worked example three more weights shown as a ladder of expressions, one row per algebraic move
\[ 25.4, \; 27.3, \; 30.0 \text{ cm} \]
Verify: check the scaling relationship
Why: Doubling a weight should multiply the length by the cube root of 2, which is about 1.26. From 200 to 450 the weight rises by a factor of 2.25, and the length from 22.9 to 30.0 rises by a factor of 1.31 — and the cube root of 2.25 is 1.31. The pattern holds exactly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 416-416
Error analysis
A student recovers a length from the coral cod model.
Annotate
On: \( 200 = 0.0167l^3 \;\Longrightarrow\; l = \frac{200}{0.0167 \cdot 3} \approx 3992 \)
A power is undone by a root of the same index. A quick sanity check settles it: a 4000-centimetre fish is forty metres long.
Prediction
Commit before reasoning.
Predict first
Why does a fish's weight depend on the cube of its length rather than on the length itself?
Correct: Weight tracks volume, and volume scales with the cube of a length.
\[ \text{length} \times k \;\Longrightarrow\; \text{volume} \times k^3 \]
Why: A fish twice as long is also roughly twice as deep and twice as wide, so its volume — and hence its weight — is about eight times as great. That is the same cubic scaling as Lesson 5.1's comparison of two stars, and it is why the model's exponent is 3 rather than something fitted. The coefficient 0.0167 encodes the fish's shape and density, and only that part is empirical.
Ranking
Recovering an input from a power model.
Put in order
Why: Step three before step four is the same isolate-then-undo order as everywhere else. Step five matters here because the model's coefficient carries three significant digits, so reporting 22.9146 centimetres would claim precision the model does not have.
Comparison
Fill the blanks. The model runs both ways.
Comparison matrix
| Question | Length to weight | Weight to length |
|---|---|---|
| What you are given | l | w |
| The operation | cube, then multiply by 0.0167 | divide by 0.0167, then take the cube root |
| Number of answers | one | one, since the index is odd |
| Doubling the input | multiplies the weight by 8 | multiplies the length by about 1.26 |
Going backwards undoes each operation in reverse order, which is exactly what an inverse function does — the subject of Lesson 6.4.
Comparison
Fill the blanks. Each lesson widened the same idea.
Comparison matrix
| Lesson | Handled | Left open |
|---|---|---|
| 4.5 | square roots of positive numbers | square roots of negatives |
| 4.6 | square roots of negatives, using i | roots of higher index |
| 6.1 | nth roots, and rational exponents | even-index roots of negatives |
| 6.2 | properties of rational exponents | nothing new about existence |
The one gap this lesson leaves is the same one Lesson 4.6 filled for index 2: an even root of a negative number is not real.
Pattern
One routine for evaluating, one for solving.
On a calculator, put brackets around the whole rational exponent, or the machine computes a power and then divides.
OpenStax Algebra and Trigonometry 2e, §1.3 Radicals and Rational Exponents §1.3
Check
Counting roots. Parity first.
Check your understanding
How many real fourth roots does 81 have, and what are they?
Answer: A
Why: The index 4 is even and 81 is positive, so there are two real roots, and both 3 and -3 raise to 81.
Check
Rational exponents. Denominator is the index.
Check your understanding
Evaluate 32^(-3/5).
Answer: A
Why: The fifth root of 32 is 2, cubed is 8, and the negative exponent gives the reciprocal.
Check
Solving. Watch the parity.
Check your understanding
Solve (x + 5)^4 = 16.
Answer: A
Why: The fourth root of 16 is +-2, so x + 5 = 2 or -2, giving x = -3 or -7.
Real world
A sphere's volume is four thirds pi r cubed. A weather balloon is inflated until it holds 900 cubic feet of helium.
Discussion prompt
Find its radius, then find how much the radius grows if the volume is doubled, and say why the second answer is not double the first.
Hint: Isolate r cubed and take a cube root.
Answer:
\[ 900 = \tfrac{4}{3}\pi r^3 \;\Longrightarrow\; r^3 = \frac{2700}{4\pi} \approx 214.9 \;\Longrightarrow\; r \approx 5.99 \text{ ft} \]
\[ 1800 = \tfrac{4}{3}\pi r^3 \;\Longrightarrow\; r^3 \approx 429.7 \;\Longrightarrow\; r \approx 7.55 \text{ ft} \]
The radius is about 6 feet, growing to about 7.55 feet when the volume doubles — an increase of about 26 percent, not 100 percent.
The reason is the cube root: doubling the volume multiplies the radius by the cube root of 2, which is about 1.26. This is Lesson 5.1's scaling rule running backwards, and it is why the coral cod's length grew so slowly with its weight. Whenever a quantity depends on a cube, recovering the length means taking a cube root, and cube roots flatten large changes into small ones.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does the cube root of negative 8 exist as a real number?
Correct: Yes — it is negative 2.
\[ (-2)^3 = -8 \;\Longrightarrow\; \sqrt[3]{-8} = -2 \]
Why: Negative 2 cubed is negative 8, since three negative factors leave the product negative. The rule that a root of a negative number fails applies only to an EVEN index, where the pairing of factors forces a non-negative result. This is one of the most useful distinctions in the lesson, because Lesson 6.5's cube-root function is defined for every real number while the square-root function is not.
Explain it
They know square roots and have just met an exponent of two thirds.
Discussion prompt
In four sentences or fewer, explain what a to the two thirds means and why the notation was chosen that way.
Hint: Use the power of a power rule.
Answer:
The denominator says which root to take and the numerator says which power to raise it to, so a to the two thirds is the cube root of a, squared. The choice is not arbitrary: if the power of a power rule is to keep working, then a to the one third cubed has to be a, which makes a to the one third a cube root.
Once that is fixed, a to the two thirds is forced to be the cube root squared, because two thirds is two times one third. Every rule you already know about exponents then applies to roots without any changes.
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 counting, read the parity of the index before anything else. For the index, remember that the denominator sits below and so does the index. For the sign, ask whether the index is even the moment you take a root. For calculators, bracket the whole exponent every time, even when it looks unnecessary. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference for nth roots. Top left: draw the six-cell table of how many real nth roots a number has, and beside each cell write one worked instance of your own. Top right: write out the derivation showing why a to the one over n must be an nth root, and then extend it to a to the m over n in two lines. Bottom left: evaluate 81 to the three quarters twice, once by taking the root first and once by taking the power first, and note how large the intermediate number gets each way. Bottom right: solve one equation with an odd index and one with an even index, showing exactly where the plus-or-minus does and does not appear. In a margin, write what the calculator does when the brackets around an exponent are left out.
If your two evaluations of 81 to the three quarters disagree, check the second: the fourth root of 531,441 really is 27, and a calculator will confirm it.
Recap
Five things, and the second explains why the notation exists at all.
| If you see | Then |
|---|---|
| An odd index | Exactly one real root, sign matching |
| An even index, positive radicand | Two real roots, plus and minus |
| An even index, negative radicand | No real root |
| An exponent m over n | Index n, power m; take the root first |
| A negative rational exponent | Take the reciprocal |
| A power to undo | Take the root of the same index |
Lesson 6.2 takes Lesson 5.1's seven exponent properties and shows that every one of them holds for rational exponents too, which is what the notation was designed to make possible.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.1 Evaluate nth Roots and Use Rational Exponents §6.1, pp. 414-419 — everything on these slides traces back here
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