Week 12 · Chapter

Chapter 12 — Radicals, Rational Exponents & Radical Equations

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 7 (analyze polynomial and rational functions (end behavior, zeros and multiplicity, domain, and asymptotes), and simplify and solve rational and radical expressions and equations, including expressions with rational exponents, checking for extraneous solutions), Objective 1 (classify real numbers, and simplify and evaluate algebraic expressions using the properties of real numbers, the order of operations, and the rules of integer exponents) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you learn to run a power backwards. If you know a garden's area, a root (the number that, multiplied by itself the right number of times, gives back what you started with) tells you the length of its side. You will learn to tidy up a root so it is easy to read, and to write a root as a fraction in an exponent, called a rational exponent (an exponent that is a fraction, such as two thirds). Then you will solve equations with the unknown hidden under a root. Squaring both sides of such an equation can hand you an answer that never worked, so you will check every answer you find.

Why you already care about this

Picture a square vegetable garden that covers 64 square feet. You want to buy edging for one side. How long is that side? You need the number that, times itself, gives 64. You know it already: 8 feet, because 8 · 8 = 64. You ran a power backwards.

Now picture the object from Weeks 3, 7, and 9, launched straight up from an 80-foot platform. It peaks at 144 feet, 2 seconds after launch, and then falls. An object dropped from rest takes √(h ⁄ 16) seconds to fall h feet, so from 144 feet the fall takes 3 seconds. That matches the landing time you found in Week 7: 2 seconds up plus 3 seconds down is 5 seconds. A safety officer on a building site, a garden designer, and a nurse reading a dosing formula with a fractional power all use this one move: undo a power.

This chapter gives you the rules for roots, a second way to write them, and a careful way to solve equations that contain them.

By the end of this chapter you will be able to:

read a radical by its index and radicand, and evaluate roots of the powers of two;
simplify a square root with the product rule and use the quotient rule on the root of a fraction;
explain, with numbers, why a root of a sum never splits into a sum of roots;
convert between a rational exponent and a radical in both directions, and evaluate it root first;
simplify expressions with rational exponents using the Week 1 exponent rules;
solve a radical equation, check every candidate, and reject an extraneous solution with its reason.

Carry this sentence with you: root first, check last — a squared answer is only a candidate.

Where we are

Already covered: Week 1 set the exponent rules; Week 3 named the radicand; Week 6 built the special products; Week 7 solved quadratics; Week 11 met extraneous solutions.
This week: roots and radicals, the product and quotient rules, rational exponents, and radical equations with their checks.
How they connect: a rational exponent obeys the exact Week 1 rules, and solving a radical equation often ends in a Week 7 quadratic.
Where we're going next: Week 13, Exponential Functions, moves the variable up into the exponent itself, so your exponent rules carry straight into the next unit.

Each earlier week supplies a tool you use here. Week 1 listed the powers of two, from 2⁰ = 1 up to 2¹⁰ = 1024, and defined the principal square root: the nonnegative number whose square is the one under the sign, so √16 = 4. Week 1 also gave you the product rule, the power rule, and the negative-exponent rule. Week 3 said the radicand of a square root must be 0 or more. Week 6 gave you the special product (a + b)² = a² + 2ab + b², so (x + 1)² = x² + 2x + 1. Week 11 showed that a correct solving step can still produce a candidate that fails the original equation. This week every one of those returns.

1. Read a root as the undo of a power

You already undo operations without thinking. Adding 5 is undone by subtracting 5. Doubling is undone by halving. A root is the move that undoes a power. If 8² = 64, then the square root of 64 is 8.

A radical is an expression that uses a root sign, such as √50 or ∛8, to name the number whose power gives the value under the sign.
The index is the small number written in the crook of a root sign, such as the 3 in ∛8. It says which power the root undoes. A square root has index 2, which is usually left unwritten.
The nth root of a, written ⁿ√a, is the number whose nth power is a. For an even index it is the nonnegative such number, and the radicand a must be 0 or more.
A cube root, written ∛a, is the number whose cube is a, so ∛8 = 2 because 2³ = 8.

The radicand keeps the meaning you learned in Week 3: it is the expression under the sign.

Worked example — the square garden. A square garden covers 64 square feet. Find the length of one side.

Name the unknown. Call the side s feet. The area of a square is side times side, so s² = 64.
Undo the square. You need the nonnegative number whose square is 64.
Find it. 8 · 8 = 64, so s = √64 = 8.

In words: each side of the garden is 8 feet long. A side cannot be negative, so the principal root is the one you want.

Worked example — the powers of two as a root table. Week 1's powers of two are 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. Read them backwards.

A square root. 2⁴ = 16, and 4² = 16, so √16 = 4.
A bigger square root. 32² = 1024, so √1024 = 32.
A cube root. 2³ = 2 · 2 · 2 = 8, so ∛8 = 2. And 8³ = 512, so ∛512 = 8.
A fourth root. 2⁴ = 2 · 2 · 2 · 2 = 16, so ⁴√16 = 2. And 4⁴ = 256, so ⁴√256 = 4.

In words: every root in this table is a power read in reverse. When a number is a power of two, its roots are easy to find.

In practice (your tools). Open the free Desmos graphing calculator and type sqrt(1024) on one line and cbrt(512) on the next. Desmos shows 32 and 8. Then type 32^2 and 8^3 to see the powers that the roots undo.

Misconception check. Students often read ⁴√16 as 16 divided by 4, which is 4. It is tempting because the index looks like a divisor. The cure: a root asks which number, used as a factor that many times, gives the radicand. Test 4: 4 · 4 · 4 · 4 = 256, not 16. Test 2: 2 · 2 · 2 · 2 = 16. So ⁴√16 = 2.

2. Simplify a square root with the product rule

Think about splitting a bill. If 4 friends each owe the same share of a total, you can split first or add first and still land on the same amount. Roots have a rule like that for products: you can take the root of each factor separately.

The product rule for radicals says that for a and b that are 0 or more, √(ab) = √a · √b. The root of a product equals the product of the roots.
A perfect square is a number that is the square of a whole number, such as 16, 25, or 64.
A square root is in simplified radical form when no perfect-square factor other than 1 is left under the root sign.

Worked example — testing the rule. Compare √(9 · 16) with √9 · √16.

Root of the product. 9 · 16 = 144, and √144 = 12.
Product of the roots. √9 = 3 and √16 = 4, so √9 · √16 = 3 · 4 = 12.

In words: both orders give 12. Your two results agree, so the rule holds here.

Worked example — simplify √50. Pull out the largest perfect-square factor.

Find the factor. The perfect squares up to 50 are 1, 4, 9, 16, 25, 36, and 49. The largest that divides 50 is 25, because 50 = 25 · 2.
Split the root. √50 = √(25 · 2) = √25 · √2.
Take the root you can. √25 = 5, so √50 = 5√2.
Check exactly. (5√2)² = 5² · (√2)² = 25 · 2 = 50.
Check with a decimal. Using the pinned √2 ≈ 1.4142, 5√2 ≈ 5 × 1.4142 ≈ 7.071.

In words: √50 = 5√2, which is about 7.071. A square garden of 50 square feet would have sides of exactly 5√2 feet, a little over 7 feet. The exact form is the answer; the decimal is your check.

Worked example — simplify √48. Use the same move on a second number.

Find the largest perfect-square factor. 48 = 16 · 3, and 16 is a perfect square.
Split and take the root. √48 = √(16 · 3) = √16 · √3 = 4√3.
Check. (4√3)² = 16 · 3 = 48.

In words: √48 = 4√3, with only 3 left under the root.

Letters work the same way. For a and b that are 0 or more, √(a²b) = √(a²) · √b = a√b. So √(25x) = 5√x when x ≥ 0. Check at x = 4: √(25 · 4) = √100 = 10, and 5√4 = 5 · 2 = 10. Both sides match.

In practice (your tools). In Desmos, type sqrt(50) on one line and 5*sqrt(2) on the next. Both lines show the same decimal, about 7.071. Two expressions that match at every digit Desmos shows are a good sign you simplified correctly; your exact check, (5√2)² = 50, is the proof.

Misconception check. Students often simplify √48 as 2√12 and stop. It is tempting because 48 = 4 · 12 is the first split that comes to mind. The cure: look under the root again. 12 = 4 · 3, so 2√12 = 2 · 2√3 = 4√3. Start with the largest perfect-square factor, here 16, and you finish in one step.

3. Use the quotient rule: the dropped object

Fractions have a matching rule. Picture a recipe card that says "split the batch into equal parts". Splitting a root works the same way for division as it did for multiplication.

The quotient rule for radicals says that for a that is 0 or more and b greater than 0, √(a ⁄ b) = √a ⁄ √b. The root of a fraction equals the root of the top over the root of the bottom.

Here is where that rule earns its keep. An object dropped from rest falls h feet in t = √(h ⁄ 16) seconds. The 16 is the same 16 as in the launched object's height h(t) = −16t² + 64t + 80: half of gravity's pull of 32 feet per second per second. The formula ignores air resistance, as the launched-object model does.

Worked example — the fall from the peak. The launched object peaks at 144 feet. How long does it fall from there?

Substitute. t = √(144 ⁄ 16).
Use the quotient rule. √(144 ⁄ 16) = √144 ⁄ √16 = 12 ⁄ 4 = 3.
Check by dividing first. 144 ⁄ 16 = 9, and √9 = 3. Same answer.
Tie it to the launched object. The object peaks at t = 2 seconds and lands at t = 5 seconds, and 5 − 2 = 3 seconds of falling. Check the landing: −16(5)² + 64(5) + 80 = −400 + 320 + 80 = 0.

In words: the fall from the peak takes 3 seconds. A site supervisor who knows that number can decide how long a drop zone must stay clear. Two different formulas, one radical and one quadratic, agree on the same 3 seconds.

Misconception check. Students often take the root of only the bottom: √(144 ⁄ 16) becomes 144 ⁄ 4 = 36. It is tempting because 16 is the familiar perfect square and the eye stops there. The cure: the root sign covers the whole fraction, so it applies to the top and the bottom. Check by dividing first: 144 ⁄ 16 = 9, and √9 = 3, not 36.

4. Never split a root over a sum

The product rule and the quotient rule make roots feel friendly. That friendliness tempts many people into a third "rule" that does not exist. A root over a sum does not split.

Test it on numbers you can check in your head.

Worked example — the sum trap. Compare √(9 + 16) with √9 + √16.

Root of the sum. Add inside first: 9 + 16 = 25. Then √25 = 5.
Sum of the roots. √9 = 3 and √16 = 4, so √9 + √16 = 3 + 4 = 7.
Compare. 5 ≠ 7.

In words: √(9 + 16) = 5, not 7. One counterexample is enough to show a "rule" is false.

The radical bar is a grouping symbol, like parentheses. Under the Week 1 order of operations, you finish everything under the bar first. Then you take the root.

Here is the contrast to keep in your notes: √(9 · 16) = 12 = 3 · 4, so roots split over a product; √(9 + 16) = 5, versus 3 + 4 = 7, so roots never split over a sum. The wrong split overshoots the true value by a factor of 7⁄5, which is 1.4 times as large.

In practice (your tools). In Desmos, type sqrt(9+16) and then sqrt(9)+sqrt(16). Desmos shows 5 and 7. Keep this pair handy; it is the fastest way to catch the error in someone else's work.

Misconception check. Students often write √(x² + 9) = x + 3. It is tempting because √(x²) = x and √9 = 3 each look fine alone. The cure: test a number. At x = 4, √(16 + 9) = √25 = 5, but 4 + 3 = 7. The two sides disagree, so the step is wrong. A root over a sum stays as it is.

5. Read a rational exponent as a root

Exponents can be fractions. That sounds strange until you ask what a fraction exponent would have to mean for the Week 1 rules to keep working.

The power rule says (aᵐ)ⁿ = aᵐⁿ. Suppose the exponent is 1⁄2. Then (a^(1⁄2))² = a^(1⁄2 · 2) = a¹ = a. So a^(1⁄2) is a number whose square is a. That is exactly what √a is. The rules force the meaning.

A rational exponent is an exponent that is a fraction, such as the 2⁄3 in 8^(2⁄3). The denominator names the root and the numerator names the power, so a^(m⁄n) = (ⁿ√a)^m, the nth root of a raised to the power m.

In particular, a^(1⁄n) = ⁿ√a. So 16^(1⁄2) = √16 = 4 and 8^(1⁄3) = ∛8 = 2. On a single line, the whole exponent sits inside the parentheses after the caret, as in 8^(2⁄3).

Worked example — 8^(2⁄3), root first. Read the exponent, then work in order.

Read the exponent. The denominator 3 means a cube root; the numerator 2 means a square.
Root first. ∛8 = 2.
Then the power. 2² = 4. So 8^(2⁄3) = (∛8)² = 4.
Check with the powers of two. 8 = 2³, so 8^(2⁄3) = (2³)^(2⁄3) = 2^(3 · 2⁄3) = 2² = 4.

In words: 8^(2⁄3) = 4.

Worked example — 16^(3⁄4) and 27^(2⁄3). Two more, with the same habit.

16^(3⁄4). Root first: ⁴√16 = 2. Then 2³ = 8. So 16^(3⁄4) = 8. Check: 16 = 2⁴, so (2⁴)^(3⁄4) = 2^(4 · 3⁄4) = 2³ = 8.
27^(2⁄3). Root first: ∛27 = 3, because 3³ = 27. Then 3² = 9. So 27^(2⁄3) = 9.

In words: 16^(3⁄4) = 8 and 27^(2⁄3) = 9.

A negative rational exponent. The Week 1 rule a⁻ⁿ = 1 ⁄ aⁿ still holds. So 8^(−2⁄3) = 1 ⁄ 8^(2⁄3) = 1 ⁄ 4. The negative sign asks for a reciprocal, never a negative number.

Converting both directions. √x = x^(1⁄2). ∛(x²) = x^(2⁄3). And x^(3⁄4) = (⁴√x)³. The denominator of the exponent is always the index of the root.

Misconception check. Students often work 16^(3⁄4) as 16 × 3⁄4 = 12. It is tempting because 3⁄4 "of" 16 is a familiar fraction move. The cure: an exponent is never a multiplier. Root first gives 8. A second slip is cubing first: 16³ = 4096, and its fourth root is still 8, because 8⁴ = 4096, but the numbers are far larger and easier to fumble. Honesty note: cubing first is not wrong, only riskier; root first is the course's habit because it keeps numbers small.

6. Simplify with rational exponents

Every Week 1 exponent rule still applies when the exponents are fractions. The only new skill is fraction arithmetic, which you have done since grade school when adding parts of a pizza.

Worked example — the product rule, x^(1⁄2) · x^(1⁄3). Two powers of the same base are multiplied.

Same base, multiplying. The product rule says add the exponents: x^(1⁄2 + 1⁄3).
Find a common denominator. It is 6. So 1⁄2 = 3⁄6 and 1⁄3 = 2⁄6.
Add. 3⁄6 + 2⁄6 = 5⁄6. The product is x^(5⁄6).
Check at a power of two, x = 64. 64^(1⁄2) · 64^(1⁄3) = 8 · 4 = 32. And 64^(5⁄6) = (⁶√64)⁵ = 2⁵ = 32, because 2⁶ = 64. Your two sides agree.

In words: x^(1⁄2) · x^(1⁄3) = x^(5⁄6).

Worked example — the power rule, (x^(3⁄2))². Here a power is raised to a power.

A power raised to a power. Multiply the exponents: 3⁄2 · 2 = 3.
Write it. (x^(3⁄2))² = x³.
Check at x = 4. 4^(3⁄2) = (√4)³ = 2³ = 8, and 8² = 64. Also 4³ = 64. They match.

In words: (x^(3⁄2))² = x³.

Worked example — the quotient rule, x^(5⁄4) ⁄ x^(1⁄4). Now one power divides another.

Same base, dividing. Subtract the exponents: 5⁄4 − 1⁄4 = 4⁄4 = 1.
Write it. x^(5⁄4) ⁄ x^(1⁄4) = x¹ = x.
Check at x = 16. 16^(5⁄4) = (⁴√16)⁵ = 2⁵ = 32, and 16^(1⁄4) = 2. Then 32 ⁄ 2 = 16, which is x.

In words: x^(5⁄4) ⁄ x^(1⁄4) = x.

In practice (your tools). In Desmos, type y = x^(1/2) * x^(1/3) and, on the next line, y = x^(5/6). The two curves sit exactly on top of each other for every x that is 0 or more. Then add y = x^(1/6) and watch it peel away; that is the wrong rule made visible.

Misconception check. Students often multiply the exponents in x^(1⁄2) · x^(1⁄3) and write x^(1⁄6). It is tempting because the power rule does multiply exponents. The cure: multiply exponents only for a power of a power; add them for a product of two powers with the same base. At x = 64 the wrong answer gives ⁶√64 = 2, not 32.

7. Solve a radical equation

A radical equation hides the unknown under a root. The plan has four steps, and the first one matters most.

A radical equation is an equation with the variable under a root sign, such as √(2x − 1) = 3.
To isolate the radical is to rearrange the equation so the root stands alone on one side before you raise both sides to a power.

The four steps: isolate the radical; raise both sides to the index (square both sides for a square root); solve; check every candidate in the original equation.

Worked example — √x = 5. The radical is already alone.

Square both sides. (√x)² = 5², so x = 25.
Check. √25 = 5. It works.

In words: x = 25. In the garden setting, a square bed with a 5-foot side has an area of 25 square feet.

Worked example — √(2x − 1) = 3. The radical is alone again.

Square both sides. 2x − 1 = 9.
Add 1 to both sides. 2x = 10.
Divide by 2. x = 5.
Check. √(2 · 5 − 1) = √(10 − 1) = √9 = 3. It works.

In words: x = 5.

Worked example — isolate first, √(x + 1) + 2 = 6. This time the 2 is in the way.

Isolate the radical. Subtract 2 from both sides: √(x + 1) = 4.
Square both sides. x + 1 = 16.
Subtract 1. x = 15.
Check in the original. √(15 + 1) + 2 = √16 + 2 = 4 + 2 = 6. It works.

In words: x = 15.

Worked example — the dropped object in reverse. How high must a drop start so that the fall lasts 3 seconds?

Set up. 3 = √(h ⁄ 16).
Square both sides. 9 = h ⁄ 16.
Multiply both sides by 16. h = 144 feet.
Check. √(144 ⁄ 16) = √9 = 3.

In words: the drop starts at 144 feet, the launched object's peak.

In practice (your tools). In Desmos, type y = sqrt(2x - 1) and y = 3 on separate lines. Click where they cross. Desmos labels the crossing point (5, 3), and its first coordinate matches the x your algebra gave.

Misconception check. Students often square √(x + 1) + 2 = 6 term by term and write (x + 1) + 4 = 36, which gives x = 31. It is tempting because squaring "both sides" feels like squaring each piece. The cure: isolate the radical first. Check x = 31 in the original: √32 is between 5 and 6, so the left side is more than 7, not 6. The term-by-term squaring broke the equation.

8. Check every candidate: extraneous solutions

Squaring is a powerful move, but it forgets signs. Both 1 and −1 square to 1. So squaring can turn a false statement into a true one, and a candidate can sneak through. In Week 11 you met an extraneous solution as a candidate from a correct step that fails the original equation. Radical equations produce them too.

Worked example — √(x + 3) = x + 1. Watch for a candidate that fails.

The radical is alone on the left. Square both sides: x + 3 = (x + 1)².
Expand with the Week 6 special product. (x + 1)² = x² + 2x + 1, so x + 3 = x² + 2x + 1.
Get 0 on one side. Subtract x and 3 from both sides: 0 = x² + x − 2.
Factor. You need two numbers with product −2 and sum 1. They are 2 and −1, so (x + 2)(x − 1) = 0.
Solve. x = −2 or x = 1. These are candidates, not yet answers.
Check x = 1. √(1 + 3) = √4 = 2, and 1 + 1 = 2. Both sides are 2, so you keep x = 1.
Check x = −2. √(−2 + 3) = √1 = 1, but −2 + 1 = −1. Since 1 ≠ −1, x = −2 fails.

In words: the only solution is x = 1. The candidate x = −2 is extraneous, because the right side would be −1, and a principal square root is never negative.

Why did −2 appear? Squaring turned the false statement 1 = −1 into the true statement 1 = 1. Picture the two graphs. The curve y = √(x + 3) starts at (−3, 0) and rises slowly to the right. The line y = x + 1 rises more steeply. They cross only once, at (1, 2). Caveat: the graphs settle which candidates are real, but your written check is still the proof the course asks for.

A radical equal to a negative number. √x = −3 has no solution, because a principal square root is never negative. If you square anyway, you get x = 9, and √9 = 3, not −3. So 9 is extraneous, and you write no solution.

In practice (your tools). In Desmos, type y = sqrt(x + 3) and y = x + 1. Click the crossing point; Desmos labels (1, 2). There is no crossing at x = −2. Now add y = -sqrt(x + 3); it meets the line at (−2, −1). That lower curve is the equation the extraneous candidate really solves.

Misconception check. Students often keep both candidates, x = 1 and x = −2, because "the algebra produced both". It is tempting because every step was correct. The cure: squaring is a one-way door, so every candidate goes back into the original equation. A second slip lives in step 2: writing (x + 1)² as x² + 1 drops the middle term 2x and leads to the wrong equation x² − x − 2 = 0. Expand with the special product every time.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

(a) Think of a place in your own life or work where you would turn an area, a time, or a size back into a length or a rate. Describe the root you would take, and say how you would check the answer makes sense.
(b) Pick the trap from this chapter you are most likely to fall into: splitting a root over a sum, multiplying exponents in a product, or keeping an extraneous candidate. Describe a short habit that would catch it before it costs you points.

Key terms

Radical — An expression that uses a root sign, such as √50 or ∛8, to name the number whose power gives the value under the sign; the root sign itself is also called the radical sign. (first used: 1. Read a root as the undo of a power)
Index — The small number written in the crook of a root sign, such as the 3 in ∛8, that says which power the root undoes; a square root has index 2, which is usually not written. (first used: 1. Read a root as the undo of a power)
Nth root — For a whole number n of 2 or more, the nth root of a, written ⁿ√a, is the number whose nth power is a; for an even n it is the nonnegative such number and needs a to be 0 or more. (first used: 1. Read a root as the undo of a power)
Cube root — The number whose cube is a given number, written ∛a, so ∛8 = 2 because 2³ = 8; a cube root has index 3. (first used: 1. Read a root as the undo of a power)
Product rule for radicals — For a and b that are 0 or more, √(ab) = √a · √b: the root of a product equals the product of the roots, which lets you pull a perfect-square factor out from under the root. (first used: 2. Simplify a square root with the product rule)
Perfect square — A number that is the square of a whole number, such as 16, 25, or 64; it is the kind of factor you pull out from under a square root. (first used: 2. Simplify a square root with the product rule)
Simplified radical form — A square root written with no perfect-square factor other than 1 left under the root sign, such as 4√3 for √48; 2√12 is not yet in this form because 12 still contains the perfect square 4. (first used: 2. Simplify a square root with the product rule)
Quotient rule for radicals — For a that is 0 or more and b greater than 0, √(a ⁄ b) = √a ⁄ √b: the root of a fraction equals the root of the numerator over the root of the denominator. (first used: 3. Use the quotient rule: the dropped object)
Rational exponent — An exponent that is a fraction, such as the 2⁄3 in 8^(2⁄3); the denominator names the root and the numerator names the power, so a^(m⁄n) means the nth root of a, raised to the power m. (first used: 5. Read a rational exponent as a root)
Radical equation — An equation with the variable under a root sign, such as √(2x − 1) = 3; it is solved by isolating the root, raising both sides to the index, and checking every candidate in the original equation. (first used: 7. Solve a radical equation)
Isolate the radical — To rearrange a radical equation so the root stands alone on one side before raising both sides to a power, as when √(x + 1) + 2 = 6 becomes √(x + 1) = 4. (first used: 7. Solve a radical equation)

Summary

A root undoes a power; the index says which power.
Simplify a square root by pulling out its largest perfect-square factor.
A root splits over a product or a quotient, never over a sum.
A rational exponent is a root: the denominator is the root, the numerator is the power.
Take the root first; it keeps the numbers small.
A negative exponent asks for a reciprocal, never a negative number.
With fractional exponents, add for a product, multiply for a power of a power, subtract for a quotient.
To solve a radical equation, isolate the radical before you square.
Squaring can create a false candidate, so check every candidate in the original equation.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

A square quilt block covers 75 square inches. Write the length of its side exactly, in simplified radical form, and check your answer by squaring it.
Evaluate 125^(2⁄3) root first, and then write ∛(y²) with a rational exponent.
Simplify y^(2⁄3) · y^(1⁄4) into one power of y.
Solve √(x + 10) = x − 2. Check both candidates and say which one, if any, is extraneous.

Answers (worked reasoning — read after attempting)

The side s satisfies s² = 75, so s = √75. The largest perfect-square factor of 75 is 25, because 75 = 25 · 3. So √75 = √25 · √3 = 5√3 inches. Check: (5√3)² = 25 · 3 = 75. Splitting 75 as 5 · 15 shows no perfect square at all, which is why you list the perfect squares up to 75 first.
The denominator 3 means a cube root, and ∛125 = 5 because 5 · 5 · 5 = 125. Then the numerator 2 means a square: 5² = 25. So 125^(2⁄3) = 25. Multiplying 125 by 2⁄3 would be the exponent-as-multiplier slip. For the second part, the index 3 becomes the denominator and the power 2 becomes the numerator: ∛(y²) = y^(2⁄3).
Same base, multiplying, so add the exponents: 2⁄3 + 1⁄4. The common denominator is 12, so 2⁄3 = 8⁄12 and 1⁄4 = 3⁄12. Then 8⁄12 + 3⁄12 = 11⁄12, and the product is y^(11⁄12). Multiplying the exponents instead would give 2⁄12 = 1⁄6, which is the power rule used in the wrong place.
The radical is alone, so square both sides: x + 10 = (x − 2)² = x² − 4x + 4. Subtract x and 10: 0 = x² − 5x − 6. Two numbers with product −6 and sum −5 are −6 and 1, so (x − 6)(x + 1) = 0, and x = 6 or x = −1. Check x = 6: √(6 + 10) = √16 = 4, and 6 − 2 = 4, so x = 6 works. Check x = −1: √(−1 + 10) = √9 = 3, but −1 − 2 = −3, and 3 ≠ −3. So x = −1 is extraneous, and the only solution is x = 6.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the power rule you used here carries into exponential functions.

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