MGU Mathematics Series — Algebra I
Part V: Quadratics, Radicals, and Data
Roots · Simplification · Pythagorean Theorem · Rational Exponents
Chapter 13 of 14
MGU Mathematics Series  ·  Algebra I  ·  Library Textbook Edition

Chapter 13:
Radicals and
Rational Exponents

Square roots, cube roots, simplification, radical operations, rational exponents, the Pythagorean theorem, the distance formula, and radical equations — the inverse of powers

■ Math Level: Algebra I ■

■ 14 Sections  •  Part V of 5 ■

Follows Chapter 12: Quadratic Functions and Equations  ·  Prepares students for Chapter 14 and Algebra II

Chapter 13 — Front Matter

Purpose and Overview

This chapter introduces square roots, cube roots, and radicals as inverse operations for powers. Every rule about exponents you have learned points toward this chapter: if squaring a number produces its square, then taking a square root reverses that process. If cubing a number means multiplying it by itself three times, then taking a cube root asks what number was cubed to produce a given result.

Radicals are not a separate topic sitting off to the side of algebra. They arise naturally from equations, geometry, measurement, physics, engineering, and finance. When you solve a quadratic equation using square roots, you are working with radicals. When you compute the length of a diagonal with the Pythagorean theorem, you are working with radicals. When a scientist computes the period of a pendulum or an engineer checks the dimensions of a structural beam, radicals appear.

This chapter will teach you to read, simplify, and manipulate radical expressions. It will show you how to add, subtract, multiply, and divide radicals. It will introduce rational exponents as a unified notation. It will connect radicals to the Pythagorean theorem and to distance on the coordinate plane. And it will prepare you for the quadratic formula and for deeper radical work in Algebra II.

Opening Question

A square garden has an area of 200 square feet. You want to know the length of one side so you can purchase fencing. You know that area equals side times side, so you need a number that, when squared, gives 200. Using a calculator you might find an approximate decimal. But is there a way to write the exact answer as a simplified expression? What does that expression mean, and why does it look the way it does? By the end of this chapter, you will be able to answer that question — and many others like it.

14Sections
12Objectives
19Key Terms
20+Worked Examples

Learning Objectives

  1. 1Evaluate and interpret square roots and cube roots.
  2. 2Identify and apply principal square roots correctly.
  3. 3Identify perfect squares and perfect cubes.
  4. 4Simplify square root and cube root expressions using the product property.
  5. 5Add and subtract like radical expressions by combining coefficients.
  6. 6Multiply radical expressions and simplify results.
  7. 7Divide radical expressions and rationalize denominators.
  8. 8Translate between radical notation and rational exponent notation.
  9. 9Apply the Pythagorean theorem to find missing side lengths of right triangles.
  10. 10Compute the distance between two points on the coordinate plane.
  11. 11Solve simple radical equations and check for extraneous solutions.
  12. 12Recognize and avoid the most common radical mistakes.
Chapter 13 — Contents

Sections at a Glance

Sections 13.1 – 13.5
Roots, Perfect Values, and Simplification
Square Roots · Cube Roots · Principal Roots · Perfect Squares and Cubes · Simplifying
Section 13.1

Square Roots and Cube Roots

An exponent tells you to multiply a number by itself a certain number of times. A root asks the reverse question. When you write 5² = 25, you are saying that five squared equals twenty-five. When you ask for the square root of 25, you are asking: what number, multiplied by itself, equals 25? The answer is 5, because 5 · 5 = 25.

When you write 3³ = 27, you are saying that three cubed equals twenty-seven. The cube root of 27 asks: what number, multiplied by itself three times, equals 27? The answer is 3, because 3 · 3 · 3 = 27.

Notation

√25 = 5   (square root of 25; index 2 is omitted by convention)

∛27 = 3   (cube root of 27; index 3 is written in the notch)

The small number in the notch of the radical sign is the index. For square roots, the index is 2 but is almost always left unwritten. When you see √ with no index, it means a square root.

Root TypeWhat It AsksExamples
Square rootWhat number squared gives the radicand?√9 = 3, √64 = 8, √0 = 0
Cube rootWhat number cubed gives the radicand?∛8 = 2, ∛125 = 5, ∛(−8) = −2
Cube Roots of Negative Numbers

Cube roots of negative numbers are real. ∛(−8) = −2 because (−2)³ = −8. A negative number raised to an odd power remains negative. Square roots of negative numbers, however, are not real numbers at this level. There is no real value whose square is −9.

Example 13.1.1
Evaluating Square Roots and Cube Roots
  • ProblemEvaluate: (a) √49   (b) ∛64   (c) ∛(−125)   (d) √100
  • ThinkFor each, identify what number times itself (or cubed) produces the radicand.
  • Solve (a) √49 = 7 because 7² = 49 (b) ∛64 = 4 because 4³ = 64 (c) ∛(−125) = −5 because (−5)³ = −125 (d) √100 = 10 because 10² = 100
  • CheckSquaring or cubing each answer returns the original radicand.
Answer: 7, 4, −5, 10.
Section 13.2

Principal Square Roots

Every positive number has two square roots: one positive and one negative. Both 5 and −5 are square roots of 25, because 5² = 25 and (−5)² = 25. To remove this ambiguity, mathematics defines the principal square root as the non-negative square root. When you write √25 using the radical sign, the answer is always the non-negative value: √25 = 5, not −5.

If you want the negative square root, write −√25 = −5. If you want both square roots simultaneously, write ±√25 = ±5. This ± notation appears frequently when solving quadratic equations by the square root method.

ExpressionValueReason
√255Principal (positive) square root
−√25−5Negative of the principal square root
±√25±5Both square roots
√(−25)Not realNo real number squared gives a negative
Student Note

The radical sign √ always means the principal (non-negative) square root unless a negative sign is explicitly written in front of it. Keeping this clear prevents many errors, especially when solving equations.

Example 13.2.1
Identifying Principal Square Roots
  • ProblemEvaluate: (a) √81   (b) −√81   (c) ±√81
  • ThinkThe principal square root of 81 is positive. A negative sign in front changes the result. The ± symbol includes both.
  • Solve(a) √81 = 9 (b) −√81 = −9 (c) ±√81 = ±9
Answer: 9, −9, ±9.
Section 13.3

Perfect Squares and Perfect Cubes

A perfect square is a whole number equal to the square of another whole number. A perfect cube is a whole number equal to the cube of another whole number. Memorizing these tables dramatically speeds up every simplification in this chapter.

nn² (Perfect Squares)nn³ (Perfect Cubes)
1111
2428
39327
416464
5255125
6366216
7497343
8648512
9819729
10100101000
11121
12144
13169
14196
15225
Student Tip

Notice that 64 appears in both lists: 64 = 8² and 64 = 4³. Numbers can be both perfect squares and perfect cubes. Most whole numbers, however, are neither — and those are the ones that require simplification rather than direct evaluation.

Section 13.4

Simplifying Square Roots

A square root is considered simplified when no perfect square factor greater than 1 remains under the radical sign, no fraction appears under the radical, and no radical expression appears in the denominator. The first condition is addressed here. The others are covered in Sections 13.8 and 13.9.

Product Property of Square Roots

√(ab) = √a · √b   where a ≥ 0 and b ≥ 0

The square root of a product equals the product of the square roots. Use this to split a radicand into a perfect square part and a remaining part: √20 = √(4 · 5) = √4 · √5 = 2√5.

Strategy: Always Choose the Largest Perfect Square Factor

If you choose a smaller factor, you may need to simplify again. For √72, choosing 4 gives 2√18 — not yet simplified since 18 = 9 · 2. Choosing 36 directly gives 6√2. Find the largest perfect square factor to finish in one step.

Example 13.4.1
Simplifying a Square Root
  • ProblemSimplify √72.
  • ThinkThe largest perfect square factor of 72 is 36, since 72 = 36 · 2.
  • Solve√72 = √(36 · 2) = √36 · √2 = 6√2
  • Check(6√2)² = 36 · 2 = 72. ✓
Answer: √72 = 6√2
Example 13.4.2
Simplifying √48
  • ProblemSimplify √48.
  • Think48 = 16 · 3, and 16 is the largest perfect square factor.
  • Solve√48 = √(16 · 3) = √16 · √3 = 4√3
  • Check(4√3)² = 16 · 3 = 48. ✓
Answer: √48 = 4√3
Example 13.4.3
Simplifying a Radical with a Variable
  • ProblemSimplify √(50x²), assuming x ≥ 0.
  • Think50 = 25 · 2; x² is itself a perfect square. Both factor out.
  • Solve√(50x²) = √(25 · 2 · x²) = √25 · √(x²) · √2 = 5 · x · √2 = 5x√2
  • Check(5x√2)² = 25x² · 2 = 50x². ✓
Answer: √(50x²) = 5x√2
Section 13.5

Simplifying Cube Roots

Cube roots simplify using the same logic: find the largest perfect cube factor and separate it from the remaining factor. The product property for cube roots is ∛(ab) = ∛a · ∛b.

To simplify ∛54: find the largest perfect cube factor of 54. Since 54 = 27 · 2 and 27 = 3³, use 27: ∛54 = ∛(27 · 2) = ∛27 · ∛2 = 3∛2.

Cube roots of negative numbers are simplified the same way: pull the negative out first, then simplify the absolute value. ∛(−54) = −∛54 = −3∛2.

Example 13.5.1
Simplifying a Cube Root
  • ProblemSimplify ∛250.
  • Think250 = 125 · 2, and 125 = 5³ is a perfect cube.
  • Solve∛250 = ∛(125 · 2) = ∛125 · ∛2 = 5∛2
  • Check(5∛2)³ = 125 · 2 = 250. ✓
Answer: ∛250 = 5∛2
Example 13.5.2
Simplifying a Negative Cube Root
  • ProblemSimplify ∛(−216).
  • Think216 = 6³ is a perfect cube, so the cube root evaluates exactly.
  • Solve∛(−216) = −∛216 = −6
  • Check(−6)³ = −216. ✓
Answer: ∛(−216) = −6
Practice 13.1 – 13.5Simplification Practice

Simplify each radical completely. Show the factoring step.

  1. √18
  2. √45
  3. √98
  4. √200
  5. √128
  6. √180
  7. √(75x²), x ≥ 0
  8. ∛40
  9. ∛(−250)
  10. ∛192
Sections 13.6 – 13.10
Operations, Rationalization, and Rational Exponents
Adding · Subtracting · Multiplying · Dividing · Rationalizing · Rational Exponents
Section 13.6

Adding and Subtracting Radicals

You can only add or subtract radicals that are like radicals — radicals with the same index and the same radicand. This is the same principle as combining like terms: three groups of √5 plus seven groups of √5 equals ten groups of √5.

When two radical terms look unlike, simplify each one first. After simplification they may become like radicals and can then be combined. Always simplify before attempting to add or subtract.

Key Principle

3√5 + 7√5 = 10√5   (like radicals — same radicand)

3√5 + 7√3   cannot be simplified — unlike radicands, just as 3x + 7y cannot be combined

Example 13.6.1
Adding Like Radicals
  • ProblemSimplify 5√3 + 9√3 − 2√3.
  • ThinkAll three terms share the same radicand √3. Combine coefficients.
  • Solve5√3 + 9√3 − 2√3 = (5 + 9 − 2)√3 = 12√3
Answer: 12√3
Example 13.6.2
Simplifying Before Adding
  • ProblemSimplify √12 + √75.
  • ThinkNeither radical is simplified. Simplify each one, then check for like terms.
  • Solve √12 = √(4 · 3) = 2√3 √75 = √(25 · 3) = 5√3 2√3 + 5√3 = 7√3
  • CheckDecimal check: √12 ≈ 3.464, √75 ≈ 8.660. Sum ≈ 12.124. And 7√3 ≈ 7(1.732) = 12.124. ✓
Answer: √12 + √75 = 7√3
Example 13.6.3
Unlike Radicands After Simplification
  • ProblemSimplify √18 + √8 + √5.
  • ThinkSimplify each radical, then combine what is possible.
  • Solve √18 = 3√2 √8 = 2√2 √5 = √5 (already simplified) 3√2 + 2√2 + √5 = 5√2 + √5
  • Note5√2 and √5 have different radicands. They cannot be combined further.
Answer: 5√2 + √5
Student Warning — Do Not Add Under the Radical

√4 + √9 ≠ √13. Evaluate: √4 + √9 = 2 + 3 = 5, but √13 ≈ 3.6. These are not equal. You cannot add numbers under separate radical signs. Radicals can only be combined under the radical sign when they are multiplied, not added.

Section 13.7

Multiplying Radicals

The product property of square roots works in both directions. To multiply two radicals, multiply their radicands, then simplify the result.

Product Property — Multiplication

√a · √b = √(ab)   where a ≥ 0 and b ≥ 0

Multiply coefficients together and radicands together separately, then simplify.

Key identity: √a · √a = a. A radical multiplied by itself equals the radicand.

Example 13.7.1
Multiplying Two Square Roots
  • ProblemMultiply and simplify √6 · √15.
  • Solve √6 · √15 = √(6 · 15) = √90 √90 = √(9 · 10) = 3√10
Answer: √6 · √15 = 3√10
Example 13.7.2
Multiplying Coefficients and Radicals
  • ProblemMultiply and simplify 3√2 · 5√8.
  • ThinkMultiply coefficients together; multiply radicands together.
  • Solve 3√2 · 5√8 = (3 · 5)(√2 · √8) = 15 · √16 = 15 · 4 = 60
  • Check3√2 ≈ 4.243, 5√8 ≈ 14.142. Product ≈ 60.0. ✓
Answer: 60
Example 13.7.3
Distributing with a Radical Factor
  • ProblemExpand and simplify √3(2 + √3).
  • ThinkDistribute √3 to each term inside the parentheses.
  • Solve √3(2 + √3) = √3 · 2 + √3 · √3 = 2√3 + √9 = 2√3 + 3
Answer: 2√3 + 3
Section 13.8

Dividing Radicals

Quotient Property of Square Roots

√(a/b) = √a / √b   where a ≥ 0 and b > 0

You can split a fraction under the radical into separate radicals, or combine a quotient of radicals under one radical. Use whichever direction simplifies more easily.

Example 13.8.1
Simplifying a Radical Quotient
  • ProblemSimplify √(49/25).
  • Solve√(49/25) = √49 / √25 = 7/5
Answer: 7/5
Example 13.8.2
Dividing Radical Expressions
  • ProblemSimplify √200 / √8.
  • ThinkCombine under one radical using the quotient property, then simplify.
  • Solve √200 / √8 = √(200/8) = √25 = 5
  • Check√200 ≈ 14.14, √8 ≈ 2.83. Quotient ≈ 5.0. ✓
Answer: 5
Example 13.8.3
Dividing with Coefficients
  • ProblemSimplify (6√50) / (3√2).
  • Solve (6√50) / (3√2) = (6/3) · (√50/√2) = 2 · √(50/2) = 2 · √25 = 2 · 5 = 10
Answer: 10
Section 13.9

Rationalizing Denominators

A standard convention in mathematics is that simplified radical expressions should not contain radical signs in the denominator. A fraction like 1/√3 is valid, but it is not in simplified form. The process of rewriting such a fraction is called rationalizing the denominator.

The method is to multiply numerator and denominator by the same radical — essentially multiplying by 1 — so that the denominator becomes a rational number. For a square root denominator, multiply by that same square root: 1/√3 · √3/√3 = √3/√9 = √3/3.

Example 13.9.1
Rationalizing a Simple Denominator
  • ProblemRationalize and simplify 5/√7.
  • ThinkMultiply numerator and denominator by √7.
  • Solve 5/√7 · √7/√7 = 5√7 / √49 = 5√7 / 7
Answer: 5√7/7
Example 13.9.2
Simplify the Denominator Before Rationalizing
  • ProblemRationalize and simplify 4/√12.
  • ThinkSimplify √12 first, then rationalize. This is more efficient.
  • Solve √12 = 2√3 4 / 2√3 = 2/√3 2/√3 · √3/√3 = 2√3/3
Answer: 2√3/3
Section 13.10

Rational Exponents as Roots

Exponents do not have to be whole numbers. When an exponent is written as a fraction, it represents a combination of a root and a power. Fractional exponents are called rational exponents.

The connection: a^(1/n) = the nth root of a. So a^(1/2) = √a, a^(1/3) = ∛a, a^(1/4) = the 4th root of a. When the numerator is not 1, the rational exponent a^(m/n) means the nth root of a raised to the mth power — or equivalently, the nth root of a^m. The denominator is always the root index; the numerator is always the power.

Rational Exponent Rule

a^(1/n) = ⁿ√a

a^(m/n) = (ⁿ√a)^m = ⁿ√(aᵐ)

All exponent rules from Chapter 9 still apply when exponents are rational fractions.

Example 13.10.1
Translating Between Notations
  • Problem(a) Write ∛(x²) using rational exponents.   (b) Write y^(3/5) in radical form.   (c) Evaluate 25^(1/2).
  • Solve (a) ∛(x²) = x^(2/3) (b) y^(3/5) = ⁵√(y³) (c) 25^(1/2) = √25 = 5
Answer: x^(2/3), ⁵√(y³), 5.
Example 13.10.2
Evaluating a Rational Exponent
  • ProblemEvaluate 32^(2/5).
  • ThinkDenominator 5 means fifth root. Numerator 2 means square. Take the root first.
  • Solve 32^(2/5) = (⁵√32)² ⁵√32 = 2 because 2⁵ = 32 (2)² = 4
Answer: 4
Example 13.10.3
Simplifying with Exponent Rules
  • ProblemSimplify x^(3/4) / x^(1/4).
  • ThinkUse the quotient rule: subtract exponents.
  • Solve x^(3/4) / x^(1/4) = x^(3/4 − 1/4) = x^(2/4) = x^(1/2) = √x
Answer: √x
MGU Library Connection
  • Rational exponents are revisited in Algebra II in the chapter on radical functions, where expressions like f(x) = x^(2/3) are graphed and analyzed as power functions.
  • They also appear in the Algebra II chapter on exponential functions and in Calculus I when derivatives of power functions are computed using the power rule.
Practice 13.6 – 13.10Radical Operations

Simplify completely. Show each step.

  1. 6√2 + 3√2
  2. √27 + √48
  3. √50 − √8
  4. 2√12 + 3√27
  5. √3 · √12
  6. 4√3 · 2√6
  7. √2(√3 + √8)
  8. √200 / √8
  9. 3/√5
  10. 4/√12
  11. Write x^(3/4) in radical form.
  12. Evaluate 8^(2/3).
Sections 13.11 – 13.12
The Pythagorean Theorem and Distance
Right Triangles · a² + b² = c² · Distance Formula · Midpoint Formula
Section 13.11

The Pythagorean Theorem

A right triangle is a triangle containing one 90° angle. The two sides forming the right angle are called legs, and the side opposite the right angle is called the hypotenuse. The hypotenuse is always the longest side.

The Pythagorean theorem states that for every right triangle:

Pythagorean Theorem

a² + b² = c²

where a and b are the lengths of the two legs and c is the length of the hypotenuse. To find the hypotenuse: c = √(a² + b²). To find a missing leg: a = √(c² − b²).

Diagram Instruction

Draw a right triangle with the right angle at the lower-left corner. Label the horizontal leg a, the vertical leg b, and the diagonal hypotenuse c. Mark the right angle with a small square symbol. Optionally, draw dashed squares on each side to illustrate the area relationship: the area of the square on the hypotenuse equals the combined area of the squares on the two legs.

Example 13.11.1
Finding the Hypotenuse
  • ProblemA right triangle has legs of 6 and 8. Find the hypotenuse.
  • Solve c² = 6² + 8² = 36 + 64 = 100 c = √100 = 10
  • Check6² + 8² = 36 + 64 = 100 = 10². ✓ The set (6, 8, 10) is a Pythagorean triple.
Answer: The hypotenuse is 10 units long.
Example 13.11.2
Finding a Missing Leg
  • ProblemA right triangle has a hypotenuse of 13 and one leg of 5. Find the other leg.
  • Solve a² + 5² = 13² a² + 25 = 169 a² = 144 a = √144 = 12
  • Check5² + 12² = 25 + 144 = 169 = 13². ✓
Answer: The other leg is 12 units long.
Example 13.11.3
Irrational Hypotenuse
  • ProblemA right triangle has legs of 5 and 7. Find the exact hypotenuse length and a decimal approximation.
  • Solve c² = 5² + 7² = 25 + 49 = 74 c = √74
  • Note74 has no perfect square factors greater than 1, so √74 is already fully simplified. As a decimal: √74 ≈ 8.60.
Answer: Exactly √74 units; approximately 8.60 units.
Example 13.11.4
Application — Ladder Problem
  • ProblemA 17-foot ladder leans against a wall. The base is 8 feet from the wall. How high up the wall does the ladder reach?
  • ThinkThe wall, ground, and ladder form a right triangle. The ladder is the hypotenuse. The base distance is one leg.
  • Solve 8² + h² = 17² 64 + h² = 289 h² = 225 h = √225 = 15
  • Check8² + 15² = 64 + 225 = 289 = 17². ✓
Answer: The ladder reaches 15 feet up the wall.
Common Pythagorean Triples

These whole-number sets satisfy a² + b² = c² and are worth recognizing: (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (7, 24, 25). Any multiple of a triple is also a triple — for example, (9, 12, 15) is a multiple of (3, 4, 5).

Example 13.11.5
Testing Whether a Triangle Is a Right Triangle
  • ProblemDo the side lengths 9, 40, and 41 form a right triangle?
  • ThinkThe converse of the Pythagorean theorem: if a² + b² = c², the triangle is a right triangle. Test with 41 as the hypotenuse.
  • Solve 9² + 40² = 81 + 1600 = 1681 41² = 1681 ✓
Answer: Yes, these lengths form a right triangle.
Section 13.12

Distance on the Coordinate Plane

When two points are plotted on the coordinate plane, you can compute the exact distance between them using a formula derived directly from the Pythagorean theorem. Draw a horizontal segment from one point and a vertical segment from the other — these two segments form the legs of a right triangle, and the straight line connecting the original two points is the hypotenuse.

For two points (x₁, y₁) and (x₂, y₂), the horizontal leg has length |x₂ − x₁| and the vertical leg has length |y₂ − y₁|. Applying the Pythagorean theorem gives the distance formula:

Distance Formula

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Since both differences are squared, the order of subtraction does not affect the result.

Diagram Instruction

Draw a coordinate plane with two points: A at (1, 2) and B at (5, 5). Draw a horizontal dashed segment from A to (5, 2) and a vertical dashed segment from (5, 2) to B. Label the horizontal leg "4 units" and the vertical leg "3 units." Draw and label the hypotenuse AB as "d = 5." Mark the right angle at (5, 2) with a small square.

Example 13.12.1
Distance Between Two Points
  • ProblemFind the distance between (1, 2) and (5, 5).
  • Solve d = √((5−1)² + (5−2)²) = √(4² + 3²) = √(16 + 9) = √25 = 5
Answer: The distance is 5 units.
Example 13.12.2
Distance Resulting in a Simplified Radical
  • ProblemFind the exact distance between (−2, 3) and (4, −1).
  • Solve d = √((4−(−2))² + (−1−3)²) = √(6² + (−4)²) = √(36 + 16) = √52 = √(4 · 13) = 2√13
Answer: 2√13 units (approximately 7.21 units).

The Midpoint Formula

The midpoint of a segment is the point exactly halfway between the two endpoints. Its coordinates are the averages of the endpoints' coordinates.

Midpoint Formula

Midpoint = ((x₁ + x₂)/2 , (y₁ + y₂)/2)

Example 13.12.3
Finding a Midpoint
  • ProblemFind the midpoint between (2, 6) and (8, 10).
  • Solve Midpoint = ((2+8)/2 , (6+10)/2) = (10/2 , 16/2) = (5, 8)
Answer: The midpoint is (5, 8).
MGU Library Connection
  • The distance formula and midpoint formula reappear in the MGU Geometry textbook in coordinate geometry chapters on circles, perpendicular bisectors, and coordinate proofs.
  • They also appear in Algebra II's chapter on conic sections, where the distance from a point to a focus is a defining property of each curve.
  • In Calculus, these ideas extend into arc length and surface area calculations.
Practice 13.11 – 13.12Geometry Applications
  1. A right triangle has legs of 9 and 12. Find the hypotenuse.
  2. A right triangle has a hypotenuse of 25 and one leg of 24. Find the other leg.
  3. A right triangle has legs of 7 and 7. Find the exact hypotenuse and a decimal approximation.
  4. Do side lengths 8, 15, and 17 form a right triangle? Show your work.
  5. A park path crosses diagonally through a rectangular block 60 m wide and 80 m long. How long is the path?
  6. Find the distance between (0, 0) and (3, 4).
  7. Find the exact distance between (2, −3) and (−4, 1).
  8. Find the midpoint between (−3, 2) and (5, −8).
Sections 13.13 – 13.14
Radical Equations and Common Mistakes
Solving Radical Equations · Extraneous Solutions · Ten Recurring Errors
Section 13.13

Simple Radical Equations

A radical equation is an equation containing a variable under a radical sign. To solve one, isolate the radical on one side, then eliminate it by raising both sides to the power matching the index: square both sides for square root equations, cube both sides for cube root equations.

This is a powerful technique, but squaring both sides can introduce extraneous solutions — values that satisfy the squared equation but not the original. Squaring erases sign information; for example, squaring √x = −3 produces x = 9, but √9 = 3 ≠ −3. You must always substitute every solution back into the original equation and reject any that do not check out.

Strategy for Solving Radical Equations

1. Isolate the radical on one side of the equation.

2. Raise both sides to the power equal to the index (square for √, cube for ∛).

3. Solve the resulting equation.

4. Check every solution in the original equation. Reject extraneous solutions.

Example 13.13.1
Solving a Simple Square Root Equation
  • ProblemSolve √x = 7.
  • ThinkThe radical is already isolated. Square both sides.
  • Solve (√x)² = 7² x = 49
  • Check√49 = 7. ✓
Answer: x = 49
Example 13.13.2
Solving with Isolation First
  • ProblemSolve √(2x − 3) = 5.
  • ThinkThe radical is isolated. Square both sides.
  • Solve (√(2x−3))² = 5² 2x − 3 = 25 2x = 28 x = 14
  • Check√(2·14 − 3) = √(28 − 3) = √25 = 5. ✓
Answer: x = 14
Example 13.13.3
Discovering an Extraneous Solution
  • ProblemSolve √(x + 4) = −3.
  • ThinkThe right side is negative. The principal square root is always non-negative. This may have no solution. Run the algebra to confirm.
  • Solve (√(x+4))² = (−3)² x + 4 = 9 x = 5
  • Check√(5 + 4) = √9 = 3, not −3. The solution x = 5 is extraneous.
Answer: No solution. The equation has no real solution.
Example 13.13.4
Solving a Cube Root Equation
  • ProblemSolve ∛(x − 4) = 3.
  • ThinkCube both sides to eliminate the cube root.
  • Solve (∛(x−4))³ = 3³ x − 4 = 27 x = 31
  • Check∛(31 − 4) = ∛27 = 3. ✓
Answer: x = 31
Example 13.13.5
Solving with Additional Steps
  • ProblemSolve 2√(x + 1) − 3 = 7.
  • ThinkIsolate the radical first: add 3, then divide by 2. Then square both sides.
  • Solve 2√(x+1) = 10 √(x+1) = 5 x + 1 = 25 x = 24
  • Check2√(24+1) − 3 = 2√25 − 3 = 10 − 3 = 7. ✓
Answer: x = 24
Student Warning — Isolate Before Squaring

Do not square both sides before isolating the radical. Squaring 2√(x+1) − 3 = 7 directly produces (2√(x+1) − 3)² = 49, which expands into a much more complicated expression. Always isolate the radical completely on one side first, then square.

Section 13.14

Common Radical Mistakes

Mistake 1 — Adding Under the Radical
√4 + √9 = √13 Correct: √4 + √9 = 2 + 3 = 5. Evaluate each radical separately, then add. You cannot add the radicands under separate radical signs.
Mistake 2 — Distributing a Root Over a Sum
√(9 + 16) = √9 + √16 = 7 Correct: √(9 + 16) = √25 = 5. The product property is valid; a sum-under-one-radical is not separable the same way.
Mistake 3 — Forgetting to Simplify After Multiplying
√6 · √15 = √90 (left unsimplified) Correct: √90 = √(9 · 10) = 3√10. After multiplying radicals, always check whether the result can be simplified further.
Mistake 4 — Skipping the Extraneous Solution Check
Solving √x = −4: square to get x = 16, and leave it at x = 16. Correct: Check in the original — √16 = 4, not −4. The value x = 16 is extraneous. There is no solution. Always check.
Mistake 5 — Finding c² but Forgetting to Take the Square Root
Legs 3 and 4: c = 3² + 4² = 25 Correct: c² = 25, so c = √25 = 5. The theorem gives c², not c. Always take the square root as the final step.
Mistake 6 — Misidentifying the Hypotenuse
Using one of the shorter sides as c in a² + b² = c². Correct: c must be the hypotenuse — the side opposite the right angle and always the longest side. c² = a² + b² only holds when c is the hypotenuse.
Mistake 7 — Misreading Rational Exponents
a^(2/3) = √(a³) or a^(2/3) = (a²)³ Correct: a^(2/3) = (∛a)². The denominator is always the root index. The numerator is the power. 27^(2/3) = (∛27)² = 9, not √(27³).
Mistake 8 — Leaving a Radical in the Denominator
3/√5 is the final answer. Correct: Rationalize — 3/√5 · √5/√5 = 3√5/5. A radical in the denominator is not a simplified form. Multiply by the appropriate radical over itself.
Mistake 9 — Squaring Before Isolating the Radical
For 2√(x+1) − 3 = 7, squaring immediately: (2√(x+1) − 3)² = 49. Correct: Add 3 to both sides first, then divide by 2, then square. Isolate the radical before squaring.
Mistake 10 — Writing √(x²) = x Without Qualification
√(x²) = x for all x. Correct: √(x²) = |x|. If x = −5, then x² = 25 and √25 = 5 = |−5|. At Algebra I level, problems usually specify x ≥ 0 to avoid this, but be aware of it when variables appear under radical signs.
Practice 13.13Radical Equations

Solve each equation. Check every solution and reject any extraneous results.

  1. √x = 9
  2. √(x + 2) = 4
  3. √(3x − 1) = 5
  4. 3√x = 15
  5. √x + 3 = 7
  6. 2√(x − 1) + 4 = 10
  7. ∛x = 4
  8. ∛(x + 5) = 3
  9. √(x + 1) = −2  (check carefully)
  10. √(4x + 1) = 9
Chapter 13 — Back Matter

Chapter Summary

Key Terms Review

radical
An expression using a root symbol, such as a square root or cube root.
radical sign
The symbol √ used to denote a root.
radicand
The number or expression under a radical sign.
index
The small number in the notch of the radical sign indicating which root is being taken. 2 for square roots (usually omitted), 3 for cube roots.
square root
A number that, when squared, produces the given value.
cube root
A number that, when cubed, produces the given value.
principal square root
The non-negative square root of a number. The radical sign √ always denotes the principal root.
perfect square
A whole number whose square root is also a whole number. Examples: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
perfect cube
A whole number whose cube root is also a whole number. Examples: 1, 8, 27, 64, 125, 216.
simplified radical
A radical expression containing no perfect-square (or perfect-cube) factors in the radicand, no fractions under the radical, and no radicals in any denominator.
like radicals
Radical terms with the same index and the same radicand. Like radicals can be combined by addition or subtraction.
product property
√(ab) = √a · √b (for a, b ≥ 0). Used both to simplify radicals and to multiply them.
quotient property
√(a/b) = √a / √b (for a ≥ 0, b > 0). Used to simplify radical fractions.
rationalizing
The process of rewriting a fraction so that no radical expression remains in the denominator.
rational exponent
A fractional exponent representing a root and a power. a^(m/n) = (ⁿ√a)ᵐ.
Pythagorean theorem
For any right triangle with legs a and b and hypotenuse c: a² + b² = c².
hypotenuse
The side of a right triangle opposite the right angle. Always the longest side.
distance formula
d = √((x₂ − x₁)² + (y₂ − y₁)²). Derived from the Pythagorean theorem.
extraneous solution
A value obtained during solving that does not satisfy the original equation. Must be identified by checking and rejected.

Concept Review Questions

  1. What does it mean to "take the square root" of a number? What question is being asked?
  2. Why is the principal square root defined as the non-negative value?
  3. Explain why √(−16) is not a real number, but ∛(−16) is.
  4. What is a perfect square? Give four examples not shown in the tables.
  5. Explain why √20 can be simplified even though 20 is not a perfect square.
  6. Why can 3√5 + 4√5 be simplified to 7√5, but 3√5 + 4√7 cannot be simplified further?
  7. State the product property of square roots as a formula. Give a numerical example showing it used to simplify a radical.
  8. Why is √(9 + 16) ≠ √9 + √16? What is the correct value of each expression?
  9. What does rationalizing the denominator mean, and why is it the conventional form?
  10. How does the notation a^(1/3) connect to cube roots? Give a numerical example.
  11. State the Pythagorean theorem. Identify what each variable represents.
  12. How is the distance formula related to the Pythagorean theorem?
  13. What is an extraneous solution, and why must you check every answer after solving a radical equation?

Skill Practice

Part AEvaluating Perfect Roots
  1. √144
  2. ∛343
  3. −√121
  4. ∛(−64)
  5. ±√169
  6. ∛512
Part BSimplifying Square Roots
  1. √18
  2. √45
  3. √98
  4. √200
  5. √75
  6. √128
  7. √180
  8. √147
  9. √288
  10. √(50x²), x ≥ 0
Part CAddition and Subtraction
  1. 10√5 − 4√5
  2. √27 + √48
  3. √50 − √8
  4. 2√12 + 3√27
  5. √75 − √12 + √3
  6. 4√18 − 2√50 + √2
  7. √20 + √80 − √5
  8. 6√2 + 3√2 − √8
Part DMultiplication
  1. √3 · √12
  2. √5 · √20
  3. 2√7 · 3√7
  4. 4√3 · 2√6
  5. √2(√3 + √8)
  6. √5(3 − √5)
  7. (2√3)²
  8. ∛9 · ∛3
Part EDivision and Rationalization
  1. √48 / √3
  2. 6√75 / 2√3
  3. 3/√5
  4. 8/√2
  5. 7/√7
  6. 4/√3
  7. 10/√50
  8. 6/√12
Part FRational Exponents
  1. Write x^(3/4) in radical form.
  2. Write ⁵√(y²) with rational exponents.
  3. Evaluate 64^(1/2).
  4. Evaluate 8^(2/3).
  5. Evaluate 27^(4/3).
  6. Simplify a^(5/6) / a^(1/6).

Real-World Applications

For each problem: define any variables, set up an equation or expression, solve, and answer in context.

  1. A square garden has an area of 150 square feet. Find the exact side length and a decimal approximation to the nearest tenth of a foot.
  2. A skateboard ramp forms a right triangle. The base is 12 feet and the ramp surface is 15 feet long. How tall is the ramp?
  3. Two libraries are located at coordinates (3, 7) and (9, 15) on a grid where each unit represents 1 mile. Find the straight-line distance between them.
  4. The time t in seconds for an object to fall a height of h feet is t = √(h/16). How long does it take a ball to fall 400 feet? Give an exact answer and a decimal approximation.
  5. A wire stretches from the top of a 24-foot pole to a point 10 feet from the base. How long is the wire?
  6. A screen is 16 inches wide and 9 inches tall. Find the exact diagonal length and a decimal approximation.
  7. A square floor tile has an area of 576 square centimeters. What is the side length?
  8. Find the exact distance from the origin to the point (5, −12). Does the result surprise you? Why?
  9. The speed of a wave in shallow water is v = √(32d), where d is depth in feet. Simplify this expression. Find the speed when depth is 18 feet.
  10. A park path runs diagonally across a rectangular city block 60 meters wide and 80 meters long. How long is the path?

Error Analysis

Each problem contains an error. Identify what went wrong and write the correct solution.

Problem 1
√16 + √9 = √25 = 5

What is the error? What is the correct answer?

Problem 2
√50 = √(25 + 25) = √25 + √25 = 5 + 5 = 10

Two errors are present. Identify both and write the correct simplified form.

Problem 3
Solving √(x + 3) = −5: square both sides → x + 3 = 25 → x = 22.

What check is missing? Is x = 22 valid? Explain.

Problem 4
√3 · √6 = √9 = 9

What is the error? What is the correct answer?

Problem 5
Simplifying √72: √72 = √(4 · 18) = 2√18

Why is this incomplete? What is the fully simplified form?

Problem 6
A right triangle with legs 5 and 12: c = 5² + 12² = 25 + 144 = 169

What step was omitted? What is the correct value of c?

Problem 7
8^(2/3) = (8²)³ = 64³

What is the correct interpretation of the rational exponent? Evaluate 8^(2/3) correctly.

Problem 8
Rationalizing 3/√5: multiply by √5/√5 to get 3√5/√25 = 3√5/25

What is the error in the final step? Write the correct simplified form.

Chapter 13 Checkpoint

Chapter 13 Checkpoint
Complete independently — assesses all major skills and concepts from Chapter 13
Section A — Evaluating Roots
  1. Evaluate √196.
  2. Evaluate ∛(−27).
  3. Evaluate −√64.
  4. Evaluate 36^(1/2).
  5. Evaluate 125^(2/3).
Section B — Simplifying Radicals
  1. Simplify √108.
  2. Simplify √180.
  3. Simplify ∛(−81).
  4. Simplify √(75x²) for x ≥ 0.
  5. Rewrite m^(3/5) in radical notation.
Section C — Radical Operations
  1. Add: 5√3 + 2√12.
  2. Subtract: √50 − √18.
  3. Multiply: √5 · √45.
  4. Multiply: 3√2(4 + √2).
  5. Divide and simplify: √72 / √2.
  6. Rationalize: 10/√5.
Section D — Pythagorean Theorem and Distance
  1. A right triangle has legs of 8 and 15. Find the hypotenuse.
  2. A right triangle has a hypotenuse of 26 and one leg of 10. Find the other leg.
  3. Find the exact distance between (2, 1) and (8, 9).
  4. Find the midpoint between (−4, 6) and (6, −2).
Section E — Radical Equations
  1. Solve: √x = 11.
  2. Solve: √(x − 5) = 4.
  3. Solve: 3√(x + 2) = 12.
  4. Solve and check: √(2x + 3) = −1.
  5. Solve: ∛(x + 1) = 4.
Section F — Applications
  1. A square lot has an area of 3600 square feet. What is the side length?
  2. A ladder 20 feet long leans against a wall. The base is 12 feet from the wall. How high up the wall does the ladder reach?
  3. The fall time formula is t = √(h/16) seconds. Find the exact time for a fall of 64 feet.
Bridge Note — Looking Ahead
Chapter 14: Data, Statistics, and Algebraic Modeling

Chapter 13 gave you precise tools for exact computation with roots and powers. You can now simplify radical expressions, apply the Pythagorean theorem, calculate distances on the coordinate plane, and solve radical equations. These skills appear throughout all of later algebra and in every applied science.

Chapter 14, the final chapter of Algebra I, shifts focus to data. Rather than working with exact expressions derived from definitions, you will work with real-world measurements — imperfect, variable, and drawn from observation. You will learn how algebra supports data analysis through scatter plots, regression lines, residuals, and model selection.

The connection to this chapter is direct. Measures of spread such as standard deviation involve square roots of averaged squared differences. The distance formula underlies some statistical methods for measuring deviation from trend lines. And the idea of "best fit" is, at its heart, a minimization problem that more advanced mathematics resolves using tools rooted in quadratic and radical algebra. As you move into Chapter 14, bring your full foundation: linear equations from Chapter 7, function notation from Chapter 5, and quadratic patterns from Chapter 12.

MGU Library Connection
  • Radical Rules Reference — Appendix L of this textbook
  • Pythagorean Theorem and distance formula — Algebra II, Chapter 12 (Conic Sections), and MGU Geometry (coordinate geometry chapters)
  • Rational exponents extended — Algebra II, Chapter 8: Radical Functions and Rational Exponents
  • Square roots in the quadratic formula — Algebra II, Chapter 3 and Appendix K of this textbook
  • MGU Physics: kinematics formulas, pendulum period, projectile motion — all involve square roots
  • MGU Statistics: standard deviation and residuals use square root operations derived from this chapter
  • MGU Library formula sheets: Radical Rules Reference, Algebra II Readiness Check