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Exercises 1 When we have a radical expression in the denominator of a fraction whose radicand is not a perfect square — for example, 21 — we can multiply by the appropriate form of 1; in this case, 22=1. This way, we won't change the value of the fraction and we will get rid of the radical term in the denominator. 21ba=b⋅2a⋅22⋅21⋅2a⋅a=a2⋅21⋅2Multiply121a=a2 Notice that the denominator changed from 2 to 1. That is, it changed from an irrational number to a rational number. That is why we call this process rationalizing the denominator. With this in mind, we can complete the exercise sentence.The process of eliminating a radical from the denominator of a radical expression is called rationalizing the denominator. | |

Exercises 2 Let's start by reviewing what the conjugate of a binomial is. The binomials ab+cd and ab−cd where a, b, c, and d are rational numbers are called conjugates. Let's take a look at the expression given in the exercise. 6+4⇔16+41 As we can see, this expression is in the format mentioned above, ab+cd, so we can write its conjugate by changing the positive sign to a negative one. Original binomialConjugate binomial6+46−4 | |

Exercises 3 The exercise asks us to tell if the expressions given are equivalent. 92x=?312x For this it will be useful to remember the property for dividing square roots.The square root of a quotient equals the quotient of the square roots of the numerator and denominator. ba=ba Now, let's see if we can use this to start from one of the given expressions and rewrite it to obtain the other one. 92xba=ba92xCalculate root32xba=b1⋅a312x We have found that 92x=312x. Therefore, the expressions given are equivalent. | |

Exercises 4 Let's start by reviewing what a like radical is. Like radicals are those which have the same index and radicals. Let's now take a look at the radicals given in the exercise. -316 6361 613 -3361 Notice that all the radical expressions involve square roots having 3 as radicand except for 316. This means that all the expressions are like radicals, except for this one. Therefore, this is the odd one out. | |

Exercises 5 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 19 to ensure it's not evenly divisible by any perfect squares by noticing that it is prime. 19=1⋅19 Since it contains no repeating roots, we cannot form any perfect squares. Therefore, the first rule is satisfied. The second and third rules are about fractions, but our expression does not contain any fractions. Therefore, all three rules are satisfied which proves that 19 is in simplest form because it cannot be simplified further. | |

Exercises 6 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 7 to ensure it's not evenly divisible by any perfect squares by noticing that it is prime. 7=1⋅7 Since it contains no repeating roots, we cannot form any perfect squares. Therefore, the first rule is satisfied. The second and third rules are about fractions, and our expression has a radical in the denominator of a fraction. Therefore, all three rules are not satisfied which proves that 71 is not in simplest form because it could be simplified further. | |

Exercises 7 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 48 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 48=2⋅2⋅2⋅2⋅3=22⋅22⋅3 Do you see how there are repeating primes allowing us to form perfect squares? This means that 48 is divisible by some perfect square which tells us that this expression could be simplified further and that 48 is not in simplest form. | |

Exercises 8 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 34 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 34=2⋅17 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions, but our expression does not contain any fractions. Therefore, all three rules are satisfied which proves that 34 is in simplest form because it cannot be simplified further. | |

Exercises 9 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 2 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 2=2⋅1 Since there are no repeating primes, the expression satisfies the first rule. The second rule is about roots being in the denominator. Since our denominator is 2, the rule is not met. Therefore, 25 is not in simplest form because it can be simplified further. | |

Exercises 10 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a square root, for the first rule, n=2. We can check 2 to ensure it's not evenly divisible by any perfect squares by showing its prime factorization. 10=2⋅5 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions. Our expression does does not contain any fractions within the radicand or any radicals within the denominator. Therefore, all rules are satisfied which proves that 4310 is in simplest form because it cannot be simplified further. | |

Exercises 11 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a cube root, for the first rule, n=3. We can check 2 to ensure it's not evenly divisible by any perfect cubes by showing its prime factorization. 2=2⋅1 Since there are no repeating primes, the expression satisfies the first rule. The second and third rules are about fractions, our expression does does not contain any fractions within the radicand. However, it does contain a radical in the denominator. Therefore, all rules are not satisfied which proves that 2+321 is not in simplest form because it can be simplified further. | |

Exercises 12 We can check if the radical is in it's simplest form using the following three rules.The radicand contains no perfect nth powers other than 1. The radicand contains no fractions. There are no radicals within the denominator of a fraction.Since we are dealing with a cube root, for the first rule, n=3. We can check 54 to ensure it's not evenly divisible by any perfect cubes by showing its prime factorization. 54=2⋅3⋅3⋅3=2⋅33 Do you see how there are repeating primes allowing us to form perfect cubes? This means the expression is divisible by some perfect cube which tells us the expression does not satisfy the first rule. Therefore, all rules are not satisfied which proves that 6−354 is not in simplest form because it can be simplified further. | |

Exercises 13 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 20Split into factors4⋅5a⋅b=a⋅b4⋅5Calculate root25 | |

Exercises 14 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 32Split into factors16⋅2a⋅b=a⋅b16⋅2Calculate root42 | |

Exercises 15 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 128Split into factors64⋅2a⋅b=a⋅b64⋅2Calculate root82 | |

Exercises 16 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -72Split into factors-36⋅2a⋅b=a⋅b-36⋅2Calculate root-62 | |

Exercises 17 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 125bSplit into factors25⋅5ba⋅b=a⋅b25⋅5bCalculate root55b | |

Exercises 18 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 4x2Split into factors4⋅x2a⋅b=a⋅b4⋅x2Calculate root2x | |

Exercises 19 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -81m3Split into factors-81⋅m2⋅ma⋅b=a⋅b-81⋅m2⋅mCalculate root-9mm | |

Exercises 20 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 48n5Split into factors16⋅3⋅n4⋅nCommutative Property of Multiplication16⋅n4⋅3⋅na⋅b=a⋅b16⋅n4⋅3nWrite as a power16⋅(n2)2⋅3nCalculate root4n23n | |

Exercises 21 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 494ba=ba494Calculate root72 | |

Exercises 22 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -817ba=ba-817Calculate root-97 | |

Exercises 23 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. -6423ba=ba-6423Calculate root-823 | |

Exercises 24 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 12165ba=ba12165Calculate root1165 | |

Exercises 25 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 49a3ba=ba49a3Split into factors49a2⋅aa⋅b=a⋅b49a2⋅aCalculate root7aa | |

Exercises 26 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. k2144ba=bak2144Calculate rootk12 | |

Exercises 27 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 4x2100ba=ba4x2100Calculate root2x10ba=b/2a/2x5 | |

Exercises 28 We can simplify this expression by rewriting the number under the square root as the product of at least one perfect square factor. 3625v2ba=ba3625v2a⋅b=a⋅b3625⋅v2Calculate root65v | |

Exercises 29 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 316Split into factors38⋅23a⋅b=3a⋅3b38⋅32Calculate root232 | |

Exercises 30 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-108Split into factors3-27⋅43a⋅b=3a⋅3b3-27⋅34Calculate root-334 | |

Exercises 31 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-64x5Split into factors3-64⋅x3⋅x23a⋅b=3a⋅3b3-64⋅3x3⋅3x2Calculate root-4x3x2 | |

Exercises 32 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. -3343n2Split into factors-3343⋅n23a⋅b=3a⋅3b-3343⋅3n2Calculate root-73n2 | |

Exercises 33 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-1256c3ba=3b3a3-12536cCalculate root-536cPut minus sign in front of fraction-536c | |

Exercises 34 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3278h43ba=3b3a32738h4Split into factors32738⋅h3⋅h3a⋅b=3a⋅3b32738⋅3h3⋅3hCalculate root32h3h | |

Exercises 35 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. -31000x381y23ba=3b3a-31000x3381y2Split into factors-31000⋅x3327⋅3⋅y23a⋅b=3a⋅3b-31000⋅3x3327⋅33⋅y2Calculate root-10x333y2 | |

Exercises 36 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. 3-64a3b6213ba=3b3a3-64a3b6321Split into factors3-64⋅a3⋅b63213a⋅b=3a⋅3b3-64⋅3a3⋅3b6321Calculate root-4ab2321Put minus sign in front of fraction-4ab2321 | |

Exercises 37 Let's simplify the expression and compare our answer to given work. 72Split into factors4⋅9⋅2a⋅b=a⋅b4⋅9⋅2Calculate root2⋅3⋅2Multiply62 Comparing our work, we can see that they didn't factor completely to find all the perfect squares. | |

Exercises 38 Let's simplify the expression and compare our answer to given work. 3125128y3ba=ba31253128y3Split into factors3125364⋅2⋅y3a⋅b=a⋅b3125364⋅32⋅3y3Calculate root54⋅32⋅3y3nan=a{n}54y32 Comparing our work, we can see that the quotient property of radicals wasn't used correctly. | |

Exercises 39 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 64ba=b⋅6a⋅66⋅646a⋅b=a⋅b6⋅646a⋅a=a26246a2=a646 To rationalize the denominator, we have expanded the expression by a factor of 66. | |

Exercises 40 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 13z1ba=b⋅13za⋅13z13z⋅13z13za⋅b=a⋅b13z⋅13z13za⋅a=a2(13z)213za2=a13z13z To rationalize the denominator, we have expanded the expression by a factor of 13z13z. | |

Exercises 41 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a cube root, we need to multiply it by a cube root that will give us a perfect cube under the radical. 3x22ba=b⋅3xa⋅3x3x2⋅3x23x3a⋅3b=3a⋅b3x2⋅x23xa⋅am=a1+m3x323xnan=a{n}x23x To rationalize the denominator, we have expanded the expression by a factor of 3x3x. | |

Exercises 42 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a cube root, we need to multiply it by a cube root that will give us a perfect cube under the radical. 343mba=b⋅316a⋅31634⋅3163m3163a⋅3b=3a⋅b34⋅163m316Multiply3643m316Calculate root43m316 To rationalize the denominator, we have expanded the expression by a factor of 316316. | |

Exercises 43 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a + ba − b a − ba + b In this case, because the expression is 5−82 the conjugate of the denominator is 5+8. Therefore, we have to multiply by 5+85+8. | |

Exercises 44 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a + ba − b a − ba + b In this case, because the expression is 3+75 the conjugate of the denominator is 3−7. Therefore, we have to multiply by 3−73−7. | |

Exercises 45 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 22ba=b⋅2a⋅22⋅222a⋅b=a⋅b2⋅222a⋅a=a22222a2=a222ba=b/2a/22 | |

Exercises 46 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 34ba=b⋅3a⋅33⋅343a⋅b=a⋅b3⋅343a⋅a=a23243a2=a343 | |

Exercises 47 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 485ba=b⋅48a⋅4848⋅485⋅48a⋅b=a⋅b48⋅485⋅48a⋅a=a24825⋅48Multiply482240a2=a48240 We know that we have successfully rationalized the denominator because the radical has been eliminated. However, our fraction can still be simplified a bit further. 48240Split into factors4816⋅15a⋅b=a⋅b4816⋅15Calculate root48415ba=b/4a/41215 | |

Exercises 48 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 524ba=b/4a/4131ba=ba131ba=b⋅13a⋅1313⋅1313a⋅b=a⋅b13⋅1313a⋅a=a213213a2=a1313 | |

Exercises 49 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. a3ba=b⋅aa⋅aa⋅a3aa⋅b=a⋅ba⋅a3aa⋅a=a2a23aa2=aa3a | |

Exercises 50 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical. 2x1ba=b⋅2xa⋅2x2x⋅2x2xa⋅b=a⋅b2x⋅2x2xa⋅a=a2(2x)22xa2=a2x2x | |

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##### Other subchapters in Solving Quadratic Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Solving Quadratic Equations by Graphing
- Solving Quadratic Equations Using Square Roots
- Quiz
- Solving Quadratic Equations by Completing the Square
- Solving Quadratic Equations Using the Quadratic Formula
- Solving Nonlinear Systems of Equations
- Chapter Review
- Chapter Test
- Cumulative Assessment