Properties of Radicals

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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, 2​1​ — we can multiply by the appropriate form of 1; in this case, 2​2​​=1. This way, we won't change the value of the fraction and we will get rid of the radical term in the denominator. 2​1​ba​=b⋅2​a⋅2​​2​⋅2​1⋅2​​a​⋅a​=a2​⋅2​1⋅2​​Multiply12​​1a​=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​​=?31​2x​​ 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​​=b​a​​​ 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. 92x​​ba​​=b​a​​9​2x​​Calculate root32x​​ba​=b1​⋅a31​2x​ We have found that 92x​​=31​2x​. 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.   -31​6​ ​ 63​61​ ​ 61​3​ ​ -33​61​ ​​ Notice that all the radical expressions involve square roots having 3 as radicand except for 31​6​. 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 7​1​ 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, 2​5​ 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+32​1​ 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. 20​Split into factors4⋅5​a⋅b​=a​⋅b​4​⋅5​Calculate 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. 32​Split into factors16⋅2​a⋅b​=a​⋅b​16​⋅2​Calculate 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. 128​Split into factors64⋅2​a⋅b​=a​⋅b​64​⋅2​Calculate 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. -72​Split into factors-36⋅2​a⋅b​=a​⋅b​-36​⋅2​Calculate 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. 125b​Split into factors25⋅5b​a⋅b​=a​⋅b​25​⋅5b​Calculate 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. 4x2​Split into factors4⋅x2​a⋅b​=a​⋅b​4​⋅x2​Calculate 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. -81m3​Split into factors-81⋅m2⋅m​a⋅b​=a​⋅b​-81​⋅m2​⋅m​Calculate 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. 48n5​Split into factors16⋅3⋅n4⋅n​Commutative Property of Multiplication16⋅n4⋅3⋅n​a⋅b​=a​⋅b​16​⋅n4​⋅3n​Write as a power16​⋅(n2)2​⋅3n​Calculate 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. 494​​ba​​=b​a​​49​4​​Calculate 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. -817​​ba​​=b​a​​-81​7​​Calculate 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. -6423​​ba​​=b​a​​-64​23​​Calculate 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. 12165​​ba​​=b​a​​121​65​​Calculate 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. 49a3​​ba​​=b​a​​49​a3​​Split into factors49​a2⋅a​​a⋅b​=a​⋅b​49​a2​⋅a​​Calculate 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. k2144​​ba​​=b​a​​k2​144​​Calculate 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. 4x2100​​ba​​=b​a​​4x2​100​​Calculate root2x10​ba​=b/2a/2​x5​
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. 3625v2​​ba​​=b​a​​36​25v2​​a⋅b​=a​⋅b​36​25​⋅v2​​Calculate 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. 316​Split into factors38⋅2​3a⋅b​=3a​⋅3b​38​⋅32​Calculate 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. 3-108​Split into factors3-27⋅4​3a⋅b​=3a​⋅3b​3-27​⋅34​Calculate 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. 3-64x5​Split into factors3-64⋅x3⋅x2​3a⋅b​=3a​⋅3b​3-64​⋅3x3​⋅3x2​Calculate 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. -3343n2​Split into factors-3343⋅n2​3a⋅b​=3a​⋅3b​-3343​⋅3n2​Calculate 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. 3-1256c​​3ba​​=3b​3a​​3-125​36c​​Calculate root-536c​​Put 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. 3278h4​​3ba​​=3b​3a​​327​38h4​​Split into factors327​38⋅h3⋅h​​3a⋅b​=3a​⋅3b​327​38​⋅3h3​⋅3h​​Calculate 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. -31000x381y2​​3ba​​=3b​3a​​-31000x3​381y2​​Split into factors-31000⋅x3​327⋅3⋅y2​​3a⋅b​=3a​⋅3b​-31000​⋅3x3​327​⋅33⋅y2​​Calculate 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. 3-64a3b621​​3ba​​=3b​3a​​3-64a3b6​321​​Split into factors3-64⋅a3⋅b6​321​​3a⋅b​=3a​⋅3b​3-64​⋅3a3​⋅3b6​321​​Calculate root-4ab2321​​Put minus sign in front of fraction-4ab2321​​
Exercises 37 Let's simplify the expression and compare our answer to given work. 72​Split into factors4⋅9⋅2​a⋅b​=a​⋅b​4​⋅9​⋅2​Calculate root2⋅3⋅2​Multiply62​ 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. 3125128y3​​ba​​=b​a​​3125​3128y3​​Split into factors3125​364⋅2⋅y3​​a⋅b​=a​⋅b​3125​364​⋅32​⋅3y3​​Calculate root54⋅32​⋅3y3​​nan​=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. 6​4​ba​=b⋅6​a⋅6​​6​⋅6​46​​a​⋅b​=a⋅b​6⋅6​46​​a⋅a=a262​46​​a2​=a646​​ To rationalize the denominator, we have expanded the expression by a factor of 6​6​​.
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. 13z​1​ba​=b⋅13z​a⋅13z​​13z​⋅13z​13z​​a​⋅b​=a⋅b​13z⋅13z​13z​​a⋅a=a2(13z)2​13z​​a2​=a13z13z​​ To rationalize the denominator, we have expanded the expression by a factor of 13z​13z​​.
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. 3x2​2​ba​=b⋅3x​a⋅3x​​3x2​⋅3x​23x​​3a​⋅3b​=3a⋅b​3x2⋅x​23x​​a⋅am=a1+m3x3​23x​​nan​=a{n}x23x​​ To rationalize the denominator, we have expanded the expression by a factor of 3x​3x​​.
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. 34​3m​ba​=b⋅316​a⋅316​​34​⋅316​3m316​​3a​⋅3b​=3a⋅b​34⋅16​3m316​​Multiply364​3m316​​Calculate root43m316​​ To rationalize the denominator, we have expanded the expression by a factor of 316​316​​.
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​+7​5​ the conjugate of the denominator is 3​−7​. Therefore, we have to multiply by 3​−7​3​−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. 2​2​ba​=b⋅2​a⋅2​​2​⋅2​22​​a​⋅b​=a⋅b​2⋅2​22​​a⋅a=a222​22​​a2​=a222​​ba​=b/2a/2​2​
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. 3​4​ba​=b⋅3​a⋅3​​3​⋅3​43​​a​⋅b​=a⋅b​3⋅3​43​​a⋅a=a232​43​​a2​=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. 48​5​​ba​=b⋅48​a⋅48​​48​⋅48​5​⋅48​​a​⋅b​=a⋅b​48⋅48​5⋅48​​a⋅a=a2482​5⋅48​​Multiply482​240​​a2​=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. 48240​​Split into factors4816⋅15​​a⋅b​=a​⋅b​4816​⋅15​​Calculate root48415​​ba​=b/4a/4​1215​​
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. 524​​ba​=b/4a/4​131​​ba​​=b​a​​13​1​ba​=b⋅13​a⋅13​​13​⋅13​13​​a​⋅b​=a⋅b​13⋅13​13​​a⋅a=a2132​13​​a2​=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. a​3​ba​=b⋅a​a⋅a​​a​⋅a​3a​​a​⋅b​=a⋅b​a⋅a​3a​​a⋅a=a2a2​3a​​a2​=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. 2x​1​ba​=b⋅2x​a⋅2x​​2x​⋅2x​2x​​a​⋅b​=a⋅b​2x⋅2x​2x​​a⋅a=a2(2x)2​2x​​a2​=a2x2x​​
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