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Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.1: A Review of the Properties of Exponents The first two sections in this chapter provide a review of several rules on how to simplify when working with exponents. The rules are as follows: Multiplication of like bases: bmbn = bm+n Division of like bases: ππ ππ = ππβπ (this rule applies as long as βbβ does not equal zero) Power rule of exponents: (ππ )π = πππ Power of a product rule: (ab)m = ambm π π ππ Power of a quotient rule: (π) = ππ (this rule applies as long as βbβ does not equal zero) Definition of πβπ : π πβπ = ππ (this rule applies as long as βaβ does not equal to zero) ___________________________________________________________________ Example 1: Simplify the expressions: This should help with #1-42 in the section. a) x2x4 b) z2z3z3 c) d) ππ ππ e) (ππ ) g) ππ π ππ ππ h) (ππππ )(ππ ππ ππ ) ππ ππ π π f) (πππ ) π ππ π i) ( ππ π π) a) x2x4 This problem is supposed to help you remember the multiplication of like bases rule. I will first solve the problem without the rule. Then I will show you how the rule could have been used to solve the problem. I could rewrite the problem without the exponents, if I didnβt remember the multiplication of like bases rule. x2x4 = (xx)(xxxx) ( I put each factor in its own parenthesis to make them stand out) x2x4 = xxxxxx (now I dropped the parenthesis, as they really were not needed) x2x4 = x6 (now I wrote the answer using an exponent, rather than repeated multiplication. 67 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.1: A Review of the Properties of Exponents The other way to solve this problem is using the multiplication of like bases rule. x2x4 = x2+4 = x6 (The rule tells me that if I am multiplying two terms with the same base, that the multiplication can be done by adding the exponent, and not changing the base. Answer: x6 b) z2z3z3 I could do this problem by writing each factor without their exponents. It is a bit more work this way. z2z3z3 =(zz)(zzz)(zzz) z2z3z3 = zzzzzzzz z2z3z3 = z8 However, the multiplication with like bases rule could also be used. The rule doesnβt state that it can be used to multiply three terms with the same base, but it can. z2z3z3 = z2+3+3 = z8 Answer: z8 c) π₯5 π₯2 I can solve this problem without the division with like bases rule. By writing out each term without the exponents, and canceling two of the xβs as follows: π₯5 π₯2 = π₯5 π₯2 = π₯3 π₯π₯π₯π₯π₯ π₯π₯ (πππ€ πΌ πππ ππππππ π‘βπ 2 π₯ β² π ππ π‘βπ πππππππππ‘ππ, π€ππ‘β 2 ππ π‘βππ₯ β² π ππ π‘βπ ππ’πππππ‘ππ. ) It would have been easier to solve this problem using the division with like bases rule. The rule tells me that I can solve this problem by subtracting the exponents. π₯5 π₯2 = π₯ 5β2 = π₯ 2 Answer: x2 68 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.1: A Review of the Properties of Exponents d) π2 π5 I can solve this problem two ways. I could rewrite the problem without the exponents and cancel two of the aβs.. π2 π5 = π2 π5 = πππ = π3 ( πΌ ππππππππ π‘βπ πππ‘πππ ππ’πππππ‘ππ πππ βππ£π π‘π ππππ£π π 1 ππ π‘βπ ππ’πππππ‘ππ. ) ππ πππππ 1 (ππππππ π‘βπ π‘π€π πβ²π ππ π‘βπ ππ’πππππ‘ππ π€ππ‘β 2 ππ π‘βπ πβ² π ππ π‘βπ πππππππππ‘ππ. 1 I will now solve this problem using the division with like bases rule. I will get an answer with a negative exponent., I need to remember how to make a negative exponent positive. This is covered in detail in section 4.2. I will use the rule to make a negative exponent positive when I solve this below. π2 π5 1 = π2β5 = πβ3 = π3 (I made the a-3 into my final answer using a rule from section 4.2) Answer: π ππ e) (π₯ 2 )3 I will only solve this problem using the power rule of exponents. I could have solved the problem without this rule, much like I solved the first few problems in this section without an exponent rule. (x2)3 = x2*3 = x6 (The power rule allows me to multiply the exponents to get my answer.) Answer: x6 f) (3π₯2 ) 3 I can still use the power rule to solve this problem. The power rule can be applied when there is an exponent outside of a parenthesis that DOES NOT HAVE AN ADDITION OR SUBTRACTION IN IT. I will need to write the 3 with an exponent to use the power rule. (3x2)3 = (31x2)3 (The exponent of the three is not usually written, but it helps to have it written when using the power rule.) = 31*2x2*3 (I applied the power rule to the 3 and the x by multiplying the exponents.) = 32x6 Answer: 9x6 69 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.1: A Review of the Properties of Exponents g) π5 π π2 π2 I will solve this problem using the division with like bases rule. π5 π π2 π2 1 = π5β2 π1β2 = π3 π β1 = π3 β π1 = positive before writing my answer.) π3 π (My answer had a negative exponent, and I needed to make it I could have avoided getting a negative exponent, by solving the problem a little differently. Basically, I will subtract the exponents of the bβs (2-1) and leave a positive exponent with the b in the denominator, because that is where the larger exponent of the b exists. This is another way to solve the problem that avoids negative exponents. π5 π π2 π2 π 5β2 = π2β1 = π3 π ππ π Answer: h) (5π₯π¦ 3 )(22 π₯ 3 π¦ 5 )3 First I need to simplify the second parenthesis, by using the power rule to get rid of the exponent of 3 outside of the second parenthesis. (5π₯π¦ 3 )(22 π₯ 3 π¦ 5 )3 = (5π₯π¦ 3 )(22β3 π₯ 3β3 π¦ 5β3 ) = (5π₯π¦ 3 )(26 π₯ 9 π¦15 ) = (5π₯π¦ 3 )(64π₯ 19 π¦15 ) Now I will multiply the 5 and 64 and add the exponents on the variables to get my answer. = 5*64x1+19y3+15 Answer: 320x20y18 ππ π π i) ( πππ ) First I will use the power rule to get the exponent of 4 outside of the parenthesis. π₯2 π¦ 4 π₯ 2β4 π¦ 1β4 π₯ 8π¦4 (2π₯ 3 ) = 21β4 π₯ 3β4 = 24 π₯ 12 (πππ‘πππ βππ€ πΌ π πππππππππ π‘βπ π¦ πππ π‘βπ 2. πΈππβ βππ ππ ππ₯ππππππ‘ ππ 1 π‘βππ‘ π€ππ πππ‘ π€πππ‘π‘ππ. Answer: ππ ππ πππππ 70 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents) This section focuses more heavily on negative exponents. It also introduces the concept of the exponent of 0. Definition of a0: a0 = 1 (this rule applies as long as βaβ does not equal zero) I will attempt to help you understand the 0 exponent rule with this problem. 23 Simplify: 23 You should be able to simplify this by replacing the 23 with 8, and reduce the fraction that remains. 23 8 =8=1 Now I will simplify the problem using the rules of exponents. The answer I get must be equivalent to 1. Since that is the answer I got by simplifying the problem in the last step. Using the rules of exponents can βt give me a different answer. 23 23 = 23β3 = 20 Since the two methods produce equivalent answers it must be the case that: 20 = 1 In general any number other than 0 raised to the 0 power equals 1. 23 Example 1: Simplify the expression. This should help with #1-12. a) 30 b) -30 c) 2x0 a) 30 = 1 This is a direct application of the 0 power rule. Answer: 1 b) -30 If I write 1 for the answer, I will have made a mistake. I need to think of the problem as: -1*30 (The negative in front of the 3 means to multiply the 3 by 1, now I will simplify by simplifying the exponent before the multiplication.) -30 = -1*30 = -1*1 = -1 Answer: -1 c) 2x0 I have to think of this problem as 2*x0, and simplify the exponent before doing the multiplication. 2x0 = 2*x0 = 2*1 = 2 ( I used the fact that x0=1) Answer: 2 Example 2: Simplify the expressions. Write the answer with positive exponents only. This should help with #13-38. π a) πβπ b) πβπ c) ππβπ ππ π βπ d) (ππ ) g) e) πβπ ππ f) πβπ ππ ππππ πβπ πππππ 71 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents) 1 1 a) 2β3 = 23 = 8 (This is a direct application of the a-n rule) Answer: π π 3 b) π₯ β4 I can solve this problem without shortcuts as follows: First rewrite the problem with a division sign, instead of a fraction bar. 3 = 3 ÷ π₯ β4 π₯ β4 Now I can use the a-n rule and make the exponent positive. 1 = 3 ÷ π₯4 Now I can change the division to a multiplication. = 3*x4 =3x4 The answer is 3x4. I could have got the answer without much algebra by moving the x-4 up to the numerator of the fraction and change its exponent from negative to positive. SHORTCTU TO MAKE A NEGATIVE EXPONENT POSITIVE: IF A PROBLEM DOES NOT HAVE ANY ADDITION OR SUBTRACTION- I can make a negative exponent positive by moving to the opposite side of the fraction. Answer: 3x4 c) 5π₯ β2 π¦ 5 I can use the rules of exponents to solve this, or I can get to the answer with a shortcut. I will first solve this using the rules of exponents. 1 The x-2 can be rewritten as π₯ 2 , the 5 and the y5 do not have negative exponents and do not need to be rewritten. 1 5π₯ β2 π¦ 5 = 5 β π₯ 2 β π¦ 5 (now I will write the 5 and the y5 as fractions to make it easier to follow along) 5 1 = 1 β π₯2 β the answer) π¦5 1 (now I will multiply across the numerators, and across the denominators to get πππ Answer: π ( I could have got the same answer by creating a fraction, moving the x term to π the denominator and changing its exponent from negative to positive.) π βπ d) (ππ ) Most of these problems can be done with some sort of shortcut. The shortcut to this problem is to make the exponent of (-3) positive by taking the reciprocal of the fraction. I will use the shortcut without showing the long way to do the problem. Feel free to ask me why the shortcut works if you are curious. 72 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents) 5 β3 π₯2 3 (π₯ 2 ) = (51 ) (I made the exponent positive by taking the reciprocal of the fraction. I also wrote the 5 with an exponent to make the next step easier to follow.) π₯ 2β3 π₯6 π₯6 =51β3 = 53 = 125 (I used the power of a quotient rule on this step.) Answer: ππ πππ e) π₯ β6 π₯ 2 = π₯ β6+2 = π₯ β4 (I used the product with like bases rule and added the exponents.) 1 = π₯ 4 (Now I used the rule to make the negative exponent positive.) Answer: f) π¦ β5 π¦6 1 π₯4 = π¦ β5β6 = π¦ β11 (I used the quotient with like bases rule and subtracted the exponents.) π = πππ (Now I used the rule to make the negative exponent positive.) Answer: g) π πππ 32π₯ 2 π¦ β6 = 10π₯π¦ 3 16π₯ 2β1 π¦ β6β3 5 (I simplified 32/10, and subtracted the exponents of the variables.) 16π₯π¦ β11 = 5 16 π₯ = 5π¦ 11 (I used a shortcut and made the exponent of the y positive by moving it to the denominator) πππ Answer: ππππ 73 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.3: Definition of nth Root The reverse operation of squaring a number is to find the numbers square root. For example, finding the square root of 25 is equivalent to asking what number squared is 25. There are two numbers that satisfy this. The numbers are 5 and (-5) because (5)2 = 25 and (-5)2 = 25. In general all positive numbers have two square roots, one positive and one negative. However, we are usually more concerned with the positive square root than the negative. Let βaβ represent some positive real number: ο· βπ is the positive square root of βaβ. The positive square root of a number is called the principal square root of βaβ. This is the positive number that when squared gives βaβ. ο· ββπ is the negative square root of βaβ. This is the negative number that when square gives βaβ. The reverse operation of cubing a number is to find the numbers cubed root. For example, finding the cubed root of 8 is equivalent to asking what number cubed is 8. Two is only one number that satisfies this, because (2)3 = 8. (-2) does not satisfy this because (-2)3 = -8, not 8. Let βaβ represent any real number: ο· πβπ is the cubed root of βaβ. This is the number that when cubed gives βaβ. The number will be positive if βaβ is positive, and negative if βaβ is negative. There is no principal cubed root, because there will only be one number that can be cubed to equal the number under the radical. The reverse operation of raising a number to the fourth power is to find the numbers fourth root. For example finding the fourth root of 81 is equivalent to asking what number raised to the fourth power is equal to 81. There are two numbers that satisfy this. The numbers are 3 and -3, since (3)4 = 81, and (3)4=81. In general all positive numbers have two fourth roots, one positive and one negative. However, we are usually more concerned with the positive fourth root than the negative. Let βaβ represent some positive real number: ο· πβπ is the positive fourth root of βaβ. This is often called the principal fourth root of βaβ. This is the positive number that when raised to the fourth power gives βaβ. ο· β πβπ is the negative fourth root of βaβ. This is the negative number that when raised to the fourth power gives βaβ. 74 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.3: Definition of nth Root I could go on for a long time discussing individual roots of a number βaβ. In general: ο· ο· ο· πΎπππ nππ ππ ππππ ππππππππ πππππππ, πππ π > 0 π€π π ππ¦ πβπ ππ πππ πππππππππ ππ ππππππππ πππ ππππ ππ "π". π πΎπππ "π" ππ ππ ππ π ππππππππ πππππππ, ππ πππ βπ ππ πππ πππ ππππ ππ "π". The βnβ is called the index of the radical, and the βaβ is called the radicand. ___________________________________________________________________ Example 1: Evaluate the roots. Identify those that are not real numbers. This should help with #1-30. π a) βππ b) ββππ c) ββππ d) βππ π e) ββππ π f) β βππ ππ g) β π h) πβππ a) β81 This is asking for the positive number that when squared is 81. The answer is 9, because (9)2 = 81. Answer: 9 b) ββ81 This is asking for the negative number that when squared is 81. The answer is (-9) because (-9)2 = 81 Answer: -9 c) ββ81 This is asking for the positive number that when squared gives -81. Any real number squared gives a positive number. There is no real number that when squared gives a -81. Answer: Not a real number. 3 d) β64 This is asking for the number that when cubed gives 64. The answer is 4, because (4)3 = 64. Answer: 4 3 e) β β64 3 I can think of this as β1 β β64. I will find the cubed root of 64, then multiply that by (-1) 3 3 β β64 = β1 β β64 = β1 β 4 = β4 Answer: -4 75 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.3: Definition of nth Root 16 f) β 9 This is asking for the positive number, that when squared is 16/9. This is not hard. I can find the square root of a fraction, by finding the square root of the numerator and the denominator. Answer: π π g) 3β16 First I will find the positive square root of 16, then I will multiply the result by 3. 3β16 = 3 β 4 = 12 Answer: 12 __________________________________________________________________ You will be asked to find the domain of functions that contain radicals in this section. The process for finding the domain of a radical function depends on whether the index is even or odd. If the index is even, the function will be a real number only when the radicand is greater than or equal to zero. If the index is odd, the function will always yield a real number. Determining the domain of a radical function: ο· If βnβ is even then the domain of π(π) = πβπ can be found be setting the radicand (expression under the radical) greater than or equal to 0. ο· If βnβ is odd the domain of π(π) = πβπ will be all real numbers. ___________________________________________________________________ Example 2: Find the domain of the following functions. Write your answer in interval notation. This should help with #43-46. π π a) π(π) = βπ + π b) π(π) = βπ + π 4 a) Find the domain of π(π₯) = βπ₯ + 2 The index is even, so I will find the domain by setting the radicand greater than or equal to 0. π₯+2β₯0 π₯ β₯ β2 This is the domain. I need to write this as an interval. Answer: domain [βπ, β) 5 b) Find the domain of β(π₯) = βπ₯ + 2 The index is odd, so the domain is all real numbers. There is no algebra needed to find the domain. Answer: Domain (ββ, β) 76 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.3: Definition of nth Root ___________________________________________________________________ Finding the nth root of a variable expression is similar to finding the nth root of a number. For principal roots with an even index care must be taken to make sure the answer is not negative. For example: βπ₯ 2 is asking what positive number squared is x2. I could be wrong if I wrote x for an answer, even though x2 = x2. I donβt know if x represents a positive or a negative number. The best answer I can give is: βπ₯ 2 = |π₯| When the index is odd, I do not have to worry about absolute values. Only even indexes require positive answers. The following box summarizes the last few comments. π Definition of βππ ο· πβππ = |π|, If n is even and a positive integer ο· πβππ = π ,If n is a odd and a positive integer ___________________________________________________________________ Example 3: Simplify the radical expressions. Use absolute values when necessary. π a) βππ b) βππ a) βπ₯ 2 = |π₯| (I need the absolute value in my answer because the index is even.) Answer: |π| 3 b) βπ 3 = π (I donβt need the absolute value in my answer because the index is odd.) Answer: b ___________________________________________________________________ For the rest of the section, we will assume that any variable in a problem represents a positive number. This allows us to write answers without absolute values. ___________________________________________________________________ Example 4: Simplify the expressions. Assume all variables represent positive real numbers, so no absolute values will be needed for any of the answers. This should help with #59-74. π a) βππππ b) βπππ c) βπππ π d) βπππ ππ e) β ππ π π ππ f) βππππ a) β16π₯ 2 This is asking what squared equals 16x2. I will solve this by taking the square root of the 16 and the x2. β16π₯ 2 = 4π₯, I can check my work by squaring my answer. (4x)2 = (4x)(4x) = 16x2 Answer: 4x 77 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.3: Definition of nth Root b) βπ10 This is asking the question what squared equals a10. The easiest way to get this answer is to divide the exponent by the index. The index in this problem is a 2, because this a square root problem. βπ10 = π10β2 = π5 I can check my work by squaring my answer. If I square my answer I should get a10. (a5)2 = a5*2 = a10 , I have confirmed my answer is correct. Answer: a5 3 c) βπ12 This is asking the question what cubed equals b12. The easiest way to get this answer is to divide the exponent by the index of 3. 3 βπ12 = π 12β3 = π 4 I can check my work by cubing my answer. If I cube my answer I should get b12. (b4)3=b4*3 = b12, I have confirmed my answer is correct. Answer: b4 4 d) βπ§12 This is asking the question what to the fourth power equals z12. The easiest way to get this answer is to divide the exponent by the index of 4. 4 12 βπ§ = π§ 12β4 = π§ 3 I can check my work by raising my answer to the fourth power. If I raise m y answer to the fourth power I should get z12. (z3)4= z3*4 = z12, I have confirmed my answer is correct. Answer: z3 81 e) βπ₯ 6 I need to find the square root of the numerator and the denominator to solve this. The square root of 81 is 9. I can find the square root of x6 by dividing the exponent by 2. Answer: 3 π ππ 8π₯ 3 f) β27π§6 I need to find the cubed root of the numerator and denominator to solve this. The cubed root of 8 is 2, because (2)3 = 8. I will divide the exponent of the x by three to find its cube root. The cubed root of 27 is 3, because (3)3 = 27. I will divide the exponent of the z by 3 to find its cubed root. Answer: ππ πππ ___________________________________________________________________ 78 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents We simplified radical expression that contained variables in section 4.3 by dividing the exponent of the variable by the index. The next rule formally states that you can simplify radical expressions by dividing exponents. It also states that you can make a problem that has a fraction exponent into and expression that contains a radical. Radical expressions are equivalent to expressions with fractional exponents. The following rule describes this relationship. π βπ = ππβπ This rule applies, provided n is a positive integer greater than 1 and βaβ is a real number. ___________________________________________________________________ Example 1: Evaluate the expression. This will give you practice using the rule, and prepare you for problems later in the section. a) πππβπ b) πππβπ c) βπππβπ a) 811β2 The rule tells me that I can rewrite this problem in radical form. 2 811β2 = β81 = β81 = 9 (The rule asks me to write the denominator of the fraction exponent as the index. However, I donβt really need to write an index when it is a 2. An index of 2 is assumed when no index is written. Answer: 9 b) 811β4 I will use the rule in the above box to write this is radical form. 4 4 811β4 = β81 = 3 (Remember β81 is asking what positive number times itself 4 time s is 81.) Answer: 3 c) β811β2 The negatives can get confusing. It might help to think of the negative sign as multiplication by negative 1. β811β2 = β1 β 811β2 ( I will now write this in radical form.) = β1 β β81 = β1 β 9 Answer: -9 _____________________________________________________________________ 79 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents We now need to work with problems that have more fractional exponents that do not have a 1 in the numerator. 84β3 is an example of such a problem. We can rewrite this problem two ways using the rules of exponents we studied earlier in the chapter. I will simplify the expression after I rewrite it. 4 4 ο· 84β3 = (81β3 ) = (β8) = 24 = 16 ο· 84β3 = (84 )1β3 = β84 = β4096 = 16 3 3 3 I was able to rewrite each problem so that I could use the rule in the above box and write the problem as a radical. I should be able to write any problem with a fractional exponent in radical form, and I should be able to write any problem in radical form with a fractional exponent. The following box tells me how this can be done. This rule explains how expressions with fractional exponents that have a numerator other than one are related to radical expressions. Notice there are two ways to write an expression with a fractional exponent as a radical expression. Definition of ππβπ π π ππβπ = (ππβπ ) = ( πβπ) and π ππβπ = (ππ )πβπ = βππ (provided n is a positive integer greater than 1, and m is an integer not equal to 0) The fractional exponent of m/n does two things. The numerator raises the base to the mth power and the denominator takes the nth root. The next few problems are designed to get you used to working with the rule. _____________________________________________________________________ Example 2: Write the expression in radical notation. Do not simplify. This should help with #1-6. a) ππβπ b) (ππ)πβπ π πβπ c) (πππ) a) 54β3 I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m, and the denominator of the exponent works like the n. π π π Answer: βππ ππ (βπ) b) (5π₯)3β5 I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m, and the denominator of the exponent works like the n. π π π Answer: β(ππ)π ππ (βππ) 80 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents 2 7β2 ) ππ2 c) ( I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m, and the denominator of the exponent works like the n. Answer: π β( π π ) ππ π ππ π (βπππ) Example 3: Write the expression using rational exponents rather than radical notation. This should help with #7-12. π π a) βππ b) πβπ c) βπππ 3 a) βπ₯ 2 I am just using the rule again. The index becomes the denominator in the exponent. Answer: ππβπ b) 5βπ¦ The 5 will stay separate from the y, and should not have a fractional exponent, or be in a parenthesis. It might help if I rewrite the problem so that the y has an exponent, and so that the index is written. 2 5βπ¦ = 5 β βπ¦1 (The fraction exponent attached to the y will have a 1 in the numerator and 2 in the denominator.) Answer: πππβπ 4 c) β5π₯ 2 I am taking the fourth root of the entire quantity 5x2. I will place the 5x2 in a parenthesis to help make this easier to follow. 4 4 β5π₯ 2 = β(5π₯2 )1 (The problem is now ready to be written with a fractional exponent. πβπ Answer: (πππ ) The next group of problems will teach you how to use multiple rules presented so far in the chapter to solve problems. 81 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents ____________________________________________________________________ Example 4: Write the expression using positive exponents and radical notation, then simplify. a) πππβπ b) βπππβπ c) ππβπβπ d) π βπβπ π e) (π) πβπβπ a) 253β2 First I will write this in radical form. 3 2 253β2 = (β25) = 53 Answer: 125 (To simplify this I will first find the square root of 25, then cube it. b) β253β2 First I will write the coefficient as a multiplication problem by -1. β253β2 = β1 β 253β2 (Now I will write the problem as a radical.) 2 3 = β1 β ( β25) (The quantity in the parenthesis is the same as in part a.) = -1*125 Answer: -125 c) 25β3β2 First, I will write the problem with a positive exponent. Remember I need to create a fraction to do this. 1 25β3β2 = 253β2 (I already know that the denominator equals 125. I computed it in part a.) Answer: π πππ 4 d) 8β2β3 First I will make the negative exponent positive. I will use a shortcut to do this. Ask me in class if you want to see the long way to make the exponent positive. 4 8β2β3 = 4 β 82β3 (Now I will write the problem with a radical.) 3 2 = 4 β (β8) (I will now find the cubed root of 8.) = 4 β 22 =4*4 Answer: 16 82 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents 16 β5β2 e) ( 9 ) I will first make the exponent positive by taking the reciprocal of the fraction. 4 β5β2 (9) 9 5β2 = (4) (I now will write the problem in radical form.) 5 9 = (β ) (Next I will find the square root of 9/4.) 4 3 5 = (2) (The last step is to raise both the 3 and the 2 to the fifth power.) Answer: πππ ππ Example 5: Simplify the expression using the properties of rational exponents. Write final answer using only positive exponents. a) πβπβπ πππβπ d) b) ππβπ πβπβπ π c) (ππβπβπ ) ππππ ππππβπ a) π₯ β2β3 π₯ 11β3 I just need to add the exponents to do this multiplication problem. π₯ β2β3 π₯ 11β3 = π₯ β2β3+11β3 = π₯ 9β3 Answer: x3 b) π¦ 3β5 π¦ β4β5 I just need to subtract the exponents to do this division problem. π¦ 3β5 π¦ β4β5 =π¦ 3β β(β4β ) 5 5 =π¦ 3β +4β 5 5 Answer: ππβπ 83 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.4: Rational Exponents 4 c) (2π₯ β1β2 ) I will first write this problem with positive exponents. I will also write the 2 with an exponent of 1 to help make things easier to follow. 21 4 4 (2π₯ β1β2 ) = (π₯ 1β2 ) (I will now multiply the exponents by 4 to clear the parenthesis.) = Answer: d) ππ ππ 21β4 π₯ 1β2β4 16π₯ 2 12π₯ 1β2 I will first reduce the fraction 16/12 by 4, and subtract the exponents of the xβs. 4π₯ 2β1β2 16π₯ 2 12π₯ 1β2 = 3 denominator) = Answer: ( I will now simplify the exponent. I will write the 2 as 4/2 so that I have a common 4π₯ 4β2β3β2 3 πππβπ π 84 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.5: Properties of Radicals This section involves simplifying radical expressions. The box below gives a formal definition as to what it means for a radical expression to be reduces. Simplified Form of a Radical Expression: One way to decide of a radical expression is simplified is to first rewrite it as a product of prime factors. A radical expression where the radicand is written as a product of prime factors is considered in a simplified form provided each condition is met. ο· The exponent of each factor is less than the index. ο· There is no radical in the denominator. We will need to use a couple of properties of radicals that have not been formally stated to simplify radical expressions. The box below states those properties. Multiplication and Division Properties of Radicals: π π a. βππ = πβπ β βπ π π π. βπ = π βπ βπ π π This theory assumes that a, b, πβπ πππ βπ are all real numbers. This means that this rule does not apply if the index is even and a or b represent negative numbers. Problems #1-8 in the section will get you used to working with the multiplication rule in the bottom box. You should notice the statement at the very top of the section. It states that all variables represent positive real numbers. This statement allows me to use the multiplication and division rules in _____________________________________________________________________ Example 1: Use the multiplication property of radicals to multiply the expressions. Then simplify the result. π a) βπππ ππ β πβπππ b) βπ + π β βπ + π 3 a) β2π₯2 π¦ 5 β 3β4π₯π¦ 3 I will multiply contents of the radicals to get: β8π₯ 3 π¦ 6 To find my answer I know that the cubed root of 8 is 2, and I will divide the exponents of the variables by 3. Answer: 2xy2 85 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.5: Properties of Radicals b) βπ₯ + 4 β βπ₯ + 4 I will multiply the contents of the radicals to get: β(π₯ + 4)2 To find my answer I will divide the exponent by 2. Answer: x + 4 _____________________________________________________________________ Example 2: Use the division property of radicals to divide the expression. Then simplify the result. This should help with #9-16. π π a) βππ b) π βπ βπππππ π βπππ 4 a) β96 4 β6 4 I will divide the contents of the radicals to get: β16 This is asking what number times itself 4 times is 16. The answer is 2. Answer: 2 3 b) β54π¦ 11 3 β2π¦ 2 3 I will divide the contents of the radicals to get: β27π¦ 9 I know that the cubed root of 27 is 3. I will divide the exponent of the y by three and write my answer. Answer: 3y3 _____________________________________________________________________ Example 3: Simplify the radicals. π a) βπππ b) βπππ d) βπππ e) βπππ g) βππππ ππ h) βπππππ ππ π π c) βπππ π f) βπππ π 86 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.5: Properties of Radicals 3 a) β120 I will first make a factor tree to find the prime factorization of 120. 3 3 β120 = β23 β 3 β 5 Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple of 3 from the other factors. 3 3 = β23 β β3 β 5 To get the answer I will divide the exponent in the left radical by three, and I will multiply the numbers in the right radical. π Answer: π βππ b) β150 I will first make a factor tree to find the prime factorization of 150. β150 = β2 β 3 β 52 Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple of 2 from the other factors. = β52 β β2 β 3 To get the answer I will divide the exponent in the left radical by 2, and I will multiply the numbers in the right radical. Answer: πβπ 4 c) β288 I will first make a factor tree to find the prime factorization of 288. 4 4 β288 = β25 β 32 The exponent on the 2 is not a multiple of 4, but it is bigger than 4. I need to rewrite the 25 as 24 * 21. To make it so I can divide the exponent of the two by 4. 4 = β24 β 2 β 32 87 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.5: Properties of Radicals Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple of 4 from the other factors. 4 4 = β24 β β2 β 32 To get the answer I will divide the exponent in the left radical by 4, and I will multiply the numbers in the right radical. π Answer: π β βππ d) β486 I will first make a factor tree to find the prime factorization of 486. β486 = β2 β 35 The exponent on the 3 is not a multiple of 2, but it is bigger than 2. I need to rewrite the 35 as 34 * 31. To make it so I can divide the exponent of the 3 by 2. = β34 β β2 β 3 To get the answer I will divide the exponent in the left radical by 2, and I will multiply the numbers in the right radical. = 32 β β6 Answer: πβπ 3 e) βπ¦10 The exponent of the y is not a multiple of 3, and it is bigger than 3. I need to rewrite the y10 as π¦ 9 β π¦1 to meet the requirement of having an exponent that is a multiple of the index and one that is less than the index. 3 3 βπ¦10 = βπ¦ 9 β 3βπ¦ To get the answer I will divide the exponent in the left radical by 3. Answer: ππ β πβπ 88 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.5: Properties of Radicals 4 f) βπ§11 The exponent of the z is not a multiple of 4, and it is bigger than 4. I need to rewrite the z11 as π§ 8 β π§ 3 to meet the requirement of having an exponent that is a multiple of the index and one that is less than the index. 4 4 4 βπ§11 = βπ§ 8 β βπ§ 3 To get the answer I will divide the exponent in the left radical by 4. π Answer: ππ β βππ g) β18π₯ 3 π¦ 2 I will work the 18, the x3 and the y2 separately. Since 18 = 32 * 2, I will put a 32 in the left radical, and a 2 in the right radical. Since x3 = x2 * x, I will put a x2 in the left radical, and a x in the right radical. Since the exponent of the y is a multiple of 2, I will place the y2 in the left radical, and no y in the right radical. β18π₯ 3 π¦ 2 = β32 π₯2 π¦ 2 β β2π₯ Now I will divide the exponents in the left radical by 2 and write my answer. Answer: πππβππ 3 h) β54π₯π¦ 6 π§ 7 I will work with the 54, x, y6 and z7 separately. Since 54 = 2*33 I will place a 33 in the left radical, and a 2 in the right radical. Since the exponent of the x is not a multiple of 3, and it is less than 3 the entire x will go in the right radical. I can put the y6 in the left radical since its exponent is a multiple of 6. Since z7 = z6 * z1, I will place a z6 in the left radical, and a z1 in the right radical. 3 3 3 β54π₯π¦ 6 π§ 7 = β33 π¦ 6 π§ 6 β β2π₯π§ Now I will divide the exponents in the left radical by 3 and write my answer. π Answer: πππ ππ β βπππ 89 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.6: Addition and Subtraction of Radicals This is a section on adding and subtracting radical expressions. βLike radicalsβ can be added or subtracted. To be βlike radicalsβ two radicals have to have the same index and same radicand. These two radicals are βlike radicalsβ and could be added or subtracted. 3 3 β3π₯ πππ 5β3π₯ These two radicals are not βlike radicalsβ and could not be added or subtracted. 3 4 βπ¦ πππ βπ¦ (These are not like, because they have a different index. _____________________________________________________________________ Example 1: Add or subtract the radical expressions if possible. This will help with #1-12 in the section. a) πβπ + πβπ b) π π πβπ β π πβπ π a) 2β7 + 9β7 These are like radicals and can be added. To add like radicals just add the coefficients and leave the rest of the problem unchanged. Answer: ππβπ b) 1 7 π₯β5 β 2 π₯β5 2 These are like radicals and can be subtracted. To subtract like radicals just subtract the coefficients and 1 7 leave the rest of the problem unchanged. I need to subtract the 2 π₯ β 2 π₯. I have to subtract the numerators of the fractions to get my answer. 1 π₯β5 2 7 6 β 2 π₯β5 = β 2 π₯β5 Answer: βππβπ ____________________________________________________________________ Example 2: Add or subtract the radical expressions if possible. This will help with #13-28 in the section. a) πβππ + πβππ π π b) π β βπππππ β ππ β βπππππ 90 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.6: Addition and Subtraction of Radicals a) 5β18 + 7β50 These radicals are not like and canβt be added as they are written. However, both radicals can be reduced. I will be able to add the radicals after they are reduced. I will follow the steps taught in section 4.5 to reduce the radicals. 5β18 + 7β50 = 5β32 β 2 + 7β2 β 52 = 5β32 β2 + 7β52 β2 = 45β2 + 35β2 Answer: ππβπ 3 3 b) 2 β β16π₯π¦ 5 β 3π¦ β β54π₯π¦ 2 These radicals are not like and canβt be subtracted as they are written. However, both radicals can be reduced. I will be able to subtract the radicals after they are reduced. I will follow the steps taught in section 4.5 to reduce the radicals. 3 3 3 3 2 β β16π₯π¦ 5 β 3π¦ β β54π₯π¦ 2 = 2 β β23 β 2 β π₯ β π¦ 3 π¦ 2 β 3π¦ β β2 β 33 π₯π¦ 2 3 3 3 3 = 2 β β23 π¦ 3 β β2π₯π¦ 2 β 3π¦ β β33 β β2π₯π¦ 2 3 3 = 2 β 2π¦ β β2π₯π¦ 2 β 3π¦ β 3 β β2π₯π¦ 2 3 3 = 4π¦ β β2π₯π¦ 2 β 9π¦ β β2π₯π¦ 2 π Answer: βππ β βππππ 91 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.7: Multiplication of Radicals. We have already multiplied a few radicals in section 4.5. This multiplication problems in this section will require more work to simplify after the multiplication is complete than the problems in section 4.5, but the idea is the same. _____________________________________________________________________ Example 1: Multiply the radicals. π π a) βπ β βππ b) (πβππ)(πβππ) d) πβπ(πβπ β πβπ) 3 π π c) βππππ β βπππ ππ e) (π β πβπ)(π + πβπ) 3 a) β4 β β18 I can multiply the 4 and the 18 without changing the problem because the indexes are the same. 3 3 3 β4 β β18 = β72 (Now I will reduce the answer by finding the prime factorization of 72.) 3 3 3 = β23 β 32 = β23 β β32 (Now I will divide the exponent in the left radical, and multiply out the right radical.) π Answer: π βπ b) (2β14)(3β21) First I will multiply the 2 with the 3, and the 14 with the 21. (2β14)(3β21) = 6β294 (Next I will find the prime factorization of 294 and reduce the radical.) = 6β2 β 3 β 72 = 6β72 β β2 β 3 = 6 β 7β6 Answer: ππβπ 4 4 c) β8π₯π¦ 3 β β6π₯ 3 π¦ 4 First I will multiply the 8 and the 6, then add the exponents of the xβs and yβs. 4 4 4 β8π₯π¦ 3 β β6π₯ 3 π¦ 4 = β48π₯ 4 π¦ 7 (Next I will get the radical ready to be simplified.) 4 4 = β24 π₯ 4 π¦ 4 β β3π¦ 3 π Answer: πππ β βπππ 92 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.7: Multiplication of Radicals. d) 2β3(5β3 β 4β6) First I will clear the parenthesis by using the distributive property. 2β3(5β3 β 4β6) = (2β3)(5β3) β (2β3)(4β6) (Next I will complete the multiplication that I started.) = 10β9 β 8β18 = 10 β 3 β 8β32 β β2 = 30 β 8 β 3β2 Answer: ππ β ππβπ e) (2 β 3βπ₯)(4 + 5βπ₯) First I will FOIL to clear the parenthesis. (2 β 3βπ₯)(4 + 5βπ₯) = 2 β 4 + 2 β 5βπ₯ β 3βπ₯ β 4 β 3βπ₯ β 5βπ₯ = 8 + 10βπ₯ β 12βπ₯ β 15βπ₯2 (Next I will combine the inners and outers, and simplify the β15βπ₯2 Answer: π β πβπ β πππ _____________________________________________________________________ Example 2: Multiply the special products. This should help with #27-36. a) (π β πβπ)(π + πβπ) π b) (π β πβπ) a) (3 β 4βπ₯)(3 + 4βπ₯) This is just a FOIL problem. The contents of the parenthesis are identical, except for the signs. The inners and outers will cancel. To save time, I shouldnβt multiply the inners and outers. I will show the inners and outers for this example, but it is not necessary. (3 β 4βπ₯)(3 + 4βπ₯) = 3 β 3 + (3)(4βπ₯) β (4βπ₯)(3) β (4βπ₯)(4βπ₯) = 9 + 12βπ₯ β 12βπ₯ β 16βπ₯2 (Next, I combine the inners and outers, and reduce the last term.) Answer: π β πππ 93 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.7: Multiplication of Radicals. 2 b) (3 β 4βπ₯) This is also a FOIL problem. The inners and outers do not cancel. I will show you a shortcut to simplify this problem in class. First I will write out the problem so that it is more obviously a FOIL problem. 2 (3 β 4βπ₯) = (3 β 4βπ₯)(3 β 4βπ₯) (Next I will FOIL.) = 9 β 12βπ₯ β 12βπ₯ + 16βπ₯2 (Next I will combine the inners and outers, and simplify the last term.) Answer: π β ππβπ + πππ 94 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.8: Rationalization A rule in section 4.5 stated that one of the conditions for a radical expression to be in its most simplified form is that the denominator canβt contain a radical. Rationalization is the process of simplifying a radical expression by getting the radical out of the denominator. It is necessary to multiply a rational expression by a carefully selected version of 1 to rationalize any radical expression that contains a fraction. ___________________________________________________________________ Example 1: Rationalize the denominator. This will help with the square root problems in #1-22. a) a) π b) βπ π βππ c) π βππ 5 β6 I can multiply by β6 β6 to rationalize the denominator. In fact if the denominator of a fraction contains a monomial that is a square root I can rationalize the fraction by multiplying by βthe denominator over the denominator.β 5 β6 = 5 β6 β β6 β6 = 5β6 β36 Answer: b) (Next I will multiply the numerators and denominators, then simplify.) πβπ π ππ π βπ π 8 β10 This is a fraction that has a monomial square root in the denominator. I just need to multiply by the denominator over the denominator. 8 β10 = 8 β10 = 8β10 β100 = 8β10 10 Answer: β β10 β10 (I need to reduce the 8/10 that is not part of the radical.) πβππ π π ππ π βππ 95 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.8: Rationalization c) π₯ β3π₯ This is a fraction that has a monomial square root in the denominator. I just need to multiply by the denominator over the denominator. π₯ β3π₯ π₯ β3π₯ = β3π₯ β3π₯ π₯ β3π₯ = = β β 9π₯ 2 π₯β3π₯ 3π₯ (Next I can cancel the xβs that are not under the square root.) π βππ π Answer: ππ π βππ ____________________________________________________________________ Example 2: Rationalize the denominator. This will help with the cube and fourth root problems in #1-22. π a) π a) 4 b) β ππ π π βπ 5 βπ₯ 5 I have to be more careful when solving problems that are not square roots. In fact IT WOULD BE WRONG π TO MULTIPLY BY βπΏπ π βπΏπ . It takes thought to determine what to multiply by in non square root problems. I 4 βπ₯ 3 need to multiply by something so that my exponent becomes a multiple of 4. I will multiply by 4 5 4 βπ₯ 5 = βπ₯ 3 4 βx3 5 β4 4 βx3 βπ₯ 5 (Next I will multiply the denominator by adding exponents.) 4 = 5β βπ₯ 3 4 βπ₯ 8 (Next I will divide the exponent in the denominator by 4 to get rid of the radical.) π Answer: πβ βππ ππ 96 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.8: Rationalization b) 2 3 β3 I have to be more careful when solving problems that are not square roots. In fact IT WOULD BE WRONG π βπ TO MULTIPLY BY π . It takes thought to determine what to multiply by in non square root problems. I βπ 3 β32 need to multiply by something so that my exponent becomes a multiple of 3. I will multiply by 3 β32 2 3 β3 = . 3 β32 2 3 β3 β3 1 (Now I will add the exponents to complete the multiplication in the denominator.) β32 3 = 2β β32 3 β33 (Next I will divide the exponent in the denominator by 3 to clear the radical, and square the 3 under the radical in the numerator.) π Answer: πβ βπ π ππ ππ βπ π _____________________________________________________________________ Example 3: Rationalize the denominators by multiplying by the conjugate. This should help with #23-34. a) βπ π+βπ a) β6 3+β6 b) π βπβπ When I denominator of a fraction contains a binomial with a square root, I have to multiply by the conjugate 3ββ6 to β6 of the denominator to rationalize the denominator. In this problem I will multiply by 3β rationalize the denominator. β6 3+β6 = (3ββ6) β6 β (3+β6) (3ββ6) (Next I will FOIL the denominator, remember the outers and inners should cancel. I also will multiply out the numerator.) = 3β6ββ36 9β3β6+3β6ββ36 = 3β6β6 9β6 = 3(β6β2) 3 (I will cancel the inners and outers in the denominator, and replace β36 with 6. (I will factor out the GCF of 3 from the numerator.) (Next I will cancel the 3βs.) Answer: βπ β π 97 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.8: Rationalization b) 5 βπ₯β3 When I denominator of a fraction contains a binomial with a square root, I have to multiply by the conjugate of the denominator to rationalize the denominator. In this problem I will multiply by βπ₯+3 to βπ₯+3 rationalize the denominator. 5 (βπ₯β3) β (βπ₯+3) (βπ₯+3) = = Answer: π(βπ+π) πβπ 5βπ₯+15 βπ₯2 +3βπ₯β3βπ₯β9 5(βπ₯+3) π₯β9 ππ πβπ+ππ πβπ _____________________________________________________________________ 98 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.9: Radical Equations An equation that contains one or more radicals is called a radical equation. We need to remember a rule from earlier in the chapter to make this section easier. The rule is: π π π ( βπ₯) = π₯, ππππ£ππππ βπ₯ ππ π ππππ ππ’ππππ These are a few examples of how we will use this rule in section 4.10. __________________________________________________________________ Example 1: Simplify the following. Assume each variable represents a positive real number. These are to familiarize you with the above rule. π π 3 3 a) ( βππ) b) (βπ) π π c) ( βπ + π) π a) ( β2π₯ ) I will do this problem and the next problem the long way. First I will write without the exponent and write it with repeated multiplication. 3 3 3 3 3 ( β2π₯) = ( β2π₯ )( β2π₯)( β2π₯) (Next I will multiply the 2xβs under the radicals.) 3 = β(2π₯)3 (Next I will divide the exponent by the index to get rid of the radical.) = (2π₯)3β3 Answer: 2x 2 b) (β3) I will do this problem and the next problem the long way. First I will write without the exponent and write it with repeated multiplication. 2 (β3) = (β3)(β3) (Next I will multiply the 3βs under the radicals.) = β9 Answer: 3 4 4 c) ( βπ₯ + 3) I hope you see that if you raise a radical to the power of the index you can just drop the radical. This assumes that the original expression represents a real number. I will write this answer down without performing any algebra. Answer: x+3 ___________________________________________________________________ 99 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.9: Radical Equations The following box contains a list of steps that will help you solve the problems in this section. Steps to solve a radical equation. 1. Rewrite the equation so that the term with the radical is isolated. If the equation has two radicals you need to isolate one of them. I usually try to isolate a radical so that is has positive coefficient. 2. Get rid of the radical that you isolated by raising each side of the equation to the power of the index of the isolated radical. Note if the side of the equation opposite the isolated radical has a binomial you will probably need to FOIL, (assuming the radical is a square root). 3. Solve the equation obtained in step 2. If the equation still has a radical repeat steps 1 and 2. 4. Check the potential solutions in the original equation. *You can generate solutions that do not check in problems that have an even index. It is imperative that you check answers to equations of problems that contain an even index. ___________________________________________________________________ Example 2: Solve the equation. Be sure to check your answers. If a solution is extraneous, say so in your solution. This should help with #1-16. π a) βπ = π b) βπ β π = π 3 a) βπ = 4 The radical is isolated. I can skip to step 2 and raise both sides of the equation to the third power. 3 3 ( βπ) = (4)3 (Example 1 tells me I can drop the radical on the left side of the equation.) π = 64 Check: I will replace the b in the original problem with 4 to check my work. 3 β64 = 4 ( I know the cubed root of 64 is 4. You could make a prime factor tree to simplify the radical if you do not know this fact.) 4 = 4, The answer checks. b = 64 is a solution. Answer: b = 64 b) βπ₯ β 4 = 3 The radical is isolated. I can skip to step 2 and raise both sides of the equation to the second power. 2 (βπ₯ β 4) = (3)2 (Example 1 tells me I can drop the radical on the left side of the equation.) π₯β4=9 π₯ = 13 Check: I will replace the x in the original problem with 13 to check my work. β13 β 4 = 3 β9 = 3 3 = 3 The answer checks. x = 13 is a solution. Answer: x = 13 100 Chapter 4: Radicals and Complex Numbers Examples and Explanations ___________________________________________________________________ Section 4.9: Radical Equations Example 3: Solve the equation. Be sure to check your answers. If a solution is extraneous, say so in your solution. This should help with #17-24. a) ππβπ = π b) (π + π)πβπ = π a) π₯ 1β3 = 2 I need to write this problem as a radical to solve it using the steps given in the section. 3 βπ₯1 = 2 (Now I will cube both sides to remove the radical.) 3 3 ( βπ₯) = (2)3 x=8 Check: 81β3 = 2 (8 to the 1/3 power is the same as the cubed root of 8. The left side of the equation equals 2.) 2=2 Answer: x = 8 b) (π₯ + 1)1β2 = 3 I need to write this problem as a radical to solve it using the steps given in the section. 2 β(π₯ + 1)1 = 3 (I will write the left side without the index or the exponent, as neither is necessary.) βπ₯ + 1 = 3 (Now I will square both sides to get rid of the radical.) 2 (βπ₯ + 1) = (3)2 x+1 = 9 x=8 Check: (8 + 1)1β2 = 3 91β2 = 3 (The ½ power is the same as square root. I will simplify the left side by finding the square root of 9.) 3 = 3 (The solution checks.) Answer: x = 8 ___________________________________________________________________ Example 4: Solve the equation. Be sure to check your answers. If a solution is extraneous, say so in your solution. This should help with #25-32. a) π + βππ = π b) βππ + π = π a) 5 + β2π₯ = 9 I need to isolate the radical by subtracting 5 from each side of the equation. β2π₯ = 4 (Next, I will square both sides of the equation to get rid of the radical.) 2 (β2π₯) = (4)2 2π₯ = 16 Answer: x = 8 101 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.9: Radical Equations Check (example 4a): 5 + β2 β 8 = 9 5 + β16 = 9 5 + 4 = 9 (The solution checks.) Answer: x = 8 b) β5π₯ + 6 = π₯ The radical is isolated. I will square both sides to get rid of the radical. 2 (β5π₯ + 6) = (π₯)2 5π₯ + 6 = π₯ 2 ( I have to set this equal to zero and factor to solve.) 0 = π₯ 2 β 5π₯ β 6 0 = (π₯ + 1)(π₯ β 6) π₯ + 1 = 0 ππ π₯ β 6 = 0 π₯ = β1 ππ π₯ = 6 Check x = -1 Check x = 6 β5(β1) + 6 = β1 β5(6) + 6 = 6 ββ5 + 6 = β1 β30 + 6 = 6 β1 = β1 β36 = 6 1 = β1 (The solution does not check and is extraneous.) 6 = 6 (The solution checks, and x = 6 is a solution.) Answer: x = 6, x = -1 is extraneous ___________________________________________________________________ Example 5: Solve the equation. Be sure to check your answers. If a solution is extraneous, say so in your solution. This should help with #33-46. a) βππ β π = βππ β ππ b) βππ + π β βπ + π = π a) β2π¦ β 4 = β3π¦ β 20 I will square both sides to get rid of the radicals. 2 2 (β2π¦ β 4) = (β3π¦ β 20) (This allows me to drop both the radicals.) 2π¦ β 4 = 3π¦ β 20 (I will add 20 to both sides, and subtract 2y from both sides to solve this.) -2y +20 -2y +20 16 = y 102 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.9: Radical Equations Check (example 5a): β2(16) β 4 = β3(16) β 20 β28 = β28 (This is a true statement. I do not need to find the decimal equivalence. y = 16 is a solution. Answer: y = -16 b) β3π§ + 1 β βπ§ + 4 = 1 First I need to isolate a radical. I will add βπ§ + 4 to both sides and isolate the β2π§ + 1. β3π§ + 1 β βπ§ + 4 = 1 +βπ§ + 4 + βπ§ + 4 β3π§ + 1 = 1 + βπ§ + 4 (Now I will square both sides. This will get rid of the radical on the right side.) 2 2 (β3π§ + 1) = (1 + βπ§ + 4) (I have to FOIL the right side.) 3π§ + 1 = (1 + βπ§ + 4)(1 + βπ§ + 4) 2 3π§ + 1 = 1 + βπ§ + 4 + βπ§ + 4 + (βπ§ + 4) 3π§ + 1 = 1 + 2βπ§ + 4 + π§ + 4 (I will combine like terms on the right side.) 3π§ + 1 = π§ + 5 + 2βπ§ + 4 (I need to isolate the radical by subtracting z and 4 from both sides.) βπ§ β 5 β π§ β 5 2π§ β 4 = 2βπ§ + 4 (I am going to divide both sides by 2 to make the numbers easier to work with.) 2π§β4 2 = 2βπ§+4 2 π§ β 2 = βπ§ + 4 (I will square both sides to get rid of the radical.) 2 (π§ β 2)2 = (βπ§ + 4) (I have to FOIL the left side.) π§ 2 β 2π§ β 2π§ + 4 = π§ + 4 π§ 2 β 4π§ + 4 = π§ + 4 (I will set this equal to zero, and factor.) -z -4 -z - 4 π§ 2 β 5π§ = 0 π§(π§ β 5) = 0 π§ = 0 ππ π§ β 5 = 0 Potential solutions: z = 0, or z = 5 103 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.9: Radical Equations Check z = 0 Check z = 5 β3(5) + 1 β β5 + 4 = 1 β3(0) + 1 β β0 + 4 = 1 β16 β β9 = 1 β1 β β4 = 1 4β3=1 1β2=1 This solution checks. z = 5 is a solution. β1 = 1 This solution does not check and is extraneous. Answer: z = 5, z = 0 is extraneous. ___________________________________________________________________ 104 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.10: Complex Numbers Earlier in the chapter we learned that there are no real-valued square roots of a negative number. For example: ββ16 is not a real number because when a real number equals a positive number when squared, so that no real number squared equals (-16). However, there is a mathematical answer to the question what is the square root of (-16). The solution is not a real number. The solution to the question what is the square root of (-16) is an imaginary number. Most of the work we will do in this section will involve the imaginary number, i. Definition of the imaginary number i: π = ββπ ππ = βπ The imaginary number, i, will help us calculate the square roots of negative numbers. Definition of the square root of a negative number: ββπ = πβπ (This rule applies when x is a greater than or equal to zero, and βx is less than 0.) _____________________________________________________________________ Example 1: Simplify the expressions. This should help with #1-16. a) ββππ b) ββππ c) ββπ β ββπ a) ββ49 The rule in above the box states: ββ49 = πβ49 Answer: 7i b) ββ24 ββ24 = πβ24 = πβ4 β β6 Answer: ππβπ 105 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.10: Complex Numbers c) ββ6 β ββ8 We have multiplied square roots before. The rule for multiplying square roots specifically states that the values under each square root must be positive before the square roots can be multiplied. This means I need to pull the iβs out before I multiply. ββ6 β ββ8 = πβ6 β πβ8 = π 2 β48 = π 2 β16 β β3 = β1 β 4 β β3 Answer: βπβπ ___________________________________________________________________ Definition of a Complex N umber: A complex number is a number written in the form a + bi, where a and b are real numbers. We call βaβ the real part of a complex number. We call b the imaginary part of a complex number. In this section we will be asked to add, subtract, and multiply complex numbers. For the most part these operations can be done by treating the i as any variable. The only exception is that i2 must be replaced with the number (-1). _____________________________________________________________________ Example 2: Perform the indicated operations, write your answer in standard form. This should help with #17-44. a) (π β ππ) β (π + ππ) b) (ππ)(βππ) c) (π β ππ)(π β ππ) d) (π + ππ)(π β ππ) a) (6 β 3π) β (4 + 5π) I will drop the parenthesis, and combine like terms. (6 β 3π) β (4 + 5π) = 6 β 3π β 4 β 5π = (6 β 4) + (β3π β 5π) = 2 β 8π Answer: π β ππ 106 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.10: Complex Numbers b) (4π)(β5π) (4π)(β5π) = β20π 2 = β20(β1) = 20 Answer: 20 c) (3 β 2π)(5 β 6π) This is a FOIL problem. (3 β 2π)(5 β 6π) = 15 β 18π β 10π + 12π 2 = 15 β 28π + 12(β1) (I combined the inners and outers, and replaced i2 with (-1).) = 15 β 28π β 12 Answer: π β πππ d) (5 + 4π)(5 β 4π) This is a special FOIL problem. The factors are conjugate pairs. The inners and outers will cancel. (5 + 4π)(5 β 4π) = 25 β 20π + 20π β 16π 2 = 25 β 16(β1) (I canceled the inners and outers, and replaced i2 with (-1).) = 25 + 16 Answer: ππ ____________________________________________________________________ Example 3: Perform the division by multiplying by a factor equivalent to 1 that will take the i out of the denominator. This will help with #45-56. a) π ππ b) π π+ππ c) βππ πβππ 107 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.10: Complex Numbers 6 a) 7π There are two kinds of problems within example 3. This problem has a monomial in the denominator. I will π multiply by to get the desired answer. All problems between #45-56 with a monomial in the denominator π π can be solved by multiplying by . The next two problems have a binomial in the denominator and they will π be solved differently. 6 7π π 6π 6π 6π β π = 7π2 = 7(β1) = β7 (I will move the negative out of the denominator when I write the answer.) βππ π Answer: π ππ β π π (The negative can be written in the numerator or as a coefficient. The i can be written in the numerator or after the fraction.) b) 2 6+5π 6β5π This problem has a binomial in the denominator. I will multiply to get the desired answer. I call this 6β5π multiplying by conjugate of the denominator, although I am really multiplying by the conjugate over the conjugate. I can solve all problems between, #45-56, by multiplying by the conjugate of the denominator. 2 6+5π 6β5π 12β10π β 6β5π = 36β30π+30πβ25π2 (I distributed the 2 in the numerator and FOILed the denominator.) Answer: = 12β10π 36β25(β1) = 12β10π 36+25 = 12β10π 61 (I canceled the inners and outers, and changed the i2 to (-1).) (I will write this as two fractions in my answer.) ππ ππ β ππ π ππ 108 Chapter 4: Radicals and Complex Numbers Examples and Explanations Section 4.10: Complex Numbers c) β3π 2β7π I need to multiply by the conjugate of the denominator to solve this problem. β3π 2β7π β6πβ21π 2 2+7π β 2+7π = 4+14πβ14πβ49π2 (I distributed the -3i in the numerator, and FOILed the denominator.) = = = Answer: β6πβ21(β1) 4β49(β1) (I replaced the i2 with (-1) and canceled the inners and outers.) β6π+21 49+4 21β6π 53 ππ π β π ππ ππ _____________________________________________________________________ 109