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1.2A β Radical Operations Example 1 β Simplify a) 42 1 b) 162 d) 23 c) β16 1 3 e) 83 f) β8 What is the relationship between a, b, c? What is the relationship between d, e, f? RADICAL EXPRESSION If π₯ π = π , then π₯ = = If β²πβ² and π₯ are real numbers and π is a positive integer, then π₯ is an 1 π ππ‘β root of β²πβ² (ie π₯ = ππ = βπ ) if π₯ π = π πππ root theorems: 1) If π is positive and π is even, then there exist TWO real ππ‘β roots. Example 2 β Solve for π₯ a) π₯ 2 = 16 b) π₯ 2 = 11 c) π₯ 4 = 81 d) π₯ 4 = 5 2) If π is negative and π is even, then there are NO real number solutions. Example 3 β Solve for π₯ a) π₯ 2 = 25 b) π₯ 4 = β7 3) If π is odd, then there is ONE real ππ‘β root of π. Example 4 β Solve for π₯ a) π₯ 3 = 8 b) π₯ 3 = β8 c) π₯ 5 = β4 4) If π is zero, then there is ONE real ππ‘β root of π, and it is ZERO. Example 5 β Solve for π₯ a) π₯ 2 = 0 b) π₯ 5 = 0 Radical Properties from Math 10: 1 π 1) ππ = βπ as discussed in above notes π π 1 π π π π 2) π = (π ) = ( βπ) π 1 Example: 16 3) πβ π = (ππ )βπ = ( βπ)βπ = π π 4) β = π π π βπ π βπ π 1 π 3 Example: β π 5) βππ = βπ β βπ π ( βπ )π 27 64 Example: β12 3 4 2 Example: 27β3 1.2B β Working with Radicals absolute value Absolute Value is βhow many jumps the number is from zeroβ. Stated another way, it is the distance from zero on the number line, regardless of direction. Distances are always POSITIVE |3| =____. β3 is 3 jumps values. 3 is 3 jumps from zero, so the absolute value of 3, or from zero, so |β3| =____. So if|π₯| = 3, π₯ could have been _____ OR _____. For every absolute value solution, there is a positive and negative possibility. Example 1 - Evaluate a) |5| b) |β7| 5 c) |β0.34| 3 d) |6| e) |β6 8| Example 2 - What are the possible values of x? a) |π₯| = 6 Radical Review b) |π₯| = 9.7 c) |π₯| = β2 Example 3 - Identify and define all parts of the radical, then simplify: 3 5 β8 Roots of Positive Powers of π: Case 1: When π β₯ π in βππ with π a positive integer. The square roots of negative numbers are undefined in the set of real numbers. Therefore, if π₯ β₯ 0, simplification is easier to realize. For example: 3 βπ₯ 2 = βπ₯ 2 = βπ₯ 3 = 3 βπ₯ 4 = 3 βπ₯ 5 = 3 βπ₯ 6 = 3 βπ₯ 3 = βπ₯ 4 = βπ₯ 5 = βπ₯ 6 = Example 4 β Simplify. Assume all variables represent positive numbers a) β16π¦ 2 b) βπ₯ 4 π¦ 3 c) β25π₯ 5 π¦ 3 π§ 2 3 d) β8π₯ 4 π¦ 5 3 e) β27π₯ 3 π¦ 6 Before Case 2 is discussed, itβs important to understand the Principal Square Root Theorem. Every positive number βnβ has two square roots. One is positive, and the other is negative. Example: π₯ 2 = 16, so π₯ = ±β16 = ±4 The PRINCIPAL SQUARE ROOT is the POSITIVE NUMBER SQUARE ROOT. Unless otherwise stated, the βsquare rootβ of a number refers ONLY to the principal square root. Thus, β16 = 4 Case 2: When π is a real number in βππ , with π a positive integer, we may require RESTRICTIONS to be placed on the expression. If the exponent of a radicand is even, then a NEGATIVE value will be changed into a POSITIVE value before the square root is taken. Example: β(β3)2 = = = When variables exist in a radicand, it is not known if the variable represents a negative number, zero, or a positive number. An ABSOLUTE VALUE is sometimes needed to ensure that the result is a positive number (required ONLY when π₯ changes from an EVEN power to an ODD power). Examples: βπ₯ 2 = βπ₯ β π₯ = |π₯| βπ₯ 4 = π₯ 2 (no absolute value needed as π₯ 2 can only result as a positive) βπ₯ 6 = |π₯ 3 | etc. If the exponent in the radicand is ODD, then a NEGATIVE value of π₯ will make the value NEGATIVE, which is UNDEFINED in the real number system. Therefore, π₯ β₯ 0 for all odd exponents. Examples: βπ₯ 3 = β(π₯ β π₯) β π₯ = π₯ βπ₯ ; π₯ β₯ 0 βπ₯ 5 = βπ₯ 7 = Example 5 β Simplify. Let the variables be any real numbers. a) β16π₯ 2 b) β25π₯ 2 π¦ 4 c) βπ₯14 d) β36π₯ 2 π¦ 5 Summary: π) ββπ is UNDEFINED (unless π β€ π) 2) βπ = βπ has no real solution AND ββπ = π has no real solution 3) ππ = βπ has no real solutions, as does πππ = βπ 1.3 β Simplifying Radicals Radicals can be written as fractional exponents, as learned in Math 10. 1 Examples: β2 = 22 1 3 1 π βπ₯ = π₯ 3 Generally: βπ = ππ Three Important Relationships of Radicals, all from Math 10: 1) π βπ π = π, π β₯ 0 because π π π βπ π = 1 (ππ )π π 2 = π Example: β52 = 1 π 1 1 = π π 2) βππ = βπ × βπ, π, π β₯ 0 because βππ = (ππ)π = ππ × π π = βπ × βπ π π 3) β = π π βπ π βπ π π π 1 , π β₯ 0, π > 0 because β = ( ) = π π π 1 ππ 1 ππ π = βπ π βπ Simplifying Expressions Containing Radicals Example 1 β Simplify β20 Two Methods: It is beneficial to know the perfect squares up to 144, perfect cubes up to 125, and perfect fourths up to 81. Example 2 β Simplify a) β8 b) β27 c) 3β52 3 d) β24 3 e) 5 β81 4 f) β32 Example 3 β Simplify (assume variables are positive) a) β18π₯ 3 π¦ 6 3 e) β54π5 π10 b) β63π7 π4 4 f) βπ 7 c) β32π₯ 8 π¦11 4 g) β162π₯ 3 π¦11 π§ 5 3 d) β40π4 π 8 π15 3 h) β π₯ 13 64 Changing Mixed Radicals to Entire Radicals Example 4 β Change to Entire (assume variables are positive) 3 a) 4β3 b) 3β5 c) 2 β7 d) β2π₯β6π₯ e) π₯ 3 βπ₯ 3 f) 3π2 π βπ2 π g) 3π₯ 2 π¦ 3 β2π₯π¦ 2 5 1.4 β Adding and Subtracting Radical Expressions Like Radicals βLike Radicalsβ work very similar to βLike Termsβ. Simplify: 3π₯ + 2π₯ Simplify 3β2 + 2β2 Like radicals have __________________________________________ Steps for adding & subtracting like radicals: Example 1 - Simplify a) 7β3 β 2β3 d) 2β75 + 3β3 3 3 b) β5 β10 β 6 β10 3 c) 4β2 β 5β2 e) ββ27 + 3β5 β β80 β 2β12 f) β9π β 3β16π, π β₯ 0 In example f, why does b have to be greater than or equal to zero? Example 2 β Simplify a) β27π₯π¦ + β8π₯π¦ c) 3π₯ β63π¦ β 5β28π₯ 2 π¦ , π¦ β₯ 0 3 3 b) 4 β16 + 3β54 d) 53 β16π₯ 4 π¦ 5 2 3 β π₯π¦ β54π₯π¦ 2 1.5A β Multiplying & Dividing Radical Expressions multiplying radicals Example 1 - Multiply 2β5 (3β5) Verify your answer: To multiply radicals: In general: Example 2 - Simplify: a) 5β3 (β6) b) 2β6 (4β8) 3 c) β3β2π₯ (4β3π₯) π₯ β₯ 0 e) (4β2 + 3)(β7 β 5β14) 3 3 d) β2β11(4β2 β 3 β3) 3 3 3 f) ( βπ₯ + 3βπ¦)(βπ₯ 2 β 3βπ₯π¦ + βπ¦ 2 ) dividing radicals Example 3 - Divide 6β12 3β6 Verify your answer: To divide radicals: In general: 3 Example 4 - Simplify: a) β24 β14 3 8 β2 b) 2β51 β3 Multiplying & Dividing Terms with Different Indices 3 Example 5 β Simplify βπ₯ 3 ( βπ₯ ) , π₯ β₯ 0 Example 6 β Simplify βπ₯ 3 3 βπ₯ ,π₯ > 0 c) β18π₯ 3 β3π₯ ,π₯ > 0 1.5B β Rationalizing the Denominator rationalizing the denominator A final answer cannot have a radical in the denominator. Therefore, you may have to βrationalize the denominatorβ β a process that will eliminate the radical from the denominator without changing the value of the expression. Example 1 - Rationalize: a) 3 b) β β5 2 7 c) 6 7 β5 If the denominator is a radical monomial, multiply the numerator and denominator by that radical. 3 4 2 e) β π¦ Example 2 β Rationalize: If the denominator is a radical binomial, multiply the numerator & denominator by its conjugate. f) a) β5π₯ 4 β10π₯ 3 3 βπ₯β2 g) b) 2 βπ₯+1 2+β2 3β5β4 d) 2 β5 3 3 β6 c) βπ+β2π βπββ2π , π, π β₯ 0 Example 3 - Simplify a) 6β 3 4π₯ ,π₯ > 0 b) β7 3 2 β9π ,π β 0 Example 4 β The surface area of a sphere is π = 4ππ 2 . If the surface area of the sphere is 144 mm2, what is the radius? 1.6 β Radical Equations Radical Equations are mathematical equations that include a radical, such as: 2β6π₯ β 1 = 11 . If the index of the radical is even, there are restrictions on the variable. Since it is not possible to find the square root of a negative number, the radicand cannot be negative. Example 1 β Find the restriction on the variable: a) 2β6π₯ β 1 = 11 c) 7ββ2π₯ + 3 = 35 b) βπ₯ + 2 = 49 d) β3π₯ + 4 = β2π₯ β 4 Steps to solving radical equations: 1. Find the restrictions on the variable in the radicand (if the index is even). Remember, the radicand must be set to β₯ 0 and then solved (if you multiply or divide by a negative number to both sides, FLIP the inequality). 2. Get the radical all by itself on one side of the equation. 3. If the index is 2, square both sides (if index is 3, cube both sides, etc.) and then solve for the variable. 4. See if the solution if affected by the restriction. 5. Check the answer using the original equation to see if solutions are valid or extraneous. Example 2 β Solve a) 2β6π₯ β 1 = 11 b) β8 + β 3π¦ 5 =2 Example 3 β Solve a) 4 + β2π₯ β 3 = 1 b) β2βπ₯ β 5 = 16 Example 4 - Solve a) β10π₯ β 7 = 3βπ₯ b) 2βπ₯ = β7π₯ + 6