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Table of Contents – Unit 1 INTRODUCTION: Why Study Algebra?………………………………..……4 . Modern Algebra Defined……………………………….....……………………5 All About Sets………………………………………..……….……….…………6 Types of Numbers………………………..…………….....……………...….…10 Properties………………………..………………………....……..….….………12 Xxxxxxxxxxxxxxxxxxxxxx…………….…….……..…....………….……….. 15 Xxxxx………………………..……………………….……....………….……….. 17 Xxxxxxxxxxxx…………………...………………………....………….……….. 18
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Introduction
Why Study Algebra? Algebra was invented as a tool to solve real-life problems in real-world situations. It has applications to virtually every human endeavor that exists -- from art to medicine to zoology. This textbook will therefore attempt to avoid treating the
Algebra II Mathematics Content Standards This discipline complements and expands the mathematical content and concepts of algebra I and geometry. Students who master algebra II will gain experience with algebraic solutions of problems in various content areas, including the solution of systems of quadratic equations, logarithmic and exponential functions, the binomial theorem, and the complex number system.
ARITHMETIC PROPERTIES
Solving Equations 1.0 Solving Equations and Inequalities Involving Absolute Value In mathematics, the absolute value of a given real number is defined as that number's distance from zero. In other words, absolute value is the number's numerical value without regard to its sign. For example, both 4 and −4 have an absolute value of 4 since, were you to plot each of them on a number line, both of them would be four units away from zero.
Of course, that means the absolute value of zero is always zero, since there is no distance between zero and itself. Consequently, absolute values are always equal to or greater than zero— but never less than zero. To designate the absolute value of a number or expression, we write that number or expression between “absolute brackets.” For example, “the absolute value of −2” would be written like this: | −2 | More generally, the absolute value of the number is denoted by | a | and is either a or – a depending on which is positive, or is zero if a = 0.
The definition may be written as
How to Solve Linear Equations in One Variable Involving Absolute Value EXERCISE 1: Solve the equation | x − 3 | = 4 When solving the solution set of the equation, make sure you consider both cases… If we remove the absolute value brackets, then x − 3 has two possibilities. 1. It is greater than or equal to zero. 2. It is less than zero.
For x − 3 to be greater than or equal to zero, x would have to be equal to or greater than three, and… |x−3|=x−3 In which case, the equation becomes x − 3 = 4 Solving for x, we have… x−3=4 x−3 +3=4 +3 x=7 Since 7 is indeed greater than zero, it is included in the solution set.
For x − 3 to be less than zero, x would have to be less than three, and… | x − 3 | = −(x − 3) = −x + 3 In this case, the equation becomes −x + 3 = 4 And solving for x, we have… −x + 3 = 4 −x + 3 − 3 = 4 – 3 −x = 1 x = −1 (both sides were multiplied by −1) Since −1 is indeed less than zero, it too is included in the solution set
The solution set of | x − 3 | = 4 is the union of the solution set in the two cases. Hence, the solution set is {−1, 7}
EXERCISE 2: Solve the equation | 2 − 3x | = 10 When solving the solution set of the equation, make sure you consider both cases… If we remove the absolute value brackets, then 2 – 3x has two possibilities. 3. It is greater than or equal to zero. 4. It is less than zero.
For 2 – 3x to be greater than or equal to zero, x would have to be equal to or less than 23 , and… | 2 – 3x | = 2 – 3x In which case, the equation becomes 2 – 3x = 10 Solving for x, we have… 2 – 3x = 10 – 3x = 8 x = − 83 Since − 83 is indeed less than 23 , it is included in the solution set.
For 2 – 3x to be less than zero, x would have to be greater than 23 , and… | 2 – 3x | = −(2 – 3x) = −2 + 3 x In this case, the equation becomes −2 + 3 x = 10 And solving for x, we have… −2 + 3x = 10 3 x = 12 x=4 Since 4 is indeed greater than
2 3
, it too is included in the solution set
The solution set of | 2 − 3x | = 10 is the union of the solution set in the two cases. Hence, the solution set is { − 83 , 4}
EXERCISE 3: Solve the equation | 3x + 1 | = 4x + 3 When solving the solution set of the equation, make sure you consider both cases… If we remove the absolute value brackets, then 3x + 1 has two possibilities. 5. It is greater than or equal to zero. 6. It is less than zero.
For 3x + 1 to be greater than or equal to zero, x would have to be greater than or equal to 13 , and… | 3x + 1 | = 3x + 1 In which case, the equation becomes 3x + 1 = 4x + 3 Solving for x, we have… 3x + 1 = 4x + 3 3x = 4x + 2 –x = 2 x = –2 Since – 2 is not greater than or equal to 13 , so it is not included in the solution set, hence the solution set is .
For 3x + 1 to be less than zero, x would have to be less than 13 , and… | 3x + 1 | = −(3x + 1) = −3x – 1 In this case, the equation becomes −3x – 1 = 4x + 3 And solving for x, we have… −3x – 1 = 4x + 3 −3x = 4 x = − 43 Since − 43 is indeed less than 13 , it is included in the solution set.
The solution set of | 3x + 1 | = 4x + 3 is the union of the solution set in the two cases. Hence, the solution set is { − 43 }
Rainbow book page 121 EXERCISE 4: Solve the equation Find the value(s) for x when | 4 − 2x | = −7. Since absolute values are never less than zero, the answer is: Ø. (The solution is a null, or empty, set.)
EXERCISE 2: Solve the equation | x − 4 | = 5. Normally, when we have the absolute value of a quantity involving a variable such as | x − 4 |, in order to find the solution we have to express that quantity as both a positive and negative value, and then apply the basic algebraic operations. Therefore, we must express x − 4 as both greater than or equal to zero and less than zero. If x − 4 ≥ 0, then x ≥ 4
And the equation is changed to... x–4 x–4+4 x x
=5 =5+4 =5+4 =9
So in this first case, the answer is the intersection of the solution sets of... x ≥ 4 and x = 9 Hence, the solution set is {9}.
On the other hand, if x − 4 < 0, then x < 4
And the equation is changed to... x–4 x–4+4 x x
=–5 =–5+4 =–5+4 =–1
So in this second case, the answer is the intersection of the solution sets of... x < 4 and x = −1 Hence, the solution set is {−1}. The solution set of | x − 4 | = 5 is the union of the solution sets in both of the two cases. Therefore, the solution set is {−1, 9}. EXERCISE 3: Find the solution set of | 2 − 3x | = 7. Remember that we have to express | 2 − 3x | as both greater than or equal to zero and less than zero. If 2 − 3x ≥ 0, then x ≤ ⅔ And the equation is changed to... 2 – 3x – 3x – 3x x
=7 =7–2 =5 = – 5/3
So in this first case, the answer is the intersection of the solution sets of... x ≤ ⅔ and x = −5/3 Hence, the solution set is {−5/3}.
On the other hand, if 2 − 3x < 0, then x > ⅔ And the equation is changed to... 2 – 3x = – 7 – 3x = – 7 – 2 – 3x = – 9
x =3 So in this second case, the answer is the intersection of the solution sets of... x > ⅔ and x = 3 Hence, the solution set is {3}.
The solution set of | 2 − 3x | = 7 is the union of the solution sets in both cases. Therefore, the solution set is {−5/3, 3}.
EXERCISE 4: Find the solution set of | 2x − 3 | = x + 5. Remember that we have to express | 2x − 3 | as both greater than or equal to zero and less than zero. If 2x − 3 ≥ 0, then x ≥ 3/2 And the equation is changed to... 2x – 3 2x – x – 3 x–3 x–3+3 x
=x+5 =x–x–5 =5 =5+3 =8
So in this first case, the answer is the intersection of the solution sets of... x ≥ 3/2 and x = 8 Hence, the solution set is {8}.
On the other hand, if 2x − 3 < 0, then x < 3/2 And the equation is changed to... 2x – 3 = – (x + 5) 2x – 3 = – x – 5
2x + x – 3 3x – 3 3x – 3 + 3 x
=–x+x–5 =–5 =–5+3 = – 2/3
So in this second case, the answer is the intersection of the solution sets of... x < 3/2 and x = −2/3 Hence, the solution set is {−⅔}.
The solution set of | 2 − 3x | = 7 is the union of the solution sets in both cases. Therefore, the solution set is {−2/3, 8}. EXERCISE 5: Find the solution set of | 3x + 1 | = 4x + 3. Remember that we have to express | 3x + 1 | as both greater than or equal to zero and less than zero. If 3x + 1 ≥ 0, then x ≥ −1/3 And the equation is changed to...
So in this first case, the answer is the intersection of the solution sets of... x ≥ −⅓ and x = −⅔ However, since −⅔ is neither greater than nor equal to −⅓, in this first case, the answer is a null set: {Ø}.
On the other hand, if 3x + 1 < 0, then x < −1/3 And the equation is changed to...
So in this second case, the answer is the intersection of the solution sets of... x < −1/3 and x = −4/7 Hence, the solution set is {−4/7}.
The solution set of | 2 − 3x | = 7 is the union of the solution sets in both cases. Therefore, the solution set is {−4/7}. EXERCISE 6: Find the solution set of | 2x − 2 | = 2 − 2x Remember that we have to express | 2x − 2 | as both greater than or equal to zero and less than zero. If 2x − 2 ≥ 0, then x ≥ 1 And the equation is changed to...
So in this first case, the answer is the intersection of the solution sets of... x ≥ 1 and x = 1 Hence, the solution set is {1}.
On the other hand, if 2x − 2 < 0, then x < 1 And the equation is changed to...
So in this second case, the answer is the intersection of the solution sets of... x < 1 and 0x = 0 Hence, the solution set is { x | x < 1}
The solution set of | 2 − 3x | = 7 is the union of the solution sets in both cases. Therefore, the solution set is {x | x ≤ 1}. EXERCISE 7: Find the solution set of | 3x − 7 | = | x + 3 | Remember that we have to express | 5x − 1 | as both greater than or equal to zero and less than zero. If 3x − 7 ≥ 0, then x ≥ 7/3 And the equation is changed to...
So in this first case, the answer is the intersection of the solution sets of... x ≥ 7/3 and x = 5 Hence, the solution set is {5}.
On the other hand, if 3x − 7 < 0, then x < 7/3 And the equation is changed to...
So in this second case, the answer is the intersection of the solution sets of... x < 7/3 and x = 1 Hence, the solution set is {1}.
The solution set of | 2 − 3x | = 7 is the union of the solution sets in both cases. Therefore, the solution set is {1, 5}. Application
1) What is the solution for this equation? | 2 − 3x | = 5 a) x = −4 or x = 4 b) x = −4 or x = 3 c) x = −1 or x = 4 d) x = −1 or x = 3
2) What is the solution set of the inequality below? 5 – | x + 4| ≤ – 3 a) –2 ≤ x ≤ 6 b) x ≤ –2 or x ≤ 6 c) –12 ≤ x ≤ 4 d) x ≤ –12 or x ≤ 4
Linear Equation and Inequalities. 2.0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
A problem that has several conditions on its variables cannot be expressed in just one equation. To satisfy all the relationships simultaneously requires a system of equations or inequalities. Purple book page 231 The general form of a linear equation in two variables isâ&#x20AC;Ś ax + by = c â&#x20AC;Świth a and b not both 0. Its graph is always a line.
Graphical Approach To solve linear equations with graphs, simply draw the graph each equation and see if the lines intersect at one point, are parallel, or are the same.
TRANSFORMING SYSTEMS OF LINEAR EQUATIONS BY> Substitution To solve a system of linear equations by substitution you proceed in a series of steps in which he replaced the system with a simpler system that has the same salute in set. This can be done I replacing one of equations withâ&#x20AC;Ś 1. an equally equation (linear operation ca) page 223, purple book 2. the sum (or difference) of the corresponding numbers of both sides of equations (substitution) 3. an equation warmed by substituting an expression in one of the original equations with a value assigned to it by the other equation. To put it another way, with this technique, a value for a variable is obtained from one equation. This value is then substituted that variable in the other equation. (Linear operation 2b, purple book page 223) 4. Linear operations: (page 233, purple book)
Formula purple book page 231
Polynomials. 3.0 Students are adept at operations on polynomials, including long division.
Factoring Polynomials .4.0 Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes.
Complex Numbers p. 380 5.0 Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane.
â&#x2C6;&#x2019; 1 is a pure imaginary number. However, when we mix real numbers with imaginary numbers we end up with an expression that is neither a real number nor a pure imaginary number. Expressions of this type are called complex numbers. A complex number is a number of the form a + bi where a and b are real numbers and i = â&#x2C6;&#x2019; 1 . The number a is called the real part of the complex number and b is called the imaginary part. The letter z is sometimes used to represent a complex number, that is, z = a + bi. When a complex number is written in the form a + bi, the complex number is said to be in simplified form, or in standard form. The form is sometimes referred to as the Cartesian or rectangular form of a complex number.
6.0 Students add, subtract, multiply, and divide complex numbers.
Rational Equations. 7.0 Students add, subtract, multiply, divide, reduce, and evaluate rational expressions with monomial and polynomial denominators and simplify complicated rational expressions, including those with negative exponents in the denominator.
Quadratic Equations. 8.0 Students solve and graph quadratic equations by factoring, completing the square, or using the quadratic formula. Students apply these techniques in solving word problems. They also solve quadratic equations in the complex number system. Purple book p. 346 We have studied systems involving to linear equations were to linear inequalities. However, we have not yet considered systems such as y = x2 – 6x + 5 2x – y = 7 because this system consists of one or tragic equation and one linear equation. We might also consider systems involving to quadratic equations.
Usually, it is easiest to solve such systems by substitution. A gravel often make it easier to understand the result. EXAMPLE: Let us solve the system y = x2 – 6x + 5 2x – y = 7 and illustrate the results graphically The first equation gives a value for y. This value may be substituted into the second equation. If we do this, the second equation becomes
Quadratic Functions. 9.0 Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a(x-b)2 + c.
Maxima, Minima and Zeros 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function.
Logarithms 13.0 Students use the definition of logarithms to translate between logarithms in any base.
It logarithm is just a different way of expressing and exponential relationship between two numbers. For instance, 23 = 8, soâ&#x20AC;Ś Log2 8 = 3 These two equations say exactly the same thing. The base of the logarithm can be any number greater than zero other than one, and by convention, if the base is 10, you don't write it. For example, log 1000 = 3 means log10 1000 = 3. Also, log base e (e â&#x2030;&#x2C6; 2.72) is written ln instead of loge. (Mathematicians use this so much that perhaps they won it a special abbreviation for it.)
Logarithm Properties 14.0 Students understand and use the properties of logarithms to simplify logarithmic numeric expressions and to identify their approximate values.
logc 1 = 0 logc c = 1 logc (ab) = logc a + logc b a log c = log c a − log c b b logc ab = b logc a log a b =
log c b log c a
With this property, you can compute something like log3 20 on your log 20 calculator by entering , using base 10 for c. log 3 loga ab = b alogab = b
The Laws of Logarithms.. 11.0 Students prove simple laws of logarithms.
Logarithms and Exponents. 11.1 Students understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.
Judging Arguments. 11.2 Students judge the validity of an argument according to whether the properties of real numbers, exponents, and logarithms have been applied correctly at each step.
The Laws of Fractional Exponents. 12.0 Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay. 15.0 Students determine whether a specific algebraic statement involving rational expressions, radical expressions, or logarithmic or exponential functions is sometimes true, always true, or never true.
16.0 Students demonstrate and explain how the geometry of the graph of a conic section (e.g., asymptotes, foci, eccentricity) depends on the coefficients of the quadratic equation representing it.
17.0 Given a quadratic equation of the form ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.
18.0 Students use fundamental counting principles to compute combinations and permutations.
19.0 Students use combinations and permutations to compute probabilities.
20.0 Students know the binomial theorem and use it to expand binomial expressions that are raised to positive integer powers.
21.0 Students apply the method of mathematical induction to prove general statements about the positive integers.
22.0 Students find the general term and the sums of arithmetic series and of both finite and infinite geometric series.
23.0 Students derive the summation formulas for arithmetic series and for both finite and infinite geometric series.
24.0 Students solve problems involving functional concepts, such as composition, defining the inverse function and performing arithmetic operations on functions.
25.0 Students use properties from number systems to justify steps in combining and simplifying functions.