We review some high school arithmetic and algebra.

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1 Chapter Algebra Review We review some high school arithmetic and algebra. 6.1 Arithmetic Operations Result 1 Let a, b, c R. Then a + b b + a (Commutative Law of Addition) ab ba (Commutative Law of Multiplication) (a + b) +c a +(b + c) (Associative Law of Addition) (ab)c a(bc) (Associative Law of Multiplication) a(b + c) (b + c)a ab + ac (Distributive Law) Note: It is important to realize that these rules only apply to the real numbers. In other number systems these rules may not apply! For example, if A and B are square matrices of the same size, we have AB 6 BA (except in special cases). 7

2 Exercise 3 (a) (3xy)( 4x) 3( 4)x y 1x y. (b) t(7x +tx 11) 14tx +4t x t. (c) 4 3(x ) 4 3x x. We often need the following formula: Formula 1 Using the Distributive Law three times yields: (a + b)(c + d) (a + b)c +(a + b)d ac + bc + ad + bd. Exercise 4 (x + 1)(3x 5) 6x +3x 10x 56x 7x 5. 8 Formula Some special cases of Formula 1 are: (i) (a + b)(a b) a b. ( Di erence of two squares ) (ii) (a + b) a +ab + b.( Aperfectsquare ) (iii) (a b) a ab + b.( Aperfectsquare ) Exercise 5 (a) (x + 6) x + 1x (b) 3(x 1)(4x + 3) (x + 6) 3(4x x 3) x 1 1x 3x 9 x 1 1x 5x 1. 9

3 . Fractions To add fractions with a common denominator we use the Distributive Law: a b + c 1 1 b a + c 1 + c (a + c) a b b b b. We have shown the following: Formula 3 a + c b a b + c b. However, note the following: Caution! Acommonmistake: a b + c 6 a b + a c. 30 To add fractions with a di erent denominator we make a common denominator: a b + c d a b c d + ad d b b d bd + bc ad + bc. bd bd Thus we have shown Formula 4 a b + c d ad + bc. bd Generally, we seek to find the least common denominator 1 of the fractions, e.g. a xy + b x y + c xy xya x y + yb x y + xc xya + yb + xc. x y x y 1 Aka, least common multiple of the denominators. 31

4 We multiply and divide fractions as follows: Formula 5 (i) a b c d ac bd, (ii) a b c d a b d c ad bc. We can also put the negative sign in a fraction in di erent places: Formula 6 a b a b a a b b. 3 Exercise 6 (a) (b) (x + 3) x 3 x 1 + x x + 3(x + ) + x(x 1) (c) s t u ut (d) 1 (x 1)(x + ) x 4 x x + 3 x 1+3 x. 3x +6+x x x + x s t u u s t. 1 x 1 4 x 4 x. x +x +6 x + x. 33

5 .3 Ratios Definition 4 A ratio is a comparison of the relative sizes of two (or more) parts of a whole. It is represented using a colon :, the word to, or using a fraction /. A proportion is a comparison of a part considered in relation to the whole. Note: The comparison between quantities is relative, so for example, the ratios :6and 1:3are equivalent (also, notice that ). Exercise 7 Suppose $50 of a $00 debt has been repaid. What is the ratio of debt paid to money owed? What is the proportion of money paid back? Proportion owed? Ratio of debt paid to money owed 50 : 150,or 1: 3,or 1/3. Proportion of debt paid 50/00 1/4. Proportion of money owed 150/00 3/4. 34 Exercise 8 The ratio of vanilla to chocolate to strawberry ice-cream cones sold is 3:5:.If 380 ice-cream cones were sold, how many of each flavour were sold? The proportions of vanilla, chocolate, and strawberry ice-cream cones are , , so, # vanilla cones #chocolatecones #strawberrycones Check: X 35

6 .4 Percentages As everyone knows, Definition 5 Percent means per 100, thus x percent, written x%, isshorthandforthe fraction x 100. Percentages can be negative, or greater than 100, e.g. 10% or 150%, thelatter referring to the improper fraction Finally, when we use the word of to express the percentage of a quantity we multiply, i.e. Definition 6 The expression y% of A isshorthandfor y 100 A. E.g., to find 0% of 50 do Exercise 9 The price of a skateboard is reduced by 5% in a sale. If the old price was $10, what is the sale price? The price reduction is $30. Thus the sale price is 4 $10 $30 $90. Exercise 10 Gas is selling at $1.15 per litre after a 5% rise in price. What was price of gas before the price increase? Method 1: 105% corresponds to $1.15, thus1% corresponds to $ ,so100% corresponds to $ $1.10 (to decimal places). 105 Method : LetP the original selling price of gas. Then, P + 5%P 1.15, so P P 1.15 ) P (1.05) 1.15 ) P or $

7 Exercise 11 A dress is selling for $100 after a 0% discount. What was the original selling price? Method 1: 80% corresponds to $100, so1% corresponds to $ $5,so100% corresponds 4 to $ $5 5 $15. 4 Method : Let SP the original selling price. Then SP 0 80 SP SP $100 ) SP $ Factoring We have used the Distributive Law to expand expressions. We also reverse this process to factor expressions as the product of simpler expressions. Expanding: x(x + 5) x + 10x. Factoring: x + 10x x(x + 5). To factor a quadratic of the form x + bx + c we note that (x + r)(x + s) x +(r + s)x + rs. Thus, when factoring a quadratic of the form x + bx + c we look for two numbers that add to give b and multiply to give c. 39

8 Exercise 1 Factor x +5x 4. The two numbers that add to give 5 and multiply to give 4 are 3 and 8. Therefore x +5x 4 (x 3)(x + 8). We can extend this method to problems where the coe cient of x is not 1. Exercise 13 Factor x 7x 4. We attempt to fill in the boxes below: x 7x 4 (x + )(x + ). After considering the factors of 4 (e.g., 1 and 4, and etc.), experimentation reveals that x 7x 4 (x + 1)(x 4)). 40 Some special cases include the di erence of two squares formula we had before: a b (a b)(a + b). Wealsohavethedi erence and sum of cubes: Formula 7 (i) a 3 b 3 (a b)(a +ab+b ) (Di erence of two cubes) (ii) a 3 + b 3 (a + b)(a ab + b ) (Sum of two cubes) Note: We can deduce (ii) from (i) by replacing b with b. Exercise 14 (a) 4x 5 (x) 5 (x 5)(x + 5) (b) x 3 +8 x (x + )(x x + 4) 41

9 To factor a polynomial of degree 3 or more, sometimes we can use the following result: Result (The Factor Theorem) Let P (x) is a polynomial. Then P (b) 0if and only if x P (x). b is a factor of Notes: If x b is a factor of P (x) then there is some polynomial Q(x) s.t. P (x) (x b)q(x). So to fully factor P (x) we now need to factor Q(x) (which is hopefully easier to factor than P (x)). 4 Exercise 15 Factor x 3 3x 10x + 4 Let P (x) x 3 3x 10x + 4. Bytrialanderrorwetryandfindb so that P (b) 0.Theintegerguessesare±1, ± etc. It is also true that b is a factor of 4 (think about it!). We find that P (1) 1, P( 1) 30, P() 0. Thus by the Factor Theorem, x is a factor. Instead of trying to use the Factor Theorem to find additional factors we use long division as follows: x x 1 x x 3 3x 10x + 4 x 3 +x x x 10x x 1x + 4 1x Therefore: x 3 3x 10x + 4 (x )(x x 1) (x )(x + 3)(x 4). Of course if the roots b are not easy integers we need another method for factorizing polynomials.

10 .6 Completing the square This is a technique for factoring a quadratic (polynomial of degree ). We rewrite a quadratic ax + bx + c in the form a(x + p) + q, whichisachievedby: 1. Factoring the number a from the terms involving x.. Adding and subtracting the square of half the coe cient of x. In general, we have apple P (x) ax + bx + c a x + b a x + c " a x + b b b a x + a a x + b + c a b 4a # + c a. (.1) Now we can solve P (x) 0for x which leads to the quadratic formula (see next section). 44 Exercise 16 Rewrite x + x +1by completing the square. The square of half the coe cient of x is 1/4.Thus x + x +1 x + x x Exercise 17 Rewrite x 1x + 11 by completing the square. x 1x + 11 [x 6x] + 11 [x 6x +9 9] + 11 [(x 3) 9] + 11 (x 3) 7. 45

11 .7 The Quadratic Formula Completing the square as above leads to the formula for the roots of a quadratic equation. Formula 8 (The Quadratic Formula) The roots of the quadratic equation ax + bx + c 0are x b ± p b 4ac. a 46 Justification: Starting from (.1) we set the quadratic function equal to zero in order to solve for x: a x + b b + c 0 a 4a ) a x + b b c b 4ac a 4a 4a ) x + b b 4ac a 4a ) x + b a ± p b 4ac a ) x b a ± p b a 4ac b ± p b 4ac a X. 47

12 Exercise 18 Solve the equation 5x +3x 30. With a 5, b 3, c 3,thequadraticformulagivesthe solutions 3 ± p 3 4(5)( 3) x 3 ± p 69. (5) 10 Definition 7 The quantity b 4ac in the quadratic formula is called the discriminant. There are three possibilities: 48 Result 3 1. If b 4ac > 0, theequationhasrealroots.. If b 4ac 0,therootsareequal. 3. If b 4ac < 0, theequationhasnorealroots(therootsarecomplex). These 3 cases correspond to the number of times the parabola y ax + bx + c crosses the x-axis, namely, 1 or 0 respectively. The three cases are illustrated graphically below: 49

13 Definition 8 If the quadratic cannot be factored using real numbers (case 3. above) then it is called irreducible. Exercise 19 Is it possible to factor x + x +using real numbers?. The discriminant is b 4ac 1 4(1)() 7 < 0, thus the answer is no.the quadratic is irreducible The Binomial Theorem We briefly cover an extension of the Perfect Square formula we had, namely (a + b) a +ab + b. If we multiply both sides by (a+b) and simplify we get another binomial expansion : (a + b) 3 a 3 +3a b +3ab + b 3. Before we give a general formula for (a + b) n we define the Factorial Notation. Definition 9 The factorial of a non-negative integer n, denoted n!, is the product of all positive integers less than or equal to n, i.e. n! n(n 1)(n ) 3 1 We also define 0! 1 (see the note below). Note:..., 4! 4 3!, 3! 3!,! 1!, soitisreasonablethat1! 1 0! 1. 51

14 Exercise 0 By definition we have: 0! 1 1! 1! 1 3! ! ! Exercise 1 Simplify the following expressions: (a) 7! 1000!,(b) 5! 998!!. (a) 7! 5! /5 /4 /3 / /1 /5 /4 /3 / / (b) 1000! 998!! ! 998!! ! 998!!

15 Result 4 (The Binomial Theorem) If n is a positive integer, then (a + b) n a n + na n 1 n(n 1) b + a n b n(n 1)(n ) + a n 3 b 3! 3! + + nab n 1 + b n. Notes: Observe the pattern in the formula! There is a lot more on this topic, for example see We will need this formula when we study Calculus. 54 Exercise Expand and simplify (x ) 5. We use the Binomial Theorem with a x, b, n 5yielding (x ) 5 x 5 +5x 4 ( ) ! x3 ( ) x ( ) 3 3! x 1 ( ) x 0 ( ) 5 4! 5! x 5 10x x 3 80x + 80x 3. 55

16 .9 Radicals The most commonly used radical is the square root. Definition 10 The square root of a number a real number x such that x a. 0, denoted p a,istheuniquenon-negative Notes: Notice that p a only makes sense if a numbers - but that is another story!) 0 (unless we are considering complex The solution of the equation x a is of course ± p a,thustakenotethat the square root is defined to be the positive square root (e.g., p 9 +3). Exponent form: p a a 1 (see next section). 56 Here are two rules for working with square roots: Formula 9 (i) p ab p a p b, (ii) p a b p a p b. Caution! p a + b 6 p a + p b For instance, with a b 1we would get p? Exercise 3 1. p 18/ p. p x y p 18/ p 93. p x p y x p y ( p x x, the absolutevalue ofx). 57

17 Let s consider the nth roots of numbers. Definition 11 The nth root (n N) of a number a, denoted np a,istheuniquerealnumber x such that x n a. Ifn is even then a must be non-negative and np a is the non-negative solution of x n a. Note: Exponent form: n p a a 1 n (see next section). Exercise 4 (a) (b) (c) (d) 4p 16 because p 16 undefined, because there is no number x such that x 4 < 0. 3p 64 4 because p 64 4 because ( 4) Thus we can see that we must be careful with roots of negative numbers: Caution! Even roots of negative numbers are undefined Odd roots of negative numbers are negative As with square roots we have the rules: Formula 10 (i) np ab np a np b, (ii) np a b np a np b. Exercise 5 Using Formula 10 we have 3 p x3 y 6 3p x 3 3 p y6 xy. 59

18 Rationalizing a Fraction To rationalize a fraction where the numerator or denominator contains an an expression like p a p b,wemultiplybothnumeratoranddenominatorby p a+ p b. Then we take advantage of the formula for a di erence of squares: ( p p p p p a b)( a + b)( a) ( p b) a b. Exercise 6 Rationalize the expression px +4 x p x +4. x p! p! x +4 x +4+ p x x +4+ (x + 4) 4 x( p x ) x x( p x ) 1 p, x 6 0. x +4+ Note: we had to exclude the possibility that x 0otherwise we are dividing by Exponents Numbers of the form b n involve two numbers, the base b and the exponent (or power ) n. Whenn is a natural number (i.e. an element of N) thenb n corresponds to repeated multiplication of the base. Definition 1 (Exponents) Let a be a positive real number and n N. Then 1. a n a a a (n factors). a a n 1 a n 4. a 1/n np a 5. a m/n np a m ( np a) m, m is an integer Note: later we discuss exponential functions. 61

19 Result 5 (Laws of exponents) Let a and b be positive numbers and let r and s be any rational numbers (that is, ratios of integers). Then 1. a r a s a r+s. ar a s a r s 3. (a r ) s a rs 4. (ab) r a r b r 5. a b r a r b r 6. a b 1 b a. Exercise 7 Simply the folllowing: (a) ( 3 ) 14. (b) 4 3/ p 4 3 p Alternatively: 4 3/ ( p 4) (c) x y x 1 + y x 1 y y x x y y + x x + 1 y xy (y x)(y + x) xy(y + x) y x x y xy y + x y x, y + x 6 0. xy (d) 1 3p 1 x 4 x x 4/3. 4/3 (e) x 3 y 4 x y z x3 y y8 x 4 x 7 y 5 z 4, y z 4 63

20 .11 Inequalities When working with inequalities note the following rules: Result 6 (Rules of inequalities) 1. Multiplication by a positive constant: If x>y,thenax > ay if a is a positive number.. Multiplication by a negative constant: If x>y,thenax < ay if a is a negative number. 3. Addition of inequalities: If x>yand u v, thenx + u>y+ v. (So if u v then x>yimplies x + u>y+ u.) Note rule., which says if we multiply both sides of an inequality by a negative number we reverse the direction of the inequality. E.g.: > 1 ) < 1 64 Things NOT to do with inequalities: Subtraction of inequalities: If x>y,andu>v,then(x u) 6> (y v). Multiplication of inequalities: If x, y, u, v are positive, then x>y, u>vimplies xu > yv. Theresult is not necessarily valid when some of the numbers are negative. Division of inequalities: If x>y,andu>v,thesedonot imply x/u > y/v. Exercise: you can find your own counter examples here. Exercise 8 Solve 1+x<7x +5: 1+x<7x +5 () 1 < 6x +5 (Rule 3.) () 4 < 6x (Rule 3.) () 6x > 4 (same thing) () x> (Rule 1.) In other words, the solution is the interval ( /3, 1). 65

21 Exercise 9 (Chart Method) Solve x 5x +6apple 0: First factor the left hand side (x )(x 3) apple 0. The equation (x )(x 3) 0 has the solutions and 3, thusthesenumbersdividethereal number line into three intervals: ( 1, ) (, 3) (3, 1). On each of these intervals we determine the sign of the factors and record this in a table: Thus from the table we see that(x )(x 3) is negative when Interval x x 3 (x )(x 3) x< + <x<3 + x> <x<3, i.e.thesolutionoftheinequality(x )(x 3) apple 0 is the interval [, 3]. 66 Another way to get the same information is to use test values. Forinstance,checking x 1for the interval ( 1, ) in x 5x +6yields 1 5(1) + 6. The polynomial doesn t change sign inside the three intervals so we conclude that it is positive on ( 1, ). Avisualmethodforsolvingtheinequalityistojustgraphtheparabolay x 5x +6and observe that the curve is on or below the x-axis when apple x apple 3. 67

22 Exercise 30 (Chart Method) Solve x 3 +3x > 4x: We can factor this as x(x +3x 4) > 0 or x(x 1)(x + 4) > 0. The equation x(x 1)(x + 4) 0 has the solutions 0, 1 and 4, thusthese numbers divide the real number line into four intervals: ( 1, 4) ( 4, 0) (0, 1) (1, 1). On each of these intervals we determine the sign of the factors and record this in a table. Thus we see from the table that x(x 1)(x + 4) is greater than zero when Interval x x 1 x +4 x(x 1)(x + 4) x< 4 4 <x< <x<1 + + x> <x<0 and x>1, thustherequiredsolutionsetis( 4, 0) [ (1, 1). 68 Amuchquickerwaytoseethisisjusttographthefunctiony x(x 1)(x + 4), and observe where the curve is above the x-axis: 69

23 .1 Absolute Value Definition 13 The Absolute Value of a number a, denoted a, isthedistancefroma to 0 on the real number line. Thus ( a if a 0 a a if a<0. Notes: Thus a 0 for every number a. E.g., 3 3, 3 3, 0 0. Sometimes we need to work out the sign of the number before we can evaluate the absolute value, e.g. p 1 p 1, 3 3, ln(0.9) ln(0.9) (ln(0.9) ). p a a. E.g., p ( 1) p as a piecewise defined function (without the absolute value sym- Exercise 31 Re-write 3x bol). 3x ( 3x if 3x 0 (3x ) if 3x < 0 ( 3x if x /3 3x if x</3. Note the following basic rules for absolute values: Result 7 Suppose a and b are real numbers and n is an integer. Then 1. ab a b,. a b a b (b 6 0), 3. a n a n. 71

24 For solving equations or inequalities involving absolute values, the following statements are helpful: Result 8 Suppose a>0. Then 4. x a if and only if x ±a, 5. x <a if and only if a<x<a, 6. x >a if and only if x>aor x< a. Notes: For instance, the statement x <asays that the distance from x to the origin is less than a, whichistrueifx lies between a and a. The expression a b ( b a ) representsthedistancebetweena and b. 7 Exercise 3 Solve x 5 3. By Property 4., x 5 3is equivalent to x 53 or x 5 3. So x 8or x.thus,x 4or x 1. Exercise 33 Solve x 5 <. 1 By Property 5., x 5 < is equivalent to <x 5 <. Therefore, adding 5 to each side, we have 3 <x<7, and the solution set is the open interval (3, 7). 73

25 Exercise 34 Solve 3x + 4. By Property 4. and 6., 3x + 4 is equivalent to 3x + 4 or 3x +apple 4. So 3x, whichgivesx /3. In the second case 3x apple 6, whichgives x apple. Sothesolutionsetis( 1, ] [ [/3, 1). 74 Chapter 3 Minimal Trigonometry and Radians 3.1 Some Trig Definitions & Identities Although we briefly reviewsome basic trigonometry, weassume students have studied trigonometry before. Trigonometry is useful for more than just calculating angles in atriangle. The trigfunctions areamongthelistof elementaryfunctions that are used to model the real world, for example, periodicity in the stock market, or economic trends using Fourier Series 1. 1 A representation of a function in terms of a sum of sine and cosine functions. 75

Arithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get

Arithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get MCA AP Calculus AB Summer Assignment The following packet is a review of many of the skills needed as we begin the study of Calculus. There two major sections to this review. Pages 2-9 are review examples