EVERYTHING MATHS




Mathematics is commonly thought of as being about numbers but mathematics is actually a language! Mathematics is the language that nature speaks to us in. As we learn to understand and speak this language, we can discover many of nature’s secrets. Just as understanding someone’s language is necessary to learn more about them, mathematics is required to learn about all aspects of the world – whether it is physical sciences, life sciences or even finance and economics. The great writers and poets of the world have the ability to draw on words and put them together in ways that can tell beautiful or inspiring stories. In a similar way, one can draw on mathematics to explain and create new things. Many of the modern technologies that have enriched our lives are greatly dependent on mathematics. DVDs, Google searches, bank cards with PIN numbers are just some examples. And just as words were not created specifically to tell a story but their existence enabled stories to be told, so the mathematics used to create these technologies was not developed for its own sake, but was available to be drawn on when the time for its application was right. There is in fact not an area of life that is not affected by mathematics. Many of the most sought after careers depend on the use of mathematics. Civil engineers use mathematics to determine how to best design new structures; economists use mathematics to describe and predict how the economy will react to certain changes; investors use mathematics to price certain types of shares or calculate how risky particular investments are; software developers use mathematics for many of the algorithms (such as Google searches and data security) that make programmes useful. But, even in our daily lives mathematics is everywhere – in our use of distance, time and money. Mathematics is even present in art, design and music as it informs proportions and musical tones. The greater our ability to understand mathematics, the greater our ability to appreciate beauty and everything in nature. Far from being just a cold and abstract discipline, mathematics embodies logic, symmetry, harmony and technological progress. More than any other language, mathematics is everywhere and universal in its application. Contents 1 Exponents and surds 4 1.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Rational exponents and surds . . . . . . . . . . . . . . . . . . . . . . . 8 1.3 Solving surd equations . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.4 Applications of exponentials . . . . . . . . . . . . . . . . . . . . . . . 23 1.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2 Equations and inequalities 30 2.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.2 Completing the square . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.3 Quadratic formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2.5 Finding the equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 2.6 Nature of roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 2.7 Quadratic inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 60 2.8 Simultaneous equations . . . . . . . . . . . . . . . . . . . . . . . . . . 67 2.9 Word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 2.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3 Number patterns 86 3.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.2 Quadratic sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 3.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 4 Analytical geometry 104 4.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 4.2 Equation of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.3 Inclination of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 4.4 Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.5 Perpendicular lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 4.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5 Functions 146 5.1 Quadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 5.2 Average gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 5.3 Hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 5.4 Exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 5.5 The sine function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 5.6 The cosine function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 5.7 The tangent function . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 5.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 6 Trigonometry 240 6.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 6.2 Trigonometric identities . . . . . . . . . . . . . . . . . . . . . . . . . . 247 6.3 Reduction formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 6.4 Trigonometric equations . . . . . . . . . . . . . . . . . . . . . . . . . 266 6.5 Area, sine, and cosine rules . . . . . . . . . . . . . . . . . . . . . . . . 280 6.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 7 Measurement 308 7.1 Area of a polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 7.2 Right prisms and cylinders . . . . . . . . . . . . . . . . . . . . . . . . 311 7.3 Right pyramids, right cones and spheres . . . . . . . . . . . . . . . . . 318 7.4 Multiplying a dimension by a constant factor . . . . . . . . . . . . . . 322 7.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 8 Euclidean geometry 332 8.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 8.2 Circle geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 8.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 9 Finance, growth and decay 374 9.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 9.2 Simple and compound depreciation . . . . . . . . . . . . . . . . . . . 377 9.3 Timelines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 9.4 Nominal and effective interest rates . . . . . . . . . . . . . . . . . . . 394 9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 10 Probability 402 10.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402 10.2 Dependent and independent events . . . . . . . . . . . . . . . . . . . 411 10.3 More Venn diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 10.4 Tree diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 10.5 Contingency tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 10.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 11 Statistics 440 11.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440 11.2 Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444 11.3 Ogives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451 11.4 Variance and standard deviation . . . . . . . . . . . . . . . . . . . . . 455 11.5 Symmetric and skewed data . . . . . . . . . . . . . . . . . . . . . . . 461 11.6 Identification of outliers . . . . . . . . . . . . . . . . . . . . . . . . . . 464 11.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 12 Linear programming 472 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 Solutions to exercises 483 2 Contents CHAPTER 1 Exponents and surds 1.1 Revision 4 1.2 Rational exponents and surds 8 1.3 Solving surd equations 19 1.4 Applications of exponentials 23 1.5 Summary 25 1 Exponents and surds 1.1 Revision EMBF2 The number system EMBF3 The diagram below shows the structure of the number system: Real R Irrational Q′ Rational Q Integer Z Natural N Whole N0 Non-real R ′ See video: 2222 at  We use the following definitions: • N: natural numbers are {1; 2; 3; . . .} • N0: whole numbers are {0; 1; 2; 3; . . .} • Z: integers are {. . . ; −3; −2; −1; 0; 1; 2; 3; . . .} • Q: rational numbers are numbers which can be written as a b where a and b are integers and b 6= 0, or as a terminating or recurring decimal number. Examples: − 7 2 ; −2,25; 0; √ 9; 0,8;˙ 23 1 • Q0 : irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator as integers. Irrational numbers also include decimal numbers that neither terminate nor recur. Examples: √ 3; √5 2; π; 1+√ 5 2 ; 1,27548 . . . • R: real numbers include all rational and irrational numbers. • R 0 : non-real numbers or imaginary numbers are numbers that are not real. Examples: √ −25; √4 −1; − q − 1 16 See video: 2223 at 4 1.1. Revision Exercise 1 – 1: The number system Use the list of words below to describe each of the following numbers (in some cases multiple words will be applicable): • Natural (N) • Whole (N0) • Integer (Z) • Rational (Q) • Irrational (Q0 ) • Real (R) • Non-real (R 0 ) 1. √ 7 2. 0,01 3. 16 2 5 4. q 6 1 4 5. 0 6. 2π 7. −5,38˙ 8. 1− √ 2 2 9. − √ −3 10. (π) 2 11. − 9 11 12. √3 −8 13. 22 7 14. 2,45897 . . . 15. 0,65 16. √5 −32 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 2224 2. 2225 3. 2226 4. 2227 5. 2228 6. 2229 7. 222B 8. 222C 9. 222D 10. 222F 11. 222G 12. 222H 13. 222J 14. 222K 15. 222M 16. 222N  Laws of exponents EMBF4 We use exponential notation to show that a number or variable is multiplied by itself a certain number of times. The exponent, also called the index or power, indicates the number of times the multiplication is repeated. base a n exponent/index a n = a × a × a × . . . × a (n times) (a ∈ R, n ∈ N) See video: 222P at  Chapter 1. Exponents and surds 5 Examples: 1. 2 × 2 × 2 × 2 = 24 2. 0,71 × 0,71 × 0,71 = (0,71) 3 3. (501) 2 = 501 × 501 4. k 6 = k × k × k × k × k × k For x 2 , we say x is squared and for y 3 , we say that y is cubed. In the last example we have k 6 ; we say that k is raised to the sixth power. We also have the following definitions for exponents. It is important to remember that we always write the final answer with a positive exponent. • a 0 = 1 (a 6= 0 because 0 0 is undefined) • a −n = 1 an (a 6= 0 because 1 0 is undefined) Examples: 1. 5 −2 = 1 5 2 = 1 25 2. (−36)0x = (1)x = x 3. 7p −1 q 3t−2 = 7t 2 pq3 We use the following laws for working with exponents: • a m × a n = a m+n • am an = a m−n • (ab) n = a n b n • a b n = a n bn • (a m) n = a mn where a > 0, b > 0 and m, n ∈ Z. Worked example 1: Laws of exponents QUESTION Simplify the following: 1. 5(m2t ) p × 2(m3p ) t 2. 8k 3x 2 (xk) 2 6 1.1. Revision 3. 2 2 × 3 × 7 4 (7 × 2)4 4. 3(3b ) a SOLUTION 1. 5(m2t ) p × 2(m3p ) t = 10m2pt+3pt = 10m5pt 2. 8k 3x 2 (xk) 2 = 8k 3x 2 x 2k 2 = 8k (3−2)x (2−2) = 8k 1x 0 = 8k 3. 2 2 × 3 × 7 4 (7 × 2)4 = 2 2 × 3 × 7 4 7 4 × 2 4 = 2(2−4) × 3 × 7 (4−4) = 2−2 × 3 = 3 4 4. 3(3b ) a = 3 × 3 ab = 3ab+1 Worked example 2: Laws of exponents QUESTION Simplify: 3 m − 3 m+1 4 × 3m − 3m SOLUTION Step 1: Simplify to a form that can be factorised 3 m − 3 m+1 4 × 3m − 3m = 3 m − (3m × 3) 4 × 3m − 3m Step 2: Take out a common factor = 3 m(1 − 3) 3m(4 − 1) Step 3: Cancel the common factor and simplify = 1 − 3 4 − 1 = − 2 3 Chapter 1. Exponents and surds 7 Exercise 1 – 2: Laws of exponents Simplify the following: 1. 4 × 4 2a × 4 2 × 4 a 2. 3 2 2−3 3. (3p 5 ) 2 4. k 2k 3x−4 k x 5. (5z−1 ) 2 + 5z 6. ( 1 4 ) 0 7. (x 2 ) 5 8. a b −2 9. (m + n) −1 10. 2(p t ) s 11. 1 1 a −1 12. k 0 k−1 13. −2 −2−a 14. −h (−h)−3 15. a 2 b 3 c 3d 2 16. 107 (70 ) × 10−6 (−6)0 − 6 17. m3n 2 ÷ nm2 × mn 2 18. (2−2 − 5 −1 ) −2 19. (y 2 ) −3 ÷ x 2 y 3 −1 20. 2 c−5 2 c−8 21. 2 9a × 4 6a × 2 2 8 5a 22. 20t 5p 10 10t 4p 9 23. 9q −2s q−3sy−4a−1 2 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 222R 2. 222S 3. 222T 4. 222V 5. 222W 6. 222X 7. 222Y 8. 222Z 9. 2232 10. 2233 11. 2234 12. 2235 13. 2236 14. 2237 15. 2238 16. 2239 17. 223B 18. 223C 19. 223D 20. 223F 21. 223G 22. 223H 23. 223J  See video: 222Q at  1.2 Rational exponents and surds EMBF5 The laws of exponents can also be extended to include the rational numbers. A rational number is any number that can be written as a fraction with an integer in the numerator and in the denominator. We also have the following definitions for working with rational exponents. 8 1.2. Rational exponents and surds • If r n = a, then r = √n a (n ≥ 2) • a 1 n = √n a • a − 1 n = (a −1 ) 1 n = n q 1 a • a m n = (a m) 1 n = √n am where a > 0, r > 0 and m, n ∈ Z, n 6= 0. For √ 25 = 5, we say that 5 is the square root of 25 and for √3 8 = 2, we say that 2 is the cube root of 8. For √5 32 = 2, we say that 2 is the fifth root of 32. When dealing with exponents, a root refers to a number that is repeatedly multiplied by itself a certain number of times to get another number. A radical refers to a number written as shown below. radical sign √n a degree radicand }radical See video: 223K at  The radical symbol and degree show which root is being determined. The radicand is the number under the radical symbol. • If n is an even natural number, then the radicand must be positive, otherwise the roots are not real. For example, √4 16 = 2 since 2 × 2 × 2 × 2 = 16, but the roots of √4 −16 are not real since (−2) × (−2) × (−2) × (−2) 6= −16. • If n is an odd natural number, then the radicand can be positive or negative. For example, √3 27 = 3 since 3 × 3 × 3 = 27 and we can also determine √3 −27 = −3 since (−3) × (−3) × (−3) = −27. It is also possible for there to be more than one n th root of a number. For example, (−2)2 = 4 and 2 2 = 4, so both −2 and 2 are square roots of 4. A surd is a radical which results in an irrational number. Irrational numbers are numbers that cannot be written as a fraction with the numerator and the denominator as integers. For example, √ 12, √3 100, √5 25 are surds. Worked example 3: Rational exponents QUESTION Write each of the following as a radical and simplify where possible: 1. 18 1 2 2. (−125) − 1 3 Chapter 1. Exponents and surds 9 3. 4 3 2 4. (−81) 1 2 5. (0,008) 1 3 SOLUTION 1. 18 1 2 = √ 18 2. (−125) − 1 3 = p3 (−125)−1 = 3 r 1 −125 = 3 r 1 (−5)3 = − 1 5 3. 4 3 2 = (43 ) 1 2 = √ 4 3 = √ 64 = 8 4. (−81) 1 2 = √ −81 = not real 5. (0,008) 1 3 = 3 r 8 1000 = 3 r 2 3 103 = 2 10 = 1 5 See video: 223M at  Worked example 4: Rational exponents QUESTION Simplify without using a calculator: 5 4−1 − 9−1 1 2 SOLUTION Step 1: Write the fraction with positive exponents in the denominator 5 1 4 − 1 9 !1 2 Step 2: Simplify the denominator 10 1.2. Rational exponents and surds = 5 9−4 36 !1 2 = 5 5 36 !1 2 = 5 ÷ 5 36 1 2 = 5 × 36 5 1 2 = (36) 1 2 Step 3: Take the square root = √ 36 = 6 Exercise 1 – 3: Rational exponents and surds 1. Simplify the following and write answers with positive exponents: a) √ 49 b) √ 36−1 c) √3 6−2 d) 3 r − 64 27 e) p4 (16x 4) 3 2. Simplify: a) s 1 2 ÷ s 1 3 b) 64m6 2 3 c) 12m 7 9 8m− 11 9 d) (5x) 0 + 5x 0 − (0,25) −0,5 + 8 2 3 3. Use the laws to re-write the following expression as a power of x: x r x q x p x √ x Think you got it? Get this answer and more practice on our Intelligent Practice Service 1a. 223P 1b. 223Q 1c. 223R 1d. 223S 1e. 223T 2a. 223V 2b. 223W 2c. 223X 2d. 223Y 3. 223Chapter 1. Exponents and surds 11 Simplification of surds EMBF6 We have seen in previous examples and exercises that rational exponents are closely related to surds. It is often useful to write a surd in exponential notation as it allows us to use the exponential laws. The additional laws listed below make simplifying surds easier: • √n a √n b = √n ab • n ra b = √n a √n b • mp√n a = mn√ a • √n am = a m n • ( √n a) m = a m n See video: 223N at  Worked example 5: Simplifying surds QUESTION Show that: 1. √n a × √n b = √n ab 2. n ra b = √n a √n b SOLUTION 1. √n a × √n b = a 1 n × b 1 n = (ab) 1 n = √n ab 2. n r a b = a b 1 n = a 1 n b 1 n = √n a √n b 12 1.2. Rational exponents and surds Examples: 1. √ 2 × √ 32 = √ 2 × 32 = √ 64 = 8 2. √3 24 √3 3 = 3 r 24 3 = √3 8 = 2 3. p√ 81 = √4 81 = √4 3 4 = 3 Like and unlike surds EMBF7 Two surds m√ a and √n b are like surds if m = n, otherwise they are called unlike surds. For example, q 1 3 and − √ 61 are like surds because m = n = 2. Examples of unlike surds are √3 5 and p5 7y 3 since m 6= n. Simplest surd form EMBF8 We can sometimes simplify surds by writing the radicand as a product of factors that can be further simplified using √n ab = √n a × √n b. See video: 2242 at Worked example 6: Simplest surd form QUESTION Write the following in simplest surd form: √ 50 SOLUTION Step 1: Write the radicand as a product of prime factors √ 50 = √ 5 × 5 × 2 = p 5 2 × 2 Step 2: Simplify using √n ab = √n a × √n b = √ 5 2 × √ 2 = 5 × √ 2 = 5√ 2 Chapter 1. Exponents and surds 13 Sometimes a surd cannot be simplified. For example, √ 6, √3 30 and √4 42 are already in their simplest form. Worked example 7: Simplest surd form QUESTION Write the following in simplest surd form: √3 54 SOLUTION Step 1: Write the radicand as a product of prime factors √3 54 = √3 3 × 3 × 3 × 2 = p3 3 3 × 2 Step 2: Simplify using √n ab = √n a × √n b = √3 3 3 × √3 2 = 3 × √3 2 = 3√3 2 Worked example 8: Simplest surd form QUESTION Simplify: √ 147 + √ 108 SOLUTION Step 1: Write the radicands as a product of prime factors √ 147 + √ 108 = √ 49 × 3 + √ 36 × 3 = p 7 2 × 3 + p 6 2 × 3 Step 2: Simplify using √n ab = √n a × √n b 14 1.2. Rational exponents and surds = √ 7 2 × √ 3 + √ 6 2 × √ 3 = 7 × √ 3 + 6 × √ 3 = 7√ 3 + 6√ 3 Step 3: Simplify and write the final answer 13√ 3 Worked example 9: Simplest surd form QUESTION Simplify: √ 20 − √ 5 2 SOLUTION Step 1: Factorise the radicands were possible √ 20 − √ 5 2 = √ 4 × 5 − √ 5 2 Step 2: Simplify using √n ab = √n a × √n b = √ 4 × √ 5 − √ 5 2 = 2 × √ 5 − √ 5 2 = 2 √ 5 − √ 5 2 Step 3: Simplify and write the final answer = √ 5 2 = 5 Chapter 1. Exponents and surds 15 Worked example 10: Simplest surd form with fractions QUESTION Write in simplest surd form: √ 75 × p3 (48)−1 SOLUTION Step 1: Factorise the radicands were possible √ 75 × p3 (48)−1 = √ 25 × 3 × 3 r 1 48 = √ 25 × 3 × 1 √3 8 × 6 Step 2: Simplify using √n ab = √n a × √n b = √ 25 × √ 3 × 1 √3 8 × √3 6 = 5 × √ 3 × 1 2 × √3 6 Step 3: Simplify and write the final answer = 5√ 3 × 1 2 √3 6 = 5 √ 3 2 √3 6 Exercise 1 – 4: Simplification of surds 1. Simplify the following and write answers with positive exponents: a) √3 16 × √3 4 b) √ a 2b 3 × √ b 5c 4 c) √ 12 √ 3 d) p x 2y 13 ÷ p y 5 16 1.2. Rational exponents and surds 2. Simplify the following: a) 1 a − 1 b −1 b) b − a a 1 2 − b 1 2 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1a. 2243 1b. 2244 1c. 2245 1d. 2246 2a. 2247 2b. 2248 www.everythingmaths.co.za Rationalising denominators EMBF9 It is often easier to work with fractions that have rational denominators instead of surd denominators. By rationalising the denominator, we convert a fraction with a surd in the denominator to a fraction that has a rational denominator. Worked example 11: Rationalising the denominator QUESTION Rationalise the denominator: 5x − 16 √ x SOLUTION Step 1: Multiply the fraction by √ x √ x Notice that √ x √ x = 1, so the value of the fraction has not been changed. 5x − 16 √ x × √ x √ x = √ x(5x − 16) √ x × √ x Step 2: Simplify the denominator = √ x(5x − 16) ( √ x) 2 = √ x(5x − 16) x The term in the denominator has changed from a surd to a rational number. Expressing the surd in the numerator is the preferred way of writing expressions. Chapter 1. Exponents and surds 17 Worked example 12: Rationalising the denominator QUESTION Write the following with a rational denominator: y − 25 √y + 5 SOLUTION Step 1: Multiply the fraction by √y−5 √y−5 To eliminate the surd from the denominator, we must multiply the fraction by an expression that will result in a difference of two squares in the denominator. y − 25 √y + 5 × √y − 5 √y − 5 Step 2: Simplify the denominator = (y − 25)(√y − 5) ( √y + 5)(√y − 5) = (y − 25)(√y − 5) ( √y) 2 − 25 = (y − 25)(√y − 5) y − 25 = √ y − 5 See video: 2249 at  Exercise 1 – 5: Rationalising the denominator Rationalise the denominator in each of the following: 1. 10 √ 5 2. 3 √ 6 3. 2 √ 3 ÷ √ 2 3 4. 3 √ 5 − 1 5. x √y 6. √ 3 + √ 7 √ 2 7. 3 √p − 4 √p 8. t − 4 √ t + 2 9. (1 + √ m) −1 10. a √ a ÷ √ b −1 18 1.2. Rational exponents and surds Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 224B 2. 224C 3. 224D 4. 224F 5. 224G 6. 224H 7. 224J 8. 224K 9. 224M 10. 224N 1.3 Solving surd equations EMBFB We also need to be able to solve equations that involve surds. See video: 224P at  Worked example 13: Surd equations QUESTION Solve for x: 5 √3 x 4 = 405 SOLUTION Step 1: Write in exponential notation 5 x 4 1 3 = 405 5x 4 3 = 405 Step 2: Divide both sides of the equation by 5 and simplify 5x 4 3 5 = 405 5 x 4 3 = 81 x 4 3 = 34 Step 3: Simplify the exponents x 4 3 3 4 = 3 4 3 4 x = 33 x = 27 Chapter 1. Exponents and surds 19 Step 4: Check the solution by substituting the answer back into the original equation LHS = 5√3 x 4 = 5(27) 4 3 = 5(33 ) 4 3 = 5(34 ) = 405 = RHS Worked example 14: Surd equations QUESTION Solve for z: z − 4 √ z + 3 = 0 SOLUTION Step 1: Factorise z − 4 √ z + 3 = 0 z − 4z 1 2 + 3 = 0 (z 1 2 − 3)(z 1 2 − 1) = 0 Step 2: Solve for both factors The zero law states: if a × b = 0, then a = 0 or b = 0. ∴ (z 1 2 − 3) = 0 or (z 1 2 − 1) = 0 Therefore z 1 2 − 3 = 0 z 1 2 = 3 z 1 2 2 = 32 z = 9 20 1.3. Solving surd equations or z 1 2 − 1 = 0 z 1 2 = 1 z 1 2 2 = 12 z = 1 Step 3: Check the solution by substituting both answers back into the original equation If z = 9: LHS = z − 4 √ z + 3 = 9 − 4 √ 9 + 3 = 12 − 12 = 0 = RHS If z = 1: LHS = z − 4 √ z + 3 = 1 − 4 √ 1 + 3 = 4 − 4 = 0 = RHS Step 4: Write the final answer The solution to z − 4 √ z + 3 = 0 is z = 9 or z = 1. Worked example 15: Surd equations QUESTION Solve for p: √ p − 2 − 3 = 0 SOLUTION Step 1: Write the equation with only the square root on the left hand side Use the additive inverse to get all other terms on the right hand side and only the Chapter 1. Exponents and surds 21 square root on the left hand side. p p − 2 = 3 Step 2: Square both sides of the equation p p − 2 2 = 32 p − 2 = 9 p = 11 Step 3: Check the solution by substituting the answer back into the original equation If p = 11: LHS = p p − 2 − 3 = √ 11 − 2 − 3 = √ 9 − 3 = 3 − 3 = 0 = RHS Step 4: Write the final answer The solution to √ p − 2 − 3 = 0 is p = 11. Exercise 1 – 6: Solving surd equations Solve for the unknown variable (remember to check that the solution is valid): 1. 2 x+1 − 32 = 0 2. 125 (3p ) = 27 (5p ) 3. 2y 1 2 − 3y 1 4 + 1 = 0 4. t − 1 = √ 7 − t 5. 2z − 7 √ z + 3 = 0 6. x 1 3 (x 1 3 + 1) = 6 7. 2 4n − 1 √4 16 = 0 8. √ 31 − 10d = 4 − d 9. y − 10√y + 9 = 0 10. f = 2 + √ 19 − 2f Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 224Q 2. 224R 3. 224S 4. 224T 5. 224V 6. 224W 7. 224X 8. 224Y 9. 224Z 10. 2252 www.everythingmaths.co.za m.everythingmaths.co.za 22 1.3. Solving surd equations 1.4 Applications of exponentials EMBFC There are many real world applications that require exponents. For example, exponentials are used to determine population growth and they are also used in finance to calculate different types of interest. Worked example 16: Applications of exponentials QUESTION A type of bacteria has a very high exponential growth rate at 80% every hour. If there are 10 bacteria, determine how many there will be in five hours, in one day and in one week? SOLUTION Step 1: Exponential formula final population = initial population × (1 + growth percentage) time period in hours Therefore, in this case: final population = 10 (1,8) n where n = number of hours. Step 2: In 5 hours final population = 10 (1,8) 5 ≈ 189 Step 3: In 1 day = 24 hours final population = 10 (1,8) 24 ≈ 13 382 588 Step 4: In 1 week = 168 hours final population = 10 (1,8) 168 ≈ 7,687 × 1043 Note this answer is given in scientific notation as it is a very big number. Chapter 1. Exponents and surds 23 Worked example 17: Applications of exponentials QUESTION A species of extremely rare deep water fish has a very long lifespan and rarely has offspring. If there are a total of 821 of this type of fish and their growth rate is 2% each month, how many will there be in half of a year? What will the population be in ten years and in one hundred years? SOLUTION Step 1: Exponential formula final population = initial population × (1 + growth percentage) time period in months Therefore, in this case: final population = 821(1,02) n where n = number of months. Step 2: In half a year = 6 months final population = 821(1,02) 6 ≈ 925 Step 3: In 10 years = 120 months final population = 821(1,02) 120 ≈ 8838 Step 4: In 100 years = 1200 months final population = 821(1,02) 1200 ≈ 1,716 × 1013 Note this answer is also given in scientific notation as it is a very big number. Exercise 1 – 7: Applications of exponentials 1. Nqobani invests R 5530 into an account which pays out a lump sum at the end of 6 years. If he gets R 9622,20 at the end of the period, what compound interest rate did the bank offer him? Give answer correct to one decimal place. 2. The current population of Johannesburg is 3 885 840 and the average rate of population growth in South Africa is 0,7% p.a. What can city planners expect the population of Johannesburg to be in 13 years time? 3. Abiona places 3 books in a stack on her desk. The next day she counts the books in the stack and then adds the same number of books to the top of the stack. After how many days will she have a stack of 192 books? 24 1.4. Applications of exponentials 4. A type of mould has a very high exponential growth rate of 40% every hour. If there are initially 45 individual mould cells in the population, determine how many there will be in 19 hours. Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 2253 2. 2254 3. 2255 4. 2256  1.5 Summary EMBFD See presentation: 2257 at  1. The number system: • N: natural numbers are {1; 2; 3; . . .} • N0: whole numbers are {0; 1; 2; 3; . . .} • Z: integers are {. . . ; −3; −2; −1; 0; 1; 2; 3; . . .} • Q: rational numbers are numbers which can be written as a b where a and b are integers and b 6= 0, or as a terminating or recurring decimal number. • Q0 : irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator as integers. Irrational numbers also include decimal numbers that neither terminate nor recur. • R: real numbers include all rational and irrational numbers. • R 0 : non-real numbers or imaginary numbers are numbers that are not real. 2. Definitions: • a n = a × a × a × · · · × a (n times) (a ∈ R, n ∈ N) • a 0 = 1 (a 6= 0 because 0 0 is undefined) • a −n = 1 an (a 6= 0 because 1 0 is undefined) 3. Laws of exponents: • a m × a n = a m+n • a m a n = a m−n • (ab) n = a n b n • a b n = a n b n • (a m) n = a mn where a > 0, b > 0 and m, n ∈ Z. Chapter 1. Exponents and surds 25 4. Rational exponents and surds: • If r n = a, then r = √n a (n ≥ 2) • a 1 n = √n a • a − 1 n = (a −1 ) 1 n = n r 1 a • a m n = (a m) 1 n = √n am where a > 0, r > 0 and m, n ∈ Z, n 6= 0. 5. Simplification of surds: • √n a √n b = √n ab • n ra b = √n a √n b • mp√n a = mn√ a Exercise 1 – 8: End of chapter exercises 1. Simplify as far as possible: a) 8 − 2 3 b) √ 16 + 8− 2 3 2. Simplify: a) x 3 4 3 b) s 2 1 2 c) m5 5 3 d) −m2 4 3 e) − m2 4 3 f) 3y 4 3 4 3. Simplify the following: a) 3a −2 b 15c −5 (a−4b 3c) −5 2 b) 9a 6 b 4 1 2 c) a 3 2 b 3 4 16 d) x 3√ x e) √3 x 4b 5 4. Re-write the following expression as a power of x: x r x q x p x √ x x 2 5. Expand: √ x − √ 2 √ x + √ 2 6. Rationalise the denominator: 10 √ x − 1 x 26 1.5. Summary 7. Write as a single term with a rational denominator: 3 2 √ x + √ x 8. Write in simplest surd form: a) √ 72 b) √ 45 + √ 80 c) √ 48 √ 12 d) √ 18 ÷ √ 72 √ 8 e) 4 √ 8 ÷ √ 2 f) 16 √ 20 ÷ √ 12 9. Expand and simplify: a) 2 + √ 2 2 b) 2 + √ 2 1 + √ 8 c) 1 + √ 3 1 + √ 8 + √ 3 10. Simplify, without use of a calculator: a) √ 5 √ 45 + 2√ 80 b) √ 98 − √ 8 √ 50 11. Simplify: √ 98x 6 + √ 128x 6 12. Rationalise the denominator: a) √ 5 + 2 √ 5 b) y − 4 √y − 2 c) 2x − 20 √ x − √ 10 13. Evaluate without using a calculator: 2 − √ 7 2 !1 2 × 2 + √ 7 2 !1 2 14. Prove (without the use of a calculator): r 8 3 + 5r 5 3 − r 1 6 = 10√ 15 + 3√ 6 6 15. Simplify completely by showing all your steps (do not use a calculator): 3 − 1 2 " √ 12 + 3 r 3 √ 3 # 16. Fill in the blank surd-form number on the right hand side of the equal sign which will make the following a true statement: −3 √ 6 × −2 √ 24 = − √ 18 × ... Chapter 1. Exponents and surds 27 17. Solve for the unknown variable: a) 3 x−1 − 27 = 0 b) 8 x − 1 √3 8 = 0 c) 27(4x ) = (64)3x d) √ 2x − 5 = 2 − x e) 2x 2 3 + 3x 1 3 − 2 = 0 18. a) Show that r 3 x+1 − 3 x 3 x−1 + 3 is equal to 3 b) Hence solve r 3 x+1 − 3 x 3 x−1 + 3 = 1 3 x−2 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1a. 2258 1b. 2259 2a. 225B 2b. 225C 2c. 225D 2d. 225F 2e. 225G 2f. 225H 3a. 225J 3b. 225K 3c. 225M 3d. 225N 3e. 225P 4. 225Q 5. 225R 6. 225S 7. 225T 8a. 225V 8b. 225W 8c. 225X 8d. 225Y 8e. 225Z 8f. 2262 9a. 2263 9b. 2264 9c. 2265 10a. 2266 10b. 2267 11. 2268 12a. 2269 12b. 226B 12c. 226C 13. 226D 14. 226F 15. 226G 16. 226H 17a. 226J 17b. 226K 17c. 226M 17d. 226N 17e. 226P 18. 226Q www.everythingmaths.co.za m.everythingmaths.co.za 28 1.5. Summary CHAPTER 2 Equations and inequalities 2.1 Revision 30 2.2 Completing the square 38 2.3 Quadratic formula 44 2.4 Substitution 48 2.5 Finding the equation 50 2.6 Nature of roots 52 2.7 Quadratic inequalities 60 2.8 Simultaneous equations 67 2.9 Word problems 74 2.10 Summary 80 2 Equations and inequalities 2.1 Revision EMBFF Solving quadratic equations using factorisation EMBFG Terminology: Expression An expression is a term or group of terms consisting of numbers, variables and the basic operators (+, −, ×, ÷, xn ). Equation A mathematical statement that asserts that two expressions are equal. Inequality An inequality states the relation between two expressions (>, <, ≥, ≤). Solution A value or set of values that satisfy the original problem statement. Root A root of an equation is the value of x such that f(x) = 0. A quadratic equation is an equation of the second degree; the exponent of one variable is 2. The following are examples of quadratic equations: 2x 2 − 5x = 12 a(a − 3) − 10 = 0 3b b + 2 + 1 = 4 b + 1 A quadratic equation has at most two solutions, also referred to as roots. There are some situations, however, in which a quadratic equation has either one solution or no solutions. 30 2.1. Revision x y y = x 2 − 4 0 −2 2 −4 x y y = x 2 0 x y y = x 2 + x + 1 0 y = (x − 2)(x + 2) = x 2 − 4 Graph of a quadratic equation with two roots: x = −2 and x = 2. y = x 2 Graph of a quadratic equation with one root: x = 0. y = x 2 + x + 1 Graph of a quadratic equation with no real roots. One method for solving quadratic equations is factorisation. The standard form of a quadratic equation is ax2 + bx + c = 0 and it is the starting point for solving any equation by factorisation. It is very important to note that one side of the equation must be equal to zero. Investigation: Zero product law Solve the following equations: 1. 6 × 0 = ? 2. −25 × 0 = ? 3. 0 × 0,69 = ? 4. 7 × ? = 0 Now solve for the variable in each of the following: 1. 6 × m = 0 2. 32 × x × 2 = 0 3. 11(z − 3) = 0 4. (k + 3)(k − 4) = 0 To obtain the two roots we use the fact that if a × b = 0, then a = 0 and/or b = 0. This is called the zero product law. Chapter 2. Equations and inequalities 31 Method for solving quadratic equations EMBFH 1. Rewrite the equation in the standard form ax2 + bx + c = 0. 2. Divide the entire equation by any common factor of the coefficients to obtain a simpler equation of the form ax2+bx+c = 0, where a, b and c have no common factors. 3. Factorise ax2 + bx + c = 0 to be of the form (rx + s) (ux + v) = 0. 4. The two solutions are (rx + s) = 0 (ux + v) = 0 So x = − s r So x = − v u 5. Always check the solution by substituting the answer back into the original equation. See video: 226R at www.everythingmaths.co.za Worked example 1: Solving quadratic equations using factorisation QUESTION Solve for x: x (x − 3) = 10 SOLUTION Step 1: Rewrite the equation in the form ax2 + bx + c = 0 Expand the brackets and subtract 10 from both sides of the equation x 2 − 3x − 10 = 0 Step 2: Factorise (x + 2) (x − 5) = 0 Step 3: Solve for both factors x + 2 = 0 x = −2 or x − 5 = 0 x = 5 32 2.1. Revision The graph shows the roots of the equation x = −2 or x = 5. This graph does not form part of the answer as the question did not ask for a sketch. It is shown here for illustration purposes only. 2 4 −2 −4 −6 −8 −10 −12 −4 −2 2 4 6 x f(x) y = x 2 − 3x − 10 0 Step 4: Check the solution by substituting both answers back into the original equation Step 5: Write the final answer Therefore x = −2 or x = 5. Worked example 2: Solving quadratic equations using factorisation QUESTION Solve the equation: 2x 2 − 5x − 12 = 0 SOLUTION Step 1: There are no common factors Step 2: The quadratic equation is already in the standard form ax2 + bx + c = 0 Step 3: Factorise We must determine the combination of factors of 2 and 12 that will give a middle term coefficient of 5. We find that 2× 1 and 3 × 4 give a middle term coefficient of 5 so we can factorise the equation as (2x + 3)(x − 4) = 0 Step 4: Solve for both roots Chapter 2. Equations and inequalities 33 We have 2x + 3 = 0 x = − 3 2 or x − 4 = 0 x = 4 Step 5: Check the solution by substituting both answers back into the original equation Step 6: Write the final answer Therefore, x = − 3 2 or x = 4. Worked example 3: Solving quadratic equations using factorisation QUESTION Solve for y: y 2 − 7 = 0 SOLUTION Step 1: Factorise as a difference of two squares We know that √ 7 2 = 7 We can write the equation as y 2 − ( √ 7)2 = 0 Step 2: Factorise (y − √ 7)(y + √ 7) = 0 Therefore y = √ 7 or y = − √ 7 34 2.1. Revision Even though the question did not ask for a sketch, it is often very useful to draw the graph. We can let f(y) = y 2 − 7 and draw a rough sketch of the graph to see where the two roots of the equation lie. 1 2 −1 −2 −3 −4 −5 −6 −7 −3 −2 −1 1 2 3 y f(y) f(y) = y 2 − 7 0 Step 3: Check the solution by substituting both answers back into the original equation Step 4: Write the final answer Therefore y = ± √ 7. Worked example 4: Solving quadratic equations using factorisation QUESTION Solve for b: 3b b + 2 + 1 = 4 b + 1 SOLUTION Step 1: Determine the restrictions The restrictions are the values for b that would result in the denominator being equal to 0, which would make the fraction undefined. Therefore b 6= − 2 and b 6= − 1. Step 2: Determine the lowest common denominator The lowest common denominator is (b + 2) (b + 1). Step 3: Multiply each term in the equation by the lowest common denominator and simplify Chapter 2. Equations and inequalities 35 3b(b + 2)(b + 1) b + 2 + (b + 2)(b + 1) = 4(b + 2)(b + 1) b + 1 3b(b + 1) + (b + 2)(b + 1) = 4(b + 2) 3b 2 + 3b + b 2 + 3b + 2 = 4b + 8 4b 2 + 2b − 6 = 0 2b 2 + b − 3 = 0 Step 4: Factorise and solve the equation (2b + 3)(b − 1) = 0 2b + 3 = 0 or b − 1 = 0 b = − 3 2 or b = 1 Step 5: Check the solution by substituting both answers back into the original equation Step 6: Write the final answer Therefore b = −1 1 2 or b = 1. Worked example 5: Squaring both sides of the equation QUESTION Solve for m: m + 2 = √ 7 + 2m SOLUTION Step 1: Square both sides of the equation Before we square both sides of the equation, we must make sure that the radical is the only term on one side of the equation and all other terms are on the other, otherwise squaring both sides will make the equation more complicated to solve. (m + 2)2 = √ 7 + 2m 2 Step 2: Expand the brackets and simplify 36 2.1. Revision (m + 2)2 = √ 7 + 2m 2 m2 + 4m + 4 = 7 + 2m m2 + 2m − 3 = 0 Step 3: Factorise and solve for m (m − 1)(m + 3) = 0 Therefore m = 1 or m = −3 Step 4: Check the solution by substituting both answers back into the original equation To find the solution we squared both sides of the equation. Squaring an expression changes negative values to positives and can therefore introduce invalid answers into the solution. Therefore it is very important to check that the answers obtained are valid. To test the answers, always substitute back into the original equation. If m = 1: RHS = p 7 + 2(1) = √ 9 = 3 LHS = 1 + 2 = 3 LHS = RHS Therefore m = 1 is valid. If m = −3: RHS = p 7 + 2(−3) = √ 1 = 1 LHS = −3 + 2 = −1 LHS 6= RHS Therefore m = −3 is not valid. Step 5: Write the final answer Therefore m = 1. See video: 226S at www.everythingmaths.co.za Chapter 2. Equations and inequalities 37 Exercise 2 – 1: Solution by factorisation Solve the following quadratic equations by factorisation. Answers may be left in surd form, where applicable. 1. 7t 2 + 14t = 0 2. 12y 2 + 24y + 12 = 0 3. 16s 2 = 400 4. y 2 − 5y + 6 = 0 5. y 2 + 5y − 36 = 0 6. 4 + p = √ p + 6 7. −y 2 − 11y − 24 = 0 8. 13y − 42 = y 2 9. (x − 1)(x + 10) = −24 10. y 2 − 5ky + 4k 2 = 0 11. 2y 2 − 61 = 101 12. 2y 2 − 10 = 0 13. −8 + h 2 = 28 14. y 2 − 4 = 10 15. √ 5 − 2p − 4 = 1 2 p 16. y 2 + 28 = 100 17. f (2f + 1) = 15 18. 2x = √ 21x − 5 19. 5y y − 2 + 3 y + 2 = −6 y 2 − 2y 20. x + 9 x 2 − 9 + 1 x + 3 = 2 x − 3 21. y − 2 y + 1 = 2y + 1 y − 7 22. 1 + t − 2 t − 1 = 5 t 2 − 4t + 3 + 10 3 − t 23. 4 m + 3 + 4 4 − m2 = 5m − 5 m2 + m − 6 24. 5 √ 5t + 1 − 4 = 5t + 1 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 226T 2. 226V 3. 226W 4. 226X 5. 226Y 6. 226Z 7. 2272 8. 2273 9. 2274 10. 2275 11. 2276 12. 2277 13. 2278 14. 2279 15. 227B 16. 227C 17. 227D 18. 227F 19. 227G 20. 227H 21. 227J 22. 227K 23. 227M 24. 227N www.everythingmaths.co.za m.everythingmaths.co.za 2.2 Completing the square EMBFJ Investigation: Completing the square Can you solve each equation using two different methods? 1. x 2 − 4 = 0 2. x 2 − 8 = 0 3. x 2 − 4x + 4 = 0 38 2.2. Completing the square 4. x 2 − 4x − 4 = 0 Factorising the last equation is quite difficult. Use the previous examples as a hint and try to create a difference of two squares. See video: 227P at www.everythingmaths.co.za We have seen that expressions of the form x 2 − b 2 are known as differences of squares and can be factorised as (x − b)(x + b). This simple factorisation leads to another technique for solving quadratic equations known as completing the square. Consider the equation x 2 − 2x − 1 = 0. We cannot easily factorise this expression. When we expand the perfect square (x − 1)2 and examine the terms we see that (x − 1)2 = x 2 − 2x + 1. We compare the two equations and notice that only the constant terms are different. We can create a perfect square by adding and subtracting the same amount to the original equation. x 2 − 2x − 1 = 0 (x 2 − 2x + 1) − 1 − 1 = 0 (x 2 − 2x + 1) − 2 = 0 (x − 1)2 − 2 = 0 Method 1: Take square roots on both sides of the equation to solve for x. (x − 1)2 − 2 = 0 (x − 1)2 = 2 p (x − 1)2 = ± √ 2 x − 1 = ± √ 2 x = 1 ± √ 2 Therefore x = 1 + √ 2 or x = 1 − √ 2 Very important: Always remember to include both a positive and a negative answer when taking the square root, since 2 2 = 4 and (−2)2 = 4. Method 2: Factorise the expression as a difference of two squares using 2 = √ 2 2 . We can write (x − 1)2 − 2 = 0 (x − 1)2 − √ 2 2 = 0 (x − 1) + √ 2 (x − 1) − √ 2 = 0 Chapter 2. Equations and inequalities 39 The solution is then (x − 1) + √ 2 = 0 x = 1 − √ 2 or (x − 1) − √ 2 = 0 x = 1 + √ 2 Method for solving quadratic equations by completing the square 1. Write the equation in the standard form ax2 + bx + c = 0. 2. Make the coefficient of the x 2 term equal to 1 by dividing the entire equation by a. 3. Take half the coefficient of the x term and square it; then add and subtract it from the equation so that the equation remains mathematically correct. In the example above, we added 1 to complete the square and then subtracted 1 so that the equation remained true. 4. Write the left hand side as a difference of two squares. 5. Factorise the equation in terms of a difference of squares and solve for x. See video: 227Q at www.everythingmaths.co.za Worked example 6: Solving quadratic equations by completing the square QUESTION Solve by completing the square: x 2 − 10x − 11 = 0 SOLUTION Step 1: The equation is already in the form ax2 + bx + c = 0 Step 2: Make sure the coefficient of the x 2 term is equal to 1 x 2 − 10x − 11 = 0 Step 3: Take half the coefficient of the x term and square it; then add and subtract it from the equation The coefficient of the x term is −10. Half of the coefficient of the x term is −5 and the square of it is 25. Therefore x 2 − 10x + 25 − 25 − 11 = 0. Step 4: Write the trinomial as a perfect square 40 2.2. Completing the square (x 2 − 10x + 25) − 25 − 11 = 0 (x − 5)2 − 36 = 0 Step 5: Method 1: Take square roots on both sides of the equation (x − 5)2 − 36 = 0 (x − 5)2 = 36 x − 5 = ± √ 36 Important: When taking a square root always remember that there is a positive and negative answer, since (6)2 = 36 and (−6)2 = 36. x − 5 = ±6 Step 6: Solve for x x = −1 or x = 11 Step 7: Method 2: Factorise equation as a difference of two squares (x − 5)2 − (6)2 = 0 [(x − 5) + 6] [(x − 5) − 6] = 0 Step 8: Simplify and solve for x (x + 1)(x − 11) = 0 ∴ x = −1 or x = 11 Step 9: Write the final answer x = −1 or x = 11 Notice that both methods produce the same answer. These roots are rational because 36 is a perfect square. Chapter 2. Equations and inequalities 41 Worked example 7: Solving quadratic equations by completing the square QUESTION Solve by completing the square: 2x 2 − 6x − 10 = 0 SOLUTION Step 1: The equation is already in standard form ax2 + bx + c = 0 Step 2: Make sure that the coefficient of the x 2 term is equal to 1 The coefficient of the x 2 term is 2. Therefore divide the entire equation by 2: x 2 − 3x − 5 = 0 Step 3: Take half the coefficient of the x term, square it; then add and subtract it from the equation The coefficient of the x term is −3, so then −3 2 2 = 9 4 : x 2 − 3x + 9 4 − 9 4 − 5 = 0 Step 4: Write the trinomial as a perfect square x − 3 2 2 − 9 4 − 20 4 = 0 x − 3 2 2 − 29 4 = 0 Step 5: Method 1: Take square roots on both sides of the equation x − 3 2 2 − 29 4 = 0 x − 3 2 2 = 29 4 x − 3 2 = ± r 29 4 Remember: When taking a square root there is a positive and a negative answer. Step 6: Solve for x 42 2.2. Completing the square x − 3 2 = ± r 29 4 x = 3 2 ± √ 29 2 = 3 ± √ 29 2 Step 7: Method 2: Factorise equation as a difference of two squares x − 3 2 2 − 29 4 = 0 x − 3 2 2 − r 29 4 !2 = 0 x − 3 2 − r 29 4 ! x − 3 2 + r 29 4 ! = 0 Step 8: Solve for x x − 3 2 − √ 29 2 ! x − 3 2 + √ 29 2 ! = 0 Therefore x = 3 2 + √ 29 2 or x = 3 2 − √ 29 2 Notice that these roots are irrational since 29 is not a perfect square. See video: 227R at www.everythingmaths.co.za Exercise 2 – 2: Solution by completing the square 1. Solve the following equations by completing the square: a) x 2 + 10x − 2 = 0 b) x 2 + 4x + 3 = 0 c) p 2 − 5 = −8p d) 2(6x + x 2 ) = −4 e) x 2 + 5x + 9 = 0 f) t 2 + 30 = 2(10 − 8t) g) 3x 2 + 6x − 2 = 0 h) z 2 + 8z − 6 = 0 i) 2z 2 = 11z j) 5 + 4z − z 2 = 0 Chapter 2. Equations and inequalities 43 2. Solve for k in terms of a: k 2 + 6k + a = 0 3. Solve for y in terms of p, q and r: py2 + qy + r = 0 Think you got it? Get this answer and more practice on our Intelligent Practice Service 1a. 227S 1b. 227T 1c. 227V 1d. 227W 1e. 227X 1f. 227Y 1g. 227Z 1h. 2282 1i. 2283 1j. 2284 2. 2285 3. 2286 www.everythingmaths.co.za m.everythingmaths.co.za 2.3 Quadratic formula EMBFK It is not always possible to solve a quadratic equation by factorisation and it can take a long time to complete the square. The method of completing the square provides a way to derive a formula that can be used to solve any quadratic equation. The quadratic formula provides an easy and fast way to solve quadratic equations. Consider the standard form of the quadratic equation ax2 + bx + c = 0. Divide both sides by a (a 6= 0) to get x 2 + bx a + c a = 0 Now using the method of completing the square, we must halve the coefficient of x and square it. We then add and subtract b 2a 2 so that the equation remains true. x 2 + bx a + b 2 4a 2 − b 2 4a 2 + c a = 0 x 2 + bx a + b 2 4a 2 − b 2 4a 2 + c a = 0 x + b 2a 2 − b 2 − 4ac 4a 2 = 0 We add the constant to both sides and take the square root of both sides of the equation, being careful to include a positive and negative answer. 44 2.3. Quadratic formula x + b 2a 2 = b 2 − 4ac 4a 2 s x + b 2a 2 = ± r b 2 − 4ac 4a 2 x + b 2a = ± √ b 2 − 4ac 2a x = − b 2a ± √ b 2 − 4ac 2a x = −b ± √ b 2 − 4ac 2a Therefore, for any quadratic equation ax2 + bx + c = 0 we can determine two roots x = −b + √ b 2 − 4ac 2a or x = −b − √ b 2 − 4ac 2a It is important to notice that the expression b 2 − 4ac must be greater than or equal to zero for the roots of the quadratic to be real. If the expression under the square root sign is less than zero, then the roots are non-real (imaginary). See video: 2287 at www.everythingmaths.co.za Worked example 8: Using the quadratic formula QUESTION Solve for x and leave your answer in simplest surd form: 2x 2 + 3x = 7 SOLUTION Step 1: Check whether the expression can be factorised The expression cannot be factorised, so the general quadratic formula must be used. Step 2: Write the equation in the standard form ax2 + bx + c = 0 2x 2 + 3x − 7 = 0 Step 3: Identify the coefficients to substitute into the formula a = 2; b = 3; c = −7 Chapter 2. Equations and inequalities 45 Step 4: Apply the quadratic formula Always write down the formula first and then substitute the values of a, b and c. x = −b ± √ b 2 − 4ac 2a = − (3) ± q (3)2 − 4 (2) (−7) 2 (2) = −3 ± √ 65 4 Step 5: Write the final answer The two roots are x = −3 + √ 65 4 or x = −3 − √ 65 4 . Worked example 9: Using the quadratic formula QUESTION Find the roots of the function f(x) = x 2 − 5x + 8. SOLUTION Step 1: Finding the roots To determine the roots of f(x), we let x 2 − 5x + 8 = 0. Step 2: Check whether the expression can be factorised The expression cannot be factorised, so the general quadratic formula must be used. Step 3: Identify the coefficients to substitute into the formula a = 1; b = −5; c = 8 Step 4: Apply the quadratic formula x = −b ± √ b 2 − 4ac 2a = − (−5) ± q (−5)2 − 4 (1) (8) 2 (1) = 5 ± √ −7 2 46 2.3. Quadratic formula