MATH 2 part 1

2. What is − 25 ?Answer: -25 does not have a real square root because there is no real number y such y2 = -25.3. Find the two square roots of 100.Answer: The square roots of 100 are 10 and –10 since, 102 = 100 and (-10)2 = 100.4. The square root of 392 = 39, you just cancel the exponent of 39 and the index.Cube Roots: In a similar way, If y is cubed and we get the value x, then y is the cube root of x.In symbol, y = 3 x (read as y is the cube root of x ) if y3 = x.Examples:1. 3 64 4 is the cube root of 64 since (4) (4) (4) = 64.2. 3 − 64 –4 is the cube root of -64 since (-4) (-4) (-4) = -64.3. 3 27 3 is the cube root of 27 since  3   3   3  = 27 125 5 125  5   5   5  1254. 3 35 35 is not a perfect cube because there is no number multiplied by itself 3 times will give you 35.Try this outA. Find the square roots of the following.1. - 25 6. 1.692. 225 7. 0.00163. 112 8. 644. − 49 121 9. 25 144 6

5. 0.0225 10. 36 49B. Find the cube root of the following.1. 3 83 6. 3 1532. 3 8 7. 3 0.0643. 3 − 27 8. 3 0.1254. 3 125 9. 3 − 512 10. 3 645. 3 343 343C. Find the indicated root of the following.1. 2500 6. 6 7292. 3 125000 7. 7 1283. 4 625 8. 5 31254. 5 − 1024 9. 60845. 4 2401 10. 7 0.0000128 Lesson 3 Naming Two Rational Numbers Between Which a Radical Expression Lies Not all radicals have radicands which are perfect squares or perfect cubes. Forexample, 2, 3, 5, 3 10, etc. These are called irrational radicals.In the case of these irrational radicals, the square roots are non-integers. On the number line, the irrationals fill up the spaces between the rationalnumbers to compose the complete set of real numbers.Example 1: 23 6 •• • 7

01 2 3 4Example 2: Some of the irrational radicals with their approximations are:1. 2 = 1.414… 2. 3 = 1.732… 3. 6 = 2.449… Square roots of these numbers can be estimated by looking for the valuesbetween which the square root lies.Examples:1. Find two consecutive integers between which 40 lies.Solution: Since 36 = 6 and 49 = 7 Then 36 < 40 < 49 Therefore 6< 40 < 7 Thus, 40 lies between 6 and 7.2. Find two consecutive integers between which 112 lies. Solution: Since 100 = 10 and 121 = 11 Then 100 < 112 < 121 Therefore 10 < 112 < 11 Thus, 112 lies between 10 and 11.3. Find two consecutive integers between which 3 12 lies.Solution: Since 3 8 = 2 and 3 27 = 3 Then 3 8 < 3 12 < 3 27 Therefore 2 < 3 12 < 3 Thus, 3 12 lies between 2 and 3.We can find approximate square roots of irrational numbers through a calculator. 8

( )Most calculators with the “square root key” easily give the square root of anumber. To find the 75 , use the keystroke sequence 7 5 the display is8.660254. You can also find approximate cube roots of irrational numbers by using a 1scientific calculator. Here’s how. Press Shift key and x y key.Examples:1. To find 3 10 , press 1 0 Shift 1 = xy 3 2.1544347 will appear on display. 2. To find 3 − 10 , press 1 0 ± Shift 1 = -2.1544347 will appear on display. xy 3Try this outA. Name two rational numbers between which each of the following radicals lie.1. 8 6. 602. 11 7. 353. 17 8. 284. 23 9. 915. 12 10. 56B. Name two rational numbers between which each of the following radicals lie.1. 15 6. 4 6002. 200 7. 5 − 2403. 3 54 8. 5 10504. 3 150 9. 6 900 9

5. 4 1500 10. 6 10000C. Use your calculator to find an approximate value to two decimal places.1. 18000 6. 5 − 2162. 5000 7. 4 25003. - 250 8. 4 50004. - 360 9. 6 150005. 3 − 8 10. 6 50,000 27D. WHO IS THIS MATHEMATICIAN This Polish mathematician was the first to use the symbol for square root.Born in 1489, he studied algebra at the University of Vienna between 1517 and 1521.He wrote Coss, The first German Algebra book in 1525.To find out: 1. Find the answer to each number. 2. Write the letter under its matching number in the DECODER.100 − 64 1 225 + 16 − 25 49 4 121 I U O C36 + 64 9 3 64 144 16 H T F D 4 169 81 400 L S R 10

_____ _____ _____ _____ _____ _____ _____ _____ _____7 10 20 2 2 4 1 12 1211 9 2_____ _____ _____ _____ _____ _____ _____20 14 3 1 13 12 12 4 2 Let’s summarize1. Perfect Squares: A number is a perfect square if it is a product of a single number multiplied twotimes by itself.2. Perfect Cubes: A number is a perfect cube if it is a product of a single number multiplied threetimes by itself.3. A radical expression is an expression using the radical sign, .4. Square Roots: Consider the expression x : a. x is a rational number, if x is a perfect square. b. x is an irrational number, if x is not a perfect square. The radical expression − x do not have real square roots.5. Cube Roots: If y is cubed and we get the value x, then y is the cube root of x. In symbol, y =3 x (read as y is the cube root of x ) if y3 = x.6. On the number line, the irrationals fill up the spaces between the rational numbers tocompose the complete set of real numbers. 11

What have you learnedA. Find the indicated square roots of the following.1. 69 6. 0.00042. 441 7. 441 1443. 1296 8. 576 5294. 784 9. 729 23045. 7.84 10. 1156 1600B. Find the cube roots of the following.1. 3 27 6. 3 64 272. 3 − 125 7. 3 125 83. 3 343 8. 3 − 1 644. 3 0.001 9. 3 343 7295. 3 0.064 10. 3 216 1000C. Name two rational numbers between which it lies.1. 5 6. 2642. 20 7. 3853. 42 8. 1794. 84 9. 2135. 145 10. 110 12

Answer KeyHow much do you KnowA. 1. 13 2. 20 3. 35 4. 1.5 5. 0.02 6. 0.009 7. 0.14 8. 1.3 9. 8 11 10. 21 12B. 1. 3 2. –5 3. 7 4. 0.1 5. 0.4 6. 4 3 7. 5 2 8. − 1 4 9. 7 9 10. 3 5 13

C. 1. 2 and 3 2. 3 and 4 3. 4 and 5 4. 5 and 6 5. 6 and 7 6. 4 and 5 7. 5 and 6 8. 6 and 7 9. 7 and 8 10. 9 and 10Try this outLesson 1A. 1. Perfect square 2. Not perfect square 3. Perfect square 4. Perfect square 5. Perfect square 6. Not perfect square 7. Not perfect square 8. Perfect square 9. Not perfect square 10. Not perfect squareB. 1. not perfect cube 2. Perfect cube 3. Perfect cube 4. Perfect cube 5. Perfect cube 6. Perfect cube 14

7. Not perfect cube 8. Perfect cube 9. Not perfect cube 10. Perfect cubeC. 1. Perfect square 2. Perfect square 3. Perfect square 4. Neither 5. Perfect square 6. Neither 7. Perfect cube 8. Perfect cube 9. Neither 10. NeitherLesson 2A. 1. –5 2. 15 3. 11 4. no real root 5. 0.15 6. 1.3 7. 0.04 8. 8 11 9. 5 12 10. 6 7B. 1. 8 2. 2 15

3. –3 4. 5 5. 7 6. 15 7. 0.4 8. 0.5 9. –8 10. 4 7C. 1. 50 2. 50 3. 5 4. –4 5. 7 6. 3 7. 2 8. 5 9. 78 10. 0.2Lesson 3A. 1. 2 < 8 < 3 2. 3 < 11 < 4 3. 4 < 17 < 5 4. 4 < 23 < 5 5. 3 < 12 < 4 6. 7 < 60 < 8 7. 5 < 35 < 6 16

8. 5 < 28 < 6 9. 9 < 91 < 10 10. 7 < 56 < 8B. 1. 3 < 15 < 4 2. 14 < 200 < 15 3. 3 < 3 54 < 4 4. 5 < 3 150 < 6 5. 6 < 4 1500 < 7 6. 4 < 4 600 < 5 7. - 2 < 5 − 240 < −3 8. 4 < 5 1050 < 5 9. 3 < 6 900 < 4 10. 4 < 6 10000 < 5C. 1. 134.16 2. 70.71 3. –15.81 4. –18.97 5. 0.67 6. –2.93 7. 7.07 8. 8.41 9. 4.97 10. 6.07D. 1. C 17

2. H 3. R 4. I 5. S 6. T 7. O 8. F 9. F 10. R 11. U 12. D 13. O 14. L 15. F 16. FHe is Christoff RudolffHow much have you learned?A. 1. 13 2. 21 3. 36 4. 28 5. 2.8 6. 0.02 7. 21 12 8. 24 23 9. 27 48 18

10. 34 40B. 1. 3 2. –5 3. 7 4. 0.1 5. 0.4 6. 4 3 7. 5 2 8. − 1 4 9. 7 9 10. 6 or 3 10 5C. 1. 2 and 3 2. 4 and 5 3. 6 and 7 4. 9 and 10 5. 12 and 13 6. 16 and 17 7. 19 and 20 8. 13 and 14 9. 14 and 15 10. 10 and 11 19

Module 1 Quadratic Equations What this module is about This module is about quadratic equations. In this module, you will be ableto develop skills in solving second-degree equations in one variable. You will findthat quadratic equations can be solved using several methods. Two of thesemethods will be discussed in the lessons. What you are expected to learn This module is designed for you to: 1. Distinguish a quadratic equation from a linear equation. 2. Write quadratic equations in standard form. 3. Solve quadratic equations by factoring. 4. Solve quadratic equations by extracting square roots. How much do you know A. Which of the following are quadratic equations? 1. 37- 46x = 2x2 2. 1 = x x2 3. 5x2 + 2x – 4 = 0 B. Write the following in the standard quadratic form ax2 + bx + c = 0 4. (x + 5)(x - 12) = 18 5. x2 – x + 14 = 2x(x - 3) 6. 4x2 – 8x + 2 = x2 - 21x + 12 C. Use the factoring method to solve each equation.

7. x2 + 3x – 10 = 0 8. x2 = 4x 9. x2 - 10x – 24 = 0 10. x2 - 6x + 5 = 0 D. Use the square root method to solve each equation. 11. x2 = 25 12. 2x2 – 10 = 0 13. (x + 3)2 = 10 14. 3(x - 4)2 = 33 15. (x + 3)2 = -9 What you will do Lesson 1 Distinguishing Quadratic Equations from Linear Equations In first year, you have learned how to solve first-degree equations in onevariable. These are equations which involve one variable. The highest exponentof the variable is 1. These first degree equations are called linear equations. Forexample, the equation 3x – 5 = 4 is a first- degree equation. A second- degree equation in one variable is an equation in which thehighest exponent of the variable is 2. These equations are called quadraticequations. The equation 3x2 -x – 5 = 0 is a second-degree equation.Example:Classify each equation as linear or quadratic: 1. x2 + 3x + 5 = 0 2. 5x – 3 = 0 3. 3x² + 12x = 0 4. 1 x + 1 = 0 2 5. x3 + 4x – 3x² = 0 Examples 2 and 4 are linear equations since the highest exponent of thevariable x is 1, while Examples 1 and 3 are quadratic equations because the 2

highest exponent of the variable x is 2. Example 5 is neither linear nor quadratic.Why?Try this outA. Identify whether the following equations are linear, quadratic or neither: 1. x2 – 4x + 7 = 0 2. 2x – 5 = 0 3. – 4 x2 – 4x + 5 = 0 4. 3 x2 - 12 = 0 5. 1 x + 1 = 0 2 6. x² - 5x = 3 7. x² - 4x – x² = 0 8. 2x – 5 = 4 9. 3x – 3x = 5 10. x2 + x = 0 24 11. 2x – 3x3 + 2x² = 5 12. 6x² = 3x +4 13. a – 5 = a² 14. 4b –25 = 3(b-5) 15. 34 = x² - 4x 16. 5t –20 –5t = 3 17. 4p² - 25 = 4p² 18. 5 x² - 4 = 3x 5 19. q² + 6q – 2( q² + 5) = (q – 2)² 20. (2x + 5)(x –3) = 0 Lesson 2 Writing Quadratic Equations in Standard Form By this time, you are already familiar with quadratic equations. Toreiterate, quadratic equations are of the form ax² + bx +c = 0, where a ≠ 0. Aquadratic equation written in this way is said to be in standard form. The equation x2 – 7x + 4 = 0 is a quadratic equation in standard formwhere, a =1, b = -7 and c = 4. 3

Examples:The following quadratic equations are written in standard form:1. x2 + 5x – 3 = 02. 3x2 + 4x +5 = 03. 4x2 – 2x = 04. x2 + 5 = 05. –3x2 = 0 Can you identify the values of a, b and c in the examples above? Noticethat in all the quadratic equations listed, there is always a value for a. Why?Study the equation below: -2x2 – 4(x – 5) = 7 – 9Did you notice that it is a quadratic equation? Why?Did you also notice that it is not written in standard form? With the use of the properties of equality, you will be able to writequadratic equations, such as the one above in standard form.Solution: Distributive property -2x2 – 4( x – 5 ) = 7- 9 Addition property of equality -2x2 – 4x +20 = 7- 9 Addition property of equality -2x2 – 4x + 20 = -2 Multiply by -1 -2x2 – 4x + 20 + 2 = 0 -1(-2x2 – 4x + 22 = 0)2x2 + 4x - 22 = 0 is now the standard form where a = 2, b = 4 and c = -22.6. Write the equation 2x² - 4x + 5 = (x −1)(x + 2) in standard form: 32 Solution: 2x² - 4x + 5 = (x −1)(x + 2) Multiply by the LCD 32 Division of numbers Distributive property6(2x² - 4x + 5 ) = (6) (x −1)(x + 2) Addition property 3212x² - 24x + 10 = 3(x² + x – 2)12x² - 24x + 10 = 3x² + 3x – 6 12x² - 24x + 10 –3x² -3x + 6 = 0 9x² -21x + 16 = 09x² -21x + 16 = 0 is now the standard form where a = 9, b = -21 and c = 22 4

Try this out Write the following in standard form. Identify the values of a, b and c.A. 1. 4x2 -11x = 7 2. 2x2 -20 = 3x 3. 6x2 = 5x - 4 4. 3x – 9 = x2 5. 5x2 = 3x 6. 4x2 = 20 7. 7x2 = 9 8. x2 = 4x –5 9. 2x = 3x2 10. 4 = - 5x2B. 1. (x + 3) (x - 2) =1 2. 2x(x - 7) = 5 3. x2 = 2(5x - 6) 4. (x - 7)2 + 3 = 0 5. (x - 4)(x + 2) = 3(x - 1) 6. 2z(z - 1) = 5z2 - 7 7. (y + 1) 2 = 2(y - 4) 8. 2p(5p – 3) = (3p + 1)2 9. 2x2 + 5x – 4 = x2 + 3x - 7 10. (y + 2)(y + 5) = (2y - 1)(y + 6)C. 1. x2 = 6x – 1 33 2. 5x2 −1 = 1 22 3. (x + 3)2 = 2x −1 43 4. x² + 1 = 5x 2 5. x² + 2x = 4 3 5

Lesson 3Solving Quadratic Equations by Factoring This time, you will learn how to solve quadratic equations. Solvingquadratic equations means solving for the value of x that will make the equationtrue. For instance, the equation x2 + x = 30 has two solutions since both x = 5and x = -6 satisfy the equation.When x = 5 x2 + x = 30 52 + 5 25 + 5 = 30 = 30 30 = 30When x = -6 x2 + x (-60)2 + (-6) = 30 36 – 6 30 = 30 = 30 = 30 Since both 5 and –6 make the equation true, then it is said that 5 and –6are the solutions or roots of the equation x2 + x = 30.It should be noted that every quadratic equation has two roots. There are many ways of solving quadratic equations. The first way ormethod you will learn is based on the following basic fact about real numbers. Zero – Product Property Thus in solving the equation ( x –4 ) ( x +3 ) = 0 If ab = 0 then either a = 0 or b = 0 or both a and b are equal to zero.(If a product is 0, then either one of the factors is 0 or both factors are 0.)Example 1 In (x – 4)(x + 3) = 0, it is clear that since the product of the two quantities(x –4) and (x +3) is equal to zero, then either the quantity (x - 4) = 0 or thequantity (x +3) = 0. 6

Thus, x - 4 = 0 or x +3 = 0.So that x = 4 or x = -3 Notice that both 4 and –3 are solutions of the original equation (x –4)(x +3)= 0. By checking, For x = 4: (x – 4)(x + 3) = 0 (4 – 4)(4 + 3) = 0 0(7) = 0 0 =0 For x = -3: (x – 4)(x + 3) = 0 (-3 – 4)(-3 + 3) = 0 (-7)(0) = 0 0=0Example 2Solve: x (2x – 9) = 0Solution: x (2x – 9) = 0 x = 0 or 2x – 9 = 0 zero product property 2x = 9 x= 9 2 The solutions are 0 and 9 . 2The checking is left for you.Try this outI. A. Solve the following:1. (x – 5)(x + 1) = 02. (x + 4)(x + 3) = 03. (x - 7)x = 04. (x + 10)(x – 3) = 05. (2x – 7)(x - 4) = 06. (3x – 4)(2x + 9) = 07. (6x + 5)(7x + 1) = 0 7

8. (x – 2 )(x + 3 ) = 09. ( x – 5)(x + 1) = 0 210. ( x – 1 )x = 0 3211. (5x + 1)(x – 7) = 012. (4t + 1)(3t –2) = 013. (x – 3)(x + 6) = 014. y(3y – 17) = 0B. What is wrong with the following solutions?a. (x – 3)(x + 4) = 0x = -3 or x = 4b. (x – 3)(x + 4) = 8 x+4=4 x –3 = 2 or x =0 x=5 In the equation x2 + x = 30, you cannot apply the zero product propertyright away because it is not in the product form, like the ones you just did. Notice,too, that it does not have a zero on one side. Well, this can be overcome bywriting the equation in standard form.So that, x2 + x = 30 becomes x2 + x –30 = 0. Still, it is not in the product form. By factoring the left side of the equation,you may eventually apply the zero - product property to finally solve the equation.Thus, x2 + x = 30 Addition property of equality x2 + x –30 = 0 Factoring (x + 6)(x - 5) = 0 x+6 = 0 or x – 5 = 0 Zero-product property x = -6 x=5 The method in finding the solution that has just been illustrated is calledFactoring. By experience, it should be noted that not all quadratic trinomials arefactorable. Thus, the factoring method does not always work. Other methods ofsolving quadratic equations will be discussed later. In the meantime, study thefollowing examples solved by factoring. 8

Example 3Solve x² + 5x + 6 = 0Solution: Since the equation is already in standard form, that is, the right-hand side of the equation is already 0, then you may do factoring right away. x² + 5x + 6 = 0 Factoring (x + 2)(x + 3) = 0 x +2 = 0 or x + 3 = 0 Zero product property x = -2 x = -3 The solutions are –2 and –3.Check: If x = -2, then x² + 5x + 6 = 0 (-2)² + 5(-2) + 6 = 0 4 – 10 + 6 = 0 -6 – 6 = 0 0=0 if x = -3, then x² + 5x + 6 = 0 (-3)² + 5(-3) + 6 = 0 9 – 15 + 6 = 0 -6 – 6 = 0 0=0Example 4Solve for x: x2 = x + 6Solution: To solve by factoring method, rewrite first the equation in standard form. x2 = x + 6 then factor; x2 - x – 6 = 0 apply the zero product property (x-3) (x+2) = 0 x–3=0 or x+2=0 x=3 x = -2 The solutions are 3 and -2.Check: If x = 3, then x2 = x + 6 (3)2 = 3 + 6 9

9=9 If x = -2, then x2 = x + 6 (-2)2 = -2 + 6 4=4 From the above examples, can you identify the steps used in solvingquadratic equations?Example 5Solve x² - 8x = -16.Solution: Again, the equation is not in standard form. So, rewrite the equation in standard form. x² - 8x = -16 Addition Property of Equality x² - 8x + 16 = 0 Factoring (x – 4)(x – 4) = 0 x – 4 = 0 or x – 4 = 0 Zero product property x=4 x=4 Since the two solutions are equal, it is not right to say that the solutionsare 4 and 4. It is alright to say that there is only one solution, 4. Observe that the left- hand side of the equation in Example 5 is a perfectsquare trinomial.Remark Keep in mind that you must have 0 on one side before you can use the zero-product property. Get all nonzero terms on one side and 0 on the other.Example 6Solve for a: (2a + 1)(a -1) = (3a -2)(2a - 4)Solution: (2a + 1)(a -1) = (3a -2)(2a - 4) 2a2 – a – 1 = 6a2 – 16a + 8 Writing in standard form 0 = 4a2 – 15a + 9 0 = (4a - 3)(a-3) Factoring 4a – 3 = 0 or a -3 = 0 Zero product property a= 3 a=3 4 10

The solutions are 3 and 3. 4Check: Substitute 3 for a in the original equation (2a + 1)(a - 1) = (3a - 2)(2a - 4) (2(3)+1) (3 - 1) = (3(3) - 2) (2(3) - 4) (6 + 1) (2) = (9 - 2) (6 - 4) 7(2) = 7(2)What about when a = 3 , will it also satisfy the original equation? 4Checking for this value is now left for you.Example 7Solve for x. (x - 6)(x + 1) = 8Solution: Be careful with an equation like this one! It might be tempting to set each factor equal to 8. You just cannot apply the zero- product property since one side of the equation is not zero. This means that the equation has to be written into the standard form just like the previous examples. (x - 6)(x + 1) = 8 By multiplication x2 - 5x - 6 = 8 Writing in standard form Factoring x2 - 5x - 14 = 0 (x - 7)(x + 2) = 0 x - 7 = 0 or x + 2 = 0 Zero product property x=7 x = -2 The solutions are 7 and –2.Check: If x = 7: (x - 6)(x +1) = 8 (7- 6)(7+1) = 8 (1)(8) = 8 8=8 If x = -2: (x -6)(x + 1) = 8 (-2 - 6)(-2+1) = 8 (-8)(-1) = 8 8=8 11

Try this outII. Solve by factoringA. 1. x2 - x - 6 = 0 2. t2 – 2t - 8 = 0 3. t2 – 7t + 10 = 0 4. x2 - x - 12 = 0 5. x2 - 3x - 10 = 0 6. x2 - 8x + 15 = 0 7. 2x2 - 3x - 9 = 0 8. x2 - 7x - 30 = 0 9. x2 +14 +9x = 0 10. x2 +33 – 14x = 0B. 1. 2a2 = 11a – 12 2. -x2 = 8 – 9x 3. 4. 2x(x+3) - 36 = 0 5. 2x (x+3) = 20 3x2 - 2x - 6 = 2x2 - 6x - 3C. 1. (x + 2)2 = 25 2. 3. (x - 4)(x + 1) = (x-3)(x-2) 4. (x + 3)2 = 3x2 + 17 5. (x - 2)2 = 2x2 – 11x + 10 (y + 2)(y + 5) = 10 Notice that all of the quadratic equations you have studied and solved inthis lesson are factorable complete forms of quadratic equations. The following incomplete form of quadratic equations may also be solvedby the factoring method:Example 8Solve for x. x2 - 3x = 0Solution: Be careful with this one, too. A common mistake is to divide both sides of the equation by the variable x. This is not allowed because you are not sure about the value of the variable x. The right way to solve this is as follows: x2 - 3x = 0 x(x -3) = 0 By factoring, x is the common monomial factor x = 0 or x - 3 = 0 Zero-product property 12

x=3 The solutions are 0 and 3.Check: If x =0 x2 - 3x = 0 (0)2 – 3(0) = 0 0=0 If x = 3 x2 - 3x = 0 (3)2 – 3(3) = 0 9 -9 = 0 0=0Example 9Solve for y: 6y2 + 8y = 0Solution: 2y(3y + 4) = 0 By factoring, y is the common monomial factor 2y = 0 or 3y +4 = 0 Zero-product property y=0 3y = -4 y=-4 3 The solutions are 0 and - 4 . 3Check: If y = 0 6y2 + 8y = 0 6(0)2 + 8(0) = 0 0=0 If y = - 4 6y2 + 8y = 0 3 6(- 4 )2 + 8(- 4 ) = 0 33 6( 16 ) – 32 = 0 93 2( 16 ) - 32 = 0 33 32 – 32 = 0 33 0=0Observe that Examples 8 and 9 are of the form ax2 + bx = 0. 13

Example 10Solve for x : x2 -144 = 0Solution: x2 -144 = 0 (x -12)(x + 12) = 0 x -12 = 0 or x + 12 = 0 By factoring x = -12 Zero-product property x = 12 The solutions are 12 and –12.Check: If x = 12: x2 -144 = 0 (12)2 -144 = 0 144 – 144 = 0 0=0 If x = -12: x2 -144 = 0 (-12)2 -144 = 0 144 – 144 = 0 0=0 Observe that Example 10 is of the form ax2 + c = 0. The method used isfactoring the difference of two squares. Finally, note that the types of factoring used in solving quadratic equationsby the factoring method are quadratic trinomials, common monomial factor, thedifference of two squares and perfect square binomials.Try this outIII. Find the solution by factoring:A. 1. 2a(a + 3) = 0 2. 3z2 – 15z = 0 3. x2 + 3x = 11x 4. 2x2 – 18x = 0 5. x2 = 14x 6. x2 - 121 = 0 7. 25x2 - 4 = 0 8. 4x2 - 9 = 0 9. x2 - 4 = 60 10. 3p2 - 5 = 7B. 1. 2x2 -8 = 0 2. x2 + x = 0 3. 2p2 = 50 14

4. 3x2 = 27 5. 6x2 –3x = 0 6. 3x² -5 = 0 7. 25x² = 4 8. x² = 0 9. 5x² = 0 10. –3x² = 0C. 1. 6x2 – x -2 = 0 2. 3x2 + x -2 = 0 3. 6x2 + 37x + 6 = 0 4. 5x2 + 13x - 6 = 0 5. x2 – 5x + 6 = 0D. What is wrong with the following solution? x² - 7x +12 = 7(x – 3)(x – 4) = 7x – 3 = 7 or x–4=7 x = 10 x = 11 Lesson 4Solving Quadratic Equations by Extracting Square Roots As mentioned in Lesson 3, the factoring works only when the left–handside of a quadratic equation in standard form is factorable. For example, thequadratic equations x2 + 5x +1 = 0 and x2 - 20 = 0 cannot be solved by factoring.There must be other methods to solve such equations. The square root method is used in solving incomplete quadraticequations of the form x2 = c, when c is a non-negative number. As you have seenin lesson 3, some quadratic equations of the form x2 = 16 can be solved byfactoring. The Square Root Property of Real Numbers If u2 = d, then u = d or u = - d for d ≥ 0. A shorter way of writing the two solutions u = d and u = - d is to writeusing double sign notation: u = ± d 15

Example 1 x2 – 5 = 0 Solve Solution: x2 – 5 = 0 x2 = 5 x=± 5 Writing in the form x2 = c Square root property x = 5 or x = - 5 The solutions are 5 and - 5 .Check: If x = 5 : x2 – 5 = 0 ( 5 )² - 5 = 0 5–5=0 0=0 If x = - 5 : x² - 5 = 0 (- 5 )² – 5 = 0 5–5=0 0=0Example 2Solve the equation x2 = 16Solution: x2 = 16 x = ± 16 x =± 4 Apply the square root property x = 4 or x = -4 The solutions are 4 and –4.Check: If x =4: x2 = 16 42 = 16 16 = 16 If x = -4: x2 = 16 (-4)2 = 16 16 = 16 16

Example 3Solve the equation 3x2 – 9 =0Solution: 3x2 – 9 = 0 Writing in the form x2 =c 3x2 = 9 By subtraction property x2 = 3 By division property x=± 3 Apply the square root property x = 3 or x = - 3 The solutions are 3 and - 3 .Check: If x = 3 : 3x2 –9 = 0 3( 3 )2 - 9 = 0 3(3) – 9 = 0 0=0 If x = - 3 : 3x2 –9 = 0 3(- 3 )2 - 9 = 0 3(3) – 9 = 0 0=0Example 4Solve for the roots 2x2 +7 = 1Solution: 2x2 = -6 x2 = -3 x = ± − 3 There are no real solutions. The equation has no real solution because − 3 does not exist in the setof real numbers.Example 5Solve (x + 1)2 = 9Solution: (x + 1)2 = 9 x +1 = ± 9 x+1 = ± 3 Extracting square roots Solving the equation x + 1 = 3 or x + 1 = -3 x=2 x = -4 17

The solutions are 2 and –4.Check: If x = 2: (x+1)2 = 9 (2+1)2 = 9 32 = 9 9=9 If x = -4: (x+1)2 = 9 (-4+1)2 = 9 (-3) 2 = 9 9=9Note that the equation can also be solved by factoring. Hence, (x + 1)2 = 9 x2 + 2x + 1 = 9 x2 + 2x + 1- 9 = 0 x2 + 2x - 8 = 0 (x - 2)(x + 4) = 0 x -2 = 0 or x + 4 = 0 x=2 x = -4 Not all equations of the form (x + p)2 = d, can also be solved by factoring,as you will see in the next example.Example 6Solve for x: (x + 3)2 = 7Solution: Using the factoring method (x + 3)2 = 7 Finding the product of the square x2 + 6x + 9 = 7 Writing in standard form x2 + 6x + 9 -7 = 0 Not factorable x2 + 6x + 2 = 0 Using the square root method: (x + 3)2 = 7 7 Apply the square root property x+3=± 7 Simplify and solve for x x = -3 ± x = -3 - 7 or x = -3 + 7 18

The solutions are –3 - 7 and –3 + 7 . Separating the equation x +3 = ± 7 into two equations did not yield anysimplification of the answers. In such case it is all right to write x = -3 ± 7 Whenever you encounter a radical that is a rational number, you may justleave the answer in this form. However, if the radical is not an irrational number,then it is expected that you give each answer individually, as in the followingexample.Example 7Solve for x. (x – 2 )² = 1 34Solution: (x – 2 )² = 1 34 Take square roots x– 2 = ± 1 Addition Property 34 x– 2 = ± 1 32 x=2± 1 32 x=2+ 1 or x=2- 1 32 32 x = 4+3 x = 4−3 6 6 x= 7 x= 1 6 6The solutions are 7 and 1 . The checking is left for you. 66Try this outSolve using the square root method. If the equation has no real solution, write noreal solution.A. 1. x2 = 25 2. b2 – 36 = 0 3. 9b2 – 16 = 0 19

4. b2 + 16 = 0 5. 25x2 = 4 6. 9b2 =4 7. 49x2 = 100 8. 11w2 = 11 9. a2 – 60 = 21 10. y2 + 144 = 0B. 1. 7y2 – 4 = 5y2 + 6 2. 3a2 – 18 = 5a2 – 10 3. (x - 2)2 = 9 4. (a + 5)2 = 7 5. (x - 6)2 = 25C. Solve each equation by either factoring method or square root method, which ever you think is easier to use. 1. 2x2 + 7x - 5 = 3x2 + 9x – 4 2. x2 + 4x + 9 = 3x2 + 4x + 1 3. (y - 2)(y + 3) = y + 10 4. (m – 2 ) 2 = 4 39 5. (d + 1 ) 2 = 9 4 16D. Justify each step in solving the equation x² + 4x – 7 = 0. x² + 4x – 7 = 0 x² + 4x = 7 x² + 4x +4 = 7 + 4 x² + 4x + 4 = 11 (x + 2)² = 11 x +2 = ± 11 x = -2 ± 11x = -2 + 11 or x = -2 - 11 20

Let’s Summarize 1. A linear equation is an equation of the form bx + c = 0, where b ≠ 0. 2. A quadratic equation is an equation of the form ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0. 3. Incomplete forms of quadratic equations are: ax² + bx = 0 ax² + c = 0 ax² = 0 4. Every quadratic equation has two solutions or roots. They may be distinct or equal. 5. Two methods of solving quadratic equations are: a. The factoring method Works for quadratic equations of the form ax2 + bx = 0 and ax2 + bx + c = 0, where the left hand side is factorable. The equation has to be in standard form or the right-hand side of the equation must be 0 before factoring is applied. b. the square root method works for quadratic equations of the form ax2 + c = 0 and (x + p)2 = d. It is not necessary for the equation to be in standard form. What have you learnedA. Which of the following are quadratic equations? If not tell why. 1. x² +4 –2x 2. 3x² -5x + 1 = 3x² 3. (x –4)(x+5) = 10x - 3B. Rewrite the following in the standard form 4. b(b + 9) = 4(5 + 2b) 5. (y +2)(y + 5) = (2y – 1)(y + 6) 6. 5(x + 1)² = 2(x – 3) +7 21

C. Use the factoring method to solve each equation 7. x2 – 9 = 0 8. x2 + 5x + 6 = 0 9. 2x2 + 19x = 0 10. x2 - 6x + 9 = 0D. Use the square root method to solve each equation 11. x2 = 1 12. 3x2 = 243 13. (x+1) 2 = 25 14. 10x2 – 70 = 0 15. (x+2) 2 = 81 22

Answer KeyHow much do you know 1. quadratic equation 2. Not quadratic equation 3. Quadratic equation 4. x² - 7x – 42 = 0 5. x² + 4x + 14 = 0 6. 3x² + 13x – 10 = 0 7. –5, 2 8. 0, 4 9. –2, 12 10. 1, 5 11. –5, 5 12. - 5 , 5 13. –3 ± 10 14. 4 ± 11 15. no real solutionTry this outLesson 1 1. quadratic equation 2. linear equation 3. neither 4. quadratic equation 5. linear equation 6. quadratic equation 7. linear equation 8. linear equation 9. neither 10. quadratic equation 11. neither 12. quadratic equation 13. quadratic equation 14. linear equation 15. quadratic equation 16. neither 17. neither 18. quadratic equation 19. quadratic equation 20. quadratic equation 23

Lesson 2A. a = 4, b = -11, c = -7 1. 4x² - 11x – 7 = 0 a = 2, b = -3, c = -20 2. 2x² - 3x – 20 = 0 a = 6, b = -5, c = 4 3. 6x² - 5x + 4 = 0 a = 1, b = -3, c = 9 4. x² - 3x + 9 = 0 a = 5, b = -3, c = 0 5. 5x² - 3x = 0 a = 4, b = 0, c = -20 6. 4x² - 20 = 0 a = 7, b = 0, c = -9 7. 7x² - 9 = 0 a = 1, b = -4, c = 5 8. x² - 4x + 5 = 0 a = 3, b = -2, c = 0 9. 3x² - 2x = 0 a = 5, b =0, c = 4 10. 5x² + 4 = 0B. a =1, b = 1, c = -7 1. x² + x – 7 = 0 a = 2, b = -14, c = -5 2. 2x² - 14x – 5 = 0 a = 1, b = -10, c = 12 3. x² - 10x + 12 = 0 a = 1, b = -14, c = 52 4. x² - 14x + 52 = 0 a = 1, b = -5, c = -5 5. x² - 5x – 5 = 0 a = 3, b = 2, c = -7 6. 3z² + 2z – 7 = 0 a = 1, b = 0, c = 9 7. y² + 9 = 0 a = 1, b = -12, c = -1 8. p² - 12p – 1 = 0 a = 1, b = 2, c = 3 9. x² + 2x +3 = 0 a = 1, b = 4, c = -16 10. y² + 4y – 16 = 0C. 1. 3x² - 6x + 1 = 0 a = 3, b = -6, c = 1 2. 5x² - 2 = 0 a = 5, b = 0, c = -2 3. 3x² + 10x + 31 = 0 a = 3, b = 10, c =31 4. 2x² - 5x + 2 = 0 a = 2, b = -5, c = 2 5. 3x² + 2x – 12 = 0 a = 3, b = 2, c = -12Lesson 3I. A. 1. –1, 5 2. –4, -3 3. 0, 7 4. –10, 3 5. 7 , 4 12 6. – 9 , 4 23 24

7. – 5 , - 1 67 8. - 3 , 2 9. –1, 10 10. 0, 3 2 11. – 1 , 7 5 12. – 1 , 2 43 13. –6, 3 14. 0, 17 3B. a. The computed solutions were taken right away from (x – 3) and (x + 4) without applying the zero-product property. The solutions should be 3 and –4. b. The right-hand side should have been set to 0.II. A. 1. –2, 3 2. –2, 4 3. 2, 5 4. –3, 4 5. –2, 5 6. 3, 5 7. – 3 , 3 2 8. –3, 10 9. –7, -2 10. 3, 11B. 1. 3 , 4 2 2. 1, 8 3. –6, 3 4. –5, 2 5. not factorable over the rationalsC. 1. –7, 3 2. not factorable over the rationals 3. not factorable over the rationals 25

4. 1, 6 5. –7, 0III.A. 1. –3, 0 2. 0, 5 3. 0, 8 4. 0, 9 5. 0, 14 6. –11, 11 7. ± 2 5 8. ± 3 2 9. ± 8 10. ± 2B. 1. ±2 2. –1, 0 3. ± 5 4. ± 3 5. 0, 1 2 6. not factorable over the rationals 7. ± 2 5 8. 0 9. 0 10. 0C. 1. 2 , - 1 32 2. –1, 2 3 3. –6, - 1 6 4. –3, 2 5 5. 2, 3D. Factoring was applied on the left-hand side of the equation while the right-hand side is not yet 0. 26

Lesson 4A. 1. ± 5 2. ± 6 3. ± 4 3 4. no real solution 5. ± 2 5 6. ± 2 3 7. ± 10 7 8. ± 1 9. ± 9 10. no real solutionB. 1. ± 5 2. no real solution 3. 5, -1 4. –5 ± 7 5. 1, 11C. 1. Factoring method; 1 2. Square root method; ± 2 3. Square root method; ± 4 4. square root method; 0, 4 3 5. square root method; -1, 1 2D. x² + 4x – 7 = 0 x² + 4x = 7 Subtract 7 on both sides x² + 4x +4 = 7 + 4 Complete the square by adding (4/2)² on both sides x² + 4x + 4 = 11 Simplify (x + 2)² = 11 Factor the left-hand side x +2 = ± 11 Apply square root property x = -2 ± 11 Subtract 2 on both sidesWhat have you learned 1. not an equation 2. linear equation 27

3. quadratic equation4. b² + b – 20 = 05. y² + 4y – 16 = 06. 5x² + 8x + 4 = 07. ± 38. –3, -29. 0, −19 210. 311. ± 112. ± 913. –6, 414. ± 715. –11, 7 28

Module 2 Quadratic Equation What this module is about This module is about quadratic equations. This is a continuation of theprevious module. In this module you will learn two more methods of solvingquadratic equations. You have to be proficient in the square root method ofsolving quadratic equations in order to understand the two methods to bediscussed. As you go over the exercises, you will develop skills in solvingquadratic equations and eventually use these skills in solving problems. What you are expected to learn This module is designed for you to: 1. Solve quadratic equations by completing the square. 2. Derive the quadratic formula. 3. Solve quadratic equations using the quadratic formula. How much do you knowA. Solve the following by completing the square: 1. x² - 8x - 15 = 0 2. x² - 2x - 7 = 0 3. x² + 3x = 1 4. x² + 2 x = 0 3 5. 6t² - 5t = t² + 3t – 1B. Solve the following by the quadratic formula. 1. x² - x – 12 = 0 2. x² -4x = 0 3. 3x² + 2 = 0 4. 3x² - x = 1 5. x² = 1 – 2x

What you will do Lesson 1 Completing the Square Completing the square is a method used to make a quadratic expression aperfect square trinomial. This method is used to solve quadratic equations whichcannot be solved by factoring.The method of completing the square is based on the special product: x2 + 2px + p2 = (x + p)2 x2 + 2px + p2 is a perfect square trinomial because it is the product of (x +p)2 , the square of a binomial. In this trinomial, notice that if you take one-half thecoefficient of the x in the middle term and square it, you get the third term. x2 + 2bx + b2 [ 1 (2b)]2 = (b)2 = b2 2 Therefore, to make an expression such as x2 + 12x into a perfect squaretrinomial, you must take one-half of 12, square it and add it to x2 + 12x.If you follow the procedure, here is how it is done. x2 + 12x [ 1 (12)]2 = (6)2 = 36 x2 + 12x + ________ 2 x2 + 12x + 36 Hence, x² + 12x + 36 is now a perfect square trinomial which whenfactored as (x + 6)(x + 6) can also be written as (x + 6)2.Let’s look at some more examples:Examples:Complete the square of the following expressions:a) x2 + 4x b) x2 - 6x c) x2 - 5x d) 2x2 - 3x 2

a) x2 + 4xSolution: x2 + 4x + ______ The coefficient of the x term is 4. x2 + 4x + [ 1 (4)]2 Take 1 of 4 or divide 4 by 2. 2 2 x2 + 4x + (2)2 Square 2. x2 + 4x + 4Note that x2 + 4x + 4 = (x + 2)2b) x2 - 6xSolution: x2 - 6x + ______ The coefficient of the x term is -6. x2 - 6x +[ 1 (-6)]2 1 of -6 is -3 2 2 x2 –6x + (-3)2 Square -3 x2 - 6x + 9Note that x² - 6x + 9 = (x - 3)2c) x2 - 5xSolution: x2 - 5x + _______ The coefficient of the x term is -5 x2 - 5x + [ 1 (-5)]2 1 of -5 is − 5 2 22 x2 –5x + ( − 5 )2 Square − 5 2 2 x2 - 5x + 25 4Note that x2 - 5x + 25 = (x - 5 )2 42 You will notice that the leading coefficient in the first three examples is 1. So,before proceeding to completing the square, transform the leading coefficient to1. The leading coefficient of the next example is not equal to 1. 3

d) 2x2 - 3x Solution: 2x2 - 3x Divide the terms by 2 to make x2 – 3 x + _____ the leading coefficient 1. 2 x2 – 3 x + [ 1 (- 3 )]2 Complete the square: 2 22 1 of - 3 is - 3 2 24 x2 – 3 x + [- 3 ]2 Square - 3 24 4 x2 – 3 x + 9 24 Note that x2 – 3 x + 9 = (x - 3 )2 24 2Try this outFind the third term to be added to make each expression a perfect squaretrinomial. Write the result as a square of a binomial.A. x2 + 4x 1. x2 + 6x 2. x2 - 10x 3. x2 - 8x 4. x2 + 12x 5.B. 1. x2 + x 2. a2 – 3a 3. b2 - 13b 4. b² + 2 b 35. c² - 5 c 2C. 2x2 + 4x 1. 3x2 - 6x 2. 3x2 - x 3. 4x2 - 3x 4. 5x2 + 2x 5. 4

Lesson 2Solution of Quadratic Equations by Completing the Square You are now ready to solve equations by completing the square. In thismethod, the left-hand side of a quadratic equation is solved by completing thesquare so that it becomes a perfect square trinomial which can be written in theform (x + p)2 = d.Example 1: Solve by completing the square:x2 + 8x – 5 = 0Solution: Note that the left member of the equation x2 + 8x – 5 = 0 is not a perfect trinomial square. Separate -5 from the left member by adding +5 to both sides of the equation.x2 + 8x – 5 = 0 You now have x² + 8x on the left-hand side of the equation. Now, makethe left-hand side a perfect square trinomial. You can do this by adding thesquare of one-half of the coefficient of x. So that,x2 + 8x + [ 1 (8)]2 . 1 of 8 is 4. 2 2x2 + 8x + (4)2 Square 4. Since you have done the first part in completing the square of the leftmember of the equation, go back to the original equation. x2 + 8x = 5 Since (4)2 or 16 is added to thex2 + 8x + (4)2 = 5 + (4)2 left member, (4)2 or 16 must alsox2 + 8x + 16 = 5 + 16 be added to the right member by Addition Property of Equality. (x + 4)2 = 21 The equation is now of the form (x + p)2 = d, and so the square rootmethod may then be used. Thus, (x+4)2 = 21 5

x+4 = ± 21 Taking the square rootx = -4 ± 21 Subtracting 4 on both sidesx = -4 + 21 or x = -4 - 21The solutions are -4 + 21 and -4 - 21 .Check:: If x = -4 + 21x² + 8x – 5 = 0 0 = x² + 8x - 5 = (-4 + 21 )² + 8(-4 + 21 ) – 5 = 16 - 8 21 + ( 21 )² - 32 + 8 21 – 5 = 16 + 21 – 32 – 5 = 37 – 37 =0If x = -4 - 21x² + 8x – 5 = 0 0 = x² + 8x - 5 = (-4 - 21 )² + 8(-4 - 21 ) – 5 = 16 + 8 21 + ( 21 )² - 32 - 8 21 – 5 = 16 + 21 – 32 – 5 = 37 – 37 =0The solutions check.Example 2: Solve by completing the square.4x2 + 4x - 3 = 0Solution: Notice that the coefficient of x² is not 1. You need to make the coefficient of x² equal to 1 first before you attempt to use completing the square. Follow the steps on the right.4x2 + 4x - 3 = 0 6

x2 + x - 3 = 0 Divide both sides by 4 4 x2 + x = 3 Add 3 to both sides 4 4x2 + x +[ 1 (1)]2 = 3 +[ 1 (1)]2 Add [ 1 (1)]2 to both sides 2 42 2x2 +x + 1 = 3 + 1 444(x + 1 )2 = 1 Factor 2x + 1=± 1 Use square root method 2x+ 1 =±1 2x = -1 ±1 2x = -1 +1 x = -1 -1 2 2x= 1 x = -3 2 2Note that this equation can be solved by factoring:4x2 + 4x - 3 = 0(2x – 1)(2x + 3) = 02x – 1 = 0 or 2x + 3 = 02x = 1 2x = -3x= 1 x=-3 2 2Check: 4x2 + 4x - 3 = 0 If x = 1 4( 1 )² + 4( 1 ) – 3 = 0 2 22 4( 1 ) + 4 – 3 = 0 42 7

1+2–3=0 3–3=0 0=0 If x = - 3 4x2 + 4x - 3 = 0 2 4(- 3 )² + 4(- 3 ) – 3 = 0 22 4( 9 ) – 12 – 3 = 0 42 9–6–3=0 3–3=0Both solutions check.Example 3:Solve by completing the square: 2x2 - 5x – 3 = 0Solution: Again, just like Example 2, you need to make the coefficient of x² equal to 1. Divide both sides of the equation by 2 to make the coefficient of x² equal to 1. Then proceed as follows: 2x2 - 5x –3 = 0 x2 – 5 x – 3 = 0 Divide both sides by 2 22 Add 3/2 to both sides x2 – 5 x = 3 22 Add [ 1 (- 5 )]2 to both sidesx2 – 5 x + [ 1 (- 5 )]2 = 3 + [ 1 (- 5 )]2 222 22 2 22 Simplifyx2 – 5 x + [- 5 ]2 = 3 + [- 5 ]2 Factor the left-hand side & get 2 4 24 the LCD of the right-hand side of the equation.x2 – 5 x + 25 = 3 + 25 Adding the fractions 2 16 2 16 (x – 5 )2 = 24 + 25 4 16 16 (x – 5 )2 = 49 4 16 8