DON ROBERT B. ESTRELLA SR. NATIONAL HIGH SCHOOL Nagsaag, San Manuel, Pangasinan. (Effective Alternative Secondary Education) MATHEMATICS II

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1 DON ROBERT B. ESTRELLA SR. NATIONAL HIGH SCHOOL Nagsaag, San Manuel, Pangasinan. (Effective Alternative Secondary Education) MATHEMATICS II Y X MODULE 1 Quadratic Equations BUREAU OF SECONDARY EDUCATION Department of Education DepEd Complex, Meralco Avenue, Pasig City

2 Module 1 Quadratic Equations What this module is about This module is about quadratic equations. In this module, you will be able to develop skills in solving second-degree equations in one variable. You will find that quadratic equations can be solved using several methods. Two of these methods 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.. 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? x = x 1. = x x 3. 5x + x 4 = 0 B. Write the following in the standard quadratic form ax + bx + c = 0 4. (x + 5)(x - 1) = x x + 14 = x(x - 3) 6. 4x 8x + = x - 1x + 1 C. Use the factoring method to solve each equation.

3 7. x + 3x 10 = 0 8. x = 4x 9. x - 10x 4 = x - 6x + 5 = 0 D. Use the square root method to solve each equation. 11. x = 5 1. x 10 = (x + 3) = (x - 4) = (x + 3) = -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 one variable. These are equations which involve one variable. The highest exponent of the variable is 1. These first degree equations are called linear equations. For example, the equation 3x 5 = 4 is a first- degree equation. A second- degree equation in one variable is an equation in which the highest exponent of the variable is. These equations are called quadratic equations. The equation 3x -x 5 = 0 is a second-degree equation. Example: Classify each equation as linear or quadratic: 1. x + 3x + 5 = 0. 5x 3 = x² + 1x = x + 1 = 0 5. x 3 + 4x 3x² = 0 Examples and 4 are linear equations since the highest exponent of the variable x is 1, while Examples 1 and 3 are quadratic equations because the 3

4 highest exponent of the variable x is. Example 5 is neither linear nor quadratic. Why? Try this out A. Identify whether the following equations are linear, quadratic or neither: 1. x 4x + 7 = 0. x 5 = x 4x + 5 = x - 1 = x + 1 = 0 6. x² - 5x = 3 7. x² - 4x x² = 0 8. x 5 = x 3x = 5 x x = x 3x 3 + x² = x² = 3x a 5 = a² 14. 4b 5 = 3(b-5) = x² - 4x 16. 5t 0 5t = p² - 5 = 4p² 3x x² - 4 = q² + 6q ( q² + 5) = (q )² 0. (x + 5)(x 3) = 0 Lesson Writing Quadratic Equations in Standard Form By this time, you are already familiar with quadratic equations. To reiterate, quadratic equations are of the form ax² + bx +c = 0, where a 0. A quadratic equation written in this way is said to be in standard form. The equation x 7x + 4 = 0 is a quadratic equation in standard form where, a =1, b = -7 and c = 4. 4

5 Examples: The following quadratic equations are written in standard form: 1. x + 5x 3 = 0. 3x + 4x +5 = x x = 0 4. x + 5 = x = 0 Can you identify the values of a, b and c in the examples above? Notice that in all the quadratic equations listed, there is always a value for a. Why? Study the equation below: -x 4(x 5) = 7 9 Did 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 write quadratic equations, such as the one above in standard form. Solution: -x 4( x 5 ) = 7-9 -x 4x +0 = 7-9 Distributive property -x 4x + 0 = - Addition property of equality -x 4x = 0 Addition property of equality -1(-x 4x + = 0) Multiply by -1 x + 4x - = 0 is now the standard form where a =, b = 4 and c = Write the equation x² - 4x + = x 1 x 3 Solution: 5 x² - 4x + = x 1 x 3 x 1 x 1x² - 4x + 10 = 3(x² + x ) 1x² - 4x + 10 = 3x² + 3x 6 1x² - 4x x² -3x + 6 = 0 9x² -1x + 16 = 0 6(x² - 4x ) = (6) in standard form: Multiply by the LCD Division of numbers Distributive property Addition property 9x² -1x + 16 = 0 is now the standard form where a = 9, b = -1 and c = 5

6 Try this out Write the following in standard form. Identify the values of a, b and c. A. 1. 4x -11x = 7. x -0 = 3x 3. 6x = 5x x 9 = x 5. 5x = 3x 6. 4x = x = 9 8. x = 4x 5 9. x = 3x = - 5x B. 1. (x + 3) (x - ) =1. x(x - 7) = 5 3. x = (5x - 6) 4. (x - 7) + 3 = 0 5. (x - 4)(x + ) = 3(x - 1) 6. z(z - 1) = 5z (y + 1) = (y - 4) 8. p(5p 3) = (3p + 1) 9. x + 5x 4 = x + 3x (y + )(y + 5) = (y - 1)(y + 6) C. 1. x 6x 1 = 3 3 5x 1 1. = ( x 3) x 1 3. = 4 3 5x 4. x² + 1 = x 5. x² + = 4 3 6

7 Lesson 3 Solving Quadratic Equations by Factoring This time, you will learn how to solve quadratic equations. Solving quadratic equations means solving for the value of x that will make the equation true. For instance, the equation x + x = 30 has two solutions since both x = 5 and x = -6 satisfy the equation. When x = 5 When x = -6 x + x = = = = 30 x + x = 30 (-6) 0 + (-6) = = = 30 Since both 5 and 6 make the equation true, then it is said that 5 and 6 are the solutions or roots of the equation x + x = 30. It should be noted that every quadratic equation has two roots. There are many ways of solving quadratic equations. The first way or method 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 the quantity (x +3) = 0. 7

8 Thus, So that x - 4 = 0 or x +3 = 0. x = 4 or x = -3 Notice that both 4 and 3 are solutions of the original equation (x 4)(x +3) = 0. By checking, Example 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 = 0 Solve: x (x 9) = 0 Solution: x (x 9) = 0 x = 0 or x 9 = 0 zero product property x = 9 9 x = The solutions are 0 and 9. The checking is left for you. Try this out I. A. Solve the following: 1. (x 5)(x + 1) = 0. (x + 4)(x + 3) = 0 3. (x - 7)x = 0 4. (x + 10)(x 3) = 0 5. (x 7)(x - 4) = 0 6. (3x 4)(x + 9) = 0 7. (6x + 5)(7x + 1) = 0 8

9 8. (x )(x + 3 ) = 0 9. ( x 5)(x + 1) = ( 3 x 1 )x = (5x + 1)(x 7) = 0 1. (4t + 1)(3t ) = (x 3)(x + 6) = y(3y 17) = 0 B. What is wrong with the following solutions? a. (x 3)(x + 4) = 0 x = -3 or x = 4 b. (x 3)(x + 4) = 8 x 3 = or x + 4 = 4 x = 5 x =0 In the equation x + x = 30, you cannot apply the zero product property right 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 by writing the equation in standard form. So that, x + x = 30 becomes x + 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, x + x = 30 x + x 30 = 0 (x + 6)(x - 5) = 0 x+6 = 0 or x 5 = 0 x = -6 x = 5 Addition property of equality Factoring Zero-product property The method in finding the solution that has just been illustrated is called Factoring. By experience, it should be noted that not all quadratic trinomials are factorable. Thus, the factoring method does not always work. Other methods of solving quadratic equations will be discussed later. In the meantime, study the following examples solved by factoring. 9

10 Example 3 Solve x² + 5x + 6 = 0 Solution: 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. Example 4 x² + 5x + 6 = 0 (x + )(x + 3) = 0 Factoring x + = 0 or x + 3 = 0 Zero product property x = - x = -3 The solutions are and 3. Check: If x = -, then x² + 5x + 6 = 0 (-)² + 5(-) + 6 = = = 0 0 = 0 Solve for x: x = x + 6 if x = -3, then x² + 5x + 6 = 0 (-3)² + 5(-3) + 6 = = = 0 0 = 0 Solution: To solve by factoring method, rewrite first the equation in standard form. x = x + 6 x - x 6 = 0 (x-3) (x+) = 0 then factor; apply the zero product property x 3 = 0 or x + = 0 x = 3 x = - The solutions are 3 and -. Check: If x = 3, then x = x + 6 (3) =

11 9 = 9 If x = -, then x = x + 6 (-) = = 4 From the above examples, can you identify the steps used in solving quadratic equations? Example 5 Solve x² - 8x = -16. Solution: Again, the equation is not in standard form. So, rewrite the equation in standard form. x² - 8x = -16 x² - 8x + 16 = 0 Addition Property of Equality (x 4)(x 4) = 0 Factoring 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 solutions are 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 perfect square 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 6 Solve for a: (a + 1)(a -1) = (3a -)(a - 4) Solution: (a + 1)(a -1) = (3a -)(a - 4) a a 1 = 6a 16a + 8 Writing in standard form 0 = 4a 15a = (4a - 3)(a-3) Factoring 4a 3 = 0 or a -3 = 0 Zero product property 3 a = 4 a = 3 11

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

13 Try this out II. Solve by factoring A. 1. x - x - 6 = 0. t t - 8 = 0 3. t 7t + 10 = 0 4. x - x - 1 = 0 5. x - 3x - 10 = 0 6. x - 8x + 15 = 0 7. x - 3x - 9 = 0 8. x - 7x - 30 = 0 9. x x = x x = 0 B. 1. a = 11a 1. -x = 8 9x 3. x(x+3) - 36 = 0 4. x (x+3) = x - x - 6 = x - 6x - 3 C. 1. (x + ) = 5. (x - 4)(x + 1) = (x-3)(x-) 3. (x + 3) = 3x (x - ) = x 11x (y + )(y + 5) = 10 Notice that all of the quadratic equations you have studied and solved in this lesson are factorable complete forms of quadratic equations. The following incomplete form of quadratic equations may also be solved by the factoring method: Example 8 Solve for x. x - 3x = 0 Solution: 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: x - 3x = 0 x(x -3) = 0 By factoring, x is the common monomial factor x = 0 or x - 3 = 0 Zero-product property 13

14 Example 9 x = 3 The solutions are 0 and 3. Check: If x =0 x - 3x = 0 (0) 3(0) = 0 0 = 0 If x = 3 x - 3x = 0 (3) 3(3) = = 0 0 = 0 Solve for y: 6y + 8y = 0 Solution: y(3y + 4) = 0 By factoring, y is the common monomial factor y = 0 or 3y +4 = 0 Zero-product property y = 0 3y = -4 4 y = - 3 The solutions are 0 and Check: If y = 0 6y + 8y = 0 6(0) + 8(0) = 0 0 = 0 If y = y + 8y = 0 6(- 3 4 ) + 8(- 3 4 ) = ( ) = ( ) - = = = 0 Observe that Examples 8 and 9 are of the form ax + bx = 0. 14

15 Example 10 Solve for x : x -144 = 0 Solution: x -144 = 0 (x -1)(x + 1) = 0 By factoring x -1 = 0 or x + 1 = 0 Zero-product property x = 1 x = -1 The solutions are 1 and 1. Check: If x = 1: x -144 = 0 (1) -144 = = 0 0 = 0 If x = -1: x -144 = 0 (-1) -144 = = 0 0 = 0 Observe that Example 10 is of the form ax + c = 0. The method used is factoring the difference of two squares. Finally, note that the types of factoring used in solving quadratic equations by the factoring method are quadratic trinomials, common monomial factor, the difference of two squares and perfect square binomials. Try this out III. Find the solution by factoring: A. 1. a(a + 3) = 0. 3z 15z = 0 3. x + 3x = 11x 4. x 18x = 0 5. x = 14x 6. x - 11 = x - 4 = x - 9 = 0 9. x - 4 = p - 5 = 7 B. 1. x -8 = 0. x + x = 0 3. p = 50 15

16 4. 3x = x 3x = x² -5 = x² = 4 8. x² = x² = x² = 0 C. 1. 6x x - = 0. 3x + x - = x + 37x + 6 = x + 13x - 6 = 0 5. x 5x + 6 = 0 D. What is wrong with the following solution? x² - 7x +1 = 7 (x 3)(x 4) = 7 x 3 = 7 or x 4 = 7 x = 10 x = 11 Lesson 4 Solving Quadratic Equations by Extracting Square Roots As mentioned in Lesson 3, the factoring works only when the left hand side of a quadratic equation in standard form is factorable. For example, the quadratic equations x + 5x +1 = 0 and x - 0 = 0 cannot be solved by factoring. There must be other methods to solve such equations. The square root method is used in solving incomplete quadratic equations of the form x = c, when c is a non-negative number. As you have seen in lesson 3, some quadratic equations of the form x = 16 can be solved by factoring. The Square Root Property of Real Numbers If u = 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 write using double sign notation: u = d 16

17 Example 1 Example Solve x 5 = 0 Solution: x 5 = 0 x = 5 Writing in the form x = c x = 5 Square root property x = 5 or x = - 5 The solutions are 5 and - 5. Check: If x = 5 : x 5 = 0 ( 5 )² - 5 = = 0 0 = 0 If x = - 5 : x² - 5 = 0 (- 5 )² 5 = = 0 0 = 0 Solve the equation x = 16 Solution: x = 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: x = 16 4 = = 16 If x = -4: x = 16 (-4) = = 16 17

18 Example 3 Example 4 Solve the equation 3x 9 =0 Solution: 3x 9 = 0 Writing in the form x =c 3x = 9 By subtraction property x = 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 : 3x 9 = 0 3( 3 ) - 9 = 0 3(3) 9 = 0 0 = 0 If x = - 3 : 3x 9 = 0 Solve for the roots x +7 = 1 3(- 3 ) - 9 = 0 3(3) 9 = 0 0 = 0 Solution: x = -6 x = -3 x = 3 There are no real solutions. The equation has no real solution because 3 does not exist in the set of real numbers. Example 5 Solve (x + 1) = 9 Solution: (x + 1) = 9 x +1 = 9 Extracting square roots x+1 = 3 Solving the equation x + 1 = 3 or x + 1 = -3 x = x = -4 18

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

20 The solutions are 3-7 and Separating the equation x +3 = 7 into two equations did not yield any simplification 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 just leave 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 following example. Example 7 Solve for x. (x 3 )² = 4 1 Solution: (x 3 )² = x = Take square roots 3 4 x 3 = 1 x = 3 1 Addition Property x = x = 6 x = 6 7 or x = 3-1 x = x = The solutions are 6 7 and 6 1. The checking is left for you. Try this out Solve using the square root method. If the equation has no real solution, write no real solution. A. 1. x = 5. b 36 = b 16 = 0 0

21 4. b + 16 = x = b = x = w = a 60 = y = 0 B. 1. 7y 4 = 5y a 18 = 5a (x - ) = 9 4. (a + 5) = 7 5. (x - 6) = 5 C. Solve each equation by either factoring method or square root method, which ever you think is easier to use. 1. x + 7x - 5 = 3x + 9x 4. x + 4x + 9 = 3x + 4x (y - )(y + 3) = y (m 3 ) = (d ) = 16 9 D. Justify each step in solving the equation x² + 4x 7 = 0. x² + 4x 7 = 0 x² + 4x = 7 x² + 4x +4 = x² + 4x + 4 = 11 (x + )² = 11 x + = 11 x = - 11 x = or x =

22 Let s Summarize 1. A linear equation is an equation of the form bx + c = 0, where b 0.. A quadratic equation is an equation of the form ax + bx + c = 0, where a, b and c are real numbers and a 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 ax + bx = 0 and ax + 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 ax + c = 0 and (x + p) = d. It is not necessary for the equation to be in standard form. What have you learned A. Which of the following are quadratic equations? If not tell why. 1. x² +4 x. 3x² -5x + 1 = 3x² 3. (x 4)(x+5) = 10x - 3 B. Rewrite the following in the standard form 4. b(b + 9) = 4(5 + b) 5. (y +)(y + 5) = (y 1)(y + 6) 6. 5(x + 1)² = (x 3) +7

23 C. Use the factoring method to solve each equation 7. x 9 = 0 8. x + 5x + 6 = 0 9. x + 19x = x - 6x + 9 = 0 D. Use the square root method to solve each equation 11. x = x = (x+1) = x 70 = (x+) = 81 3

24 Answer Key How much do you know 1. quadratic equation. Not quadratic equation 3. Quadratic equation 4. x² - 7x 4 = 0 5. x² -5x - 14 = x² + 13x 10 = , 8. 0, 4 9., , , , no real solution Try this out Lesson 1 1. quadratic equation. 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 1. quadratic equation 13. quadratic equation 14. linear equation 15. quadratic equation 16. neither 17. neither 18. quadratic equation 19. quadratic equation 0. quadratic equation 4

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

26 7. 6 5, , 9. 1, , , , , , 3 B. 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., 3., 4 3., , 4 5., , , , , , 11 3 B. 1., 4. 1, , , 5. not factorable over the rationals C. 1. 7, 3. not factorable over the rationals 3. not factorable over the rationals 6

27 4. 1, , 0 III. A. 1. 3, 0. 0, , , , , B , , 1 6. not factorable over the rationals C. 1., , , , 5 5., 3 D. Factoring was applied on the left-hand side of the equation while the righthand side is not yet 0. 7

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

29 3. quadratic equation 4. b² + b 0 = 0 5. y² + 4y 16 = x² + 8x + 4 = , , , , 7 9

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