A mathematical statement that asserts that two quantities are equal is called an equation. Examples: x Mathematics Division, IMSP, UPLB

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1 EQUATIONS

2 A mathematical statement that asserts that two quantities are equal is called an equation. Examples: x 3 11 x 4xy y 0

3 Consider the following equations: = 1. x + 9 = 1 Equation (1) is always true while equation () is true for a particular value of the variable x. An equation involving a variable is called an open equation. 3

4 A solution (root) of an open equation in one variable is the value of the variable for which the equation is true. The set of all solutions of an equation is called the solution set of the equation. In Math 17 we will consider complex solutions. If the solution of an open equation is the entire set of real numbers, the equation is said to be an identity. Otherwise, the equation is called a conditional equation. 4

5 Example 1: The solution of the equation x + 9 = 1 is x = 1. The solution set is { 1 }. This is a conditional equation. 5

6 Example : x + x 6 = 0 The solution of the equation x + x 6 = 0 are x = -3 or x =. The solution set is { -3, }. This is a conditional equation. 6

7 Example 3: 3(x + ) = 3x + 6 The equation 3(x + ) = 3x + 6 is true for any real number x. The solution set is R (set of real numbers). What kind of equation is this? 7

8 Two equations are said to be equivalent if we can obtain one from the other by applying the properties of operations on the real numbers or the properties of equality. Examples: 5x 3 = 0 is equivalent to 5x = 3. x 4 = 5 is equivalent to (x + 3)(x - 3)=0 8

9 LINEAR AND QUADRATIC EQUATIONS

10 At the end of this section, you should be able to: 1. Solve linear equations in one variable. Solve quadratic equations in one variable by: Factoring Completing the Squares Using the Quadratic Formula 3. Determine the nature of the solutions of a quadratic equation 10

11 An equation in one variable, say x, that can be written in the form ax where a and b are real numbers with is called a linear equation in the variable x. b 0 a 0 Examples: 1. x. 3 6 x3 0 7 x 0 7x14 0 standard form of a linear equation. 11

12 From the linear equation a x + b = 0, We obtain the following equivalent equations: (ax + b) + (-b ) = 0 + (-b) by APE ax + [b + (-b )] = 0 + (-b) by Asso + ax + 0 = 0 + (-b) by Inv + ax = -b by Ident + (1/a) ax = (1/a) (-b) by MPE [(1/a) a] x = (1/a) (-b) by Asso 1 x = (1/a) (-b) by Inv x = (1/a) (-b) by Ident x = -(b/a) by Defn of Div. 1

13 Thus, the solution of the linear equation is given by ax b 0, a 0 x. The solution b set of the linear equation is. a b a 13

14 Example 1: Find the solution set of the following linear equations: 1. 7x x 3 3x 10 4x x 1 3. x

15 Example : Solve for the indicated variable in terms of the other variables: 1. ax b cx d; a 3xy 4. 1 x; x y 3 3. rs at bt s; s 15

16 Any equation in one variable, say x, that can be written in the form ax bx c where a, b and c are real numbers,, is called a quadratic equation in the variable x. 0 a 0 standard form of a quadratic equation. 16

17 Examples: Write the ff. in standard form: 1. 3x 7 5x x. t 5t 6 3. t 5t t 1 t 7 17

18 To solve quadratic equations, we use any of the following methods: 1. By factoring. By completing the squares 3. By using the Quadratic Formula. 18

19 In this method, we use the following rule: Let u and v be real numbers or quantities, u v = 0 if and only if u = 0 or v = 0. Thus, if we can factor the quadratic equation ax + bx + c = 0, a 0, into two linear factors say (px + q) and (sx + t), then the solutions of this quadratic equation are the solutions of the linear factors: (px + q) = 0 or (sx + t) = 0 19

20 Example 1: Solve x x 6 = 0. Solution: x x 6 = (x 3)(x + ) Hence, x x 6 = 0 will imply that (x 3)(x + ) = 0 x 3 = 0 or x + = 0 x = 3 or x = -. SS 3, 0

21 WARNING!!! The following is NOT always true: If ab = k (where k 0) then a=k or b=k Kaya walang gagawa nito!!! 1

22 Example : Solve t(5t + ) = t(1 t ) + 7. Solution: t(5t + ) = t(1 t ) + 7 7t 7 = 0 t 1 = 0 why? (t 1)(t + 1) = 0 t 1 = 0 or t + 1 = 0 t = 1 or t = 1. SS 1, 1

23 Exercise: A. Solve the following quadratic equations by factoring: 1. 15y 8 6y. 3x x 1 x 3. a a 3 a 1 a B. Solve for the indicated variable: 1. 18a x abx 15b 0; x. 3 k t 5s 4 kst; t 3

24 Note: a = b a = b or a = b (why?) Recall: (x a) = x ax + a We use the above results in solving quadratic equations by completing the squares. 4

25 Solve: x + 4x = 0. Solution: Transpose the constant to the right: x + 4x = Complete the squares on the left side by adding 4 to both sides of the equation (why?): x + 4x + 4 = + 4 Factor the left side completely and simplify: (x + ) = 6 We obtain x + = 6 to get x = 6. SS 6, 6 5

26 Solve: x 6x + 7 = 0. Solution: Transpose the constant 7 to the right: x 6x = 7 Complete the square on the left side by adding 9 to both sides of the equation (why?): x 6x + 9 = Factor the left side completely and simplify: (x 3) = We obtain x 3 =, to get x = 3 SS 3,3 6

27 In general, if you have ax + bx + c = 0 then factor out a: a[x + (b/a)x + c/a] = 0. Hence, you will have: x + (b/a)x + c/a = 0. To complete the square, transpose the constant to the right: x + (b/a)x = c/a Then, add [(b/a)/] to both sides: x + (b/a)x + [(b/a)/] = c/a + [(b/a)/] The left hand side is already a perfect square: [x + (b/a)/] = c/a + [(b/a)/] 7

28 Exercise: Solve the following by completing the square: 1. 4x 8x + 7 = 0. x 5x = 4 3. x 7x + 6 = 0. 8

29 RECALL: From ax + bx + c = 0, a 0, We get ax + bx = c. Dividing both sides of the equation by a 0: b c x x a a Completing the squares on the left side: x b a x b a c a b a 9

30 x Simplifying: b c b b 4ac a a 4a 4a b b 4ac x a 4a b b 4 ac x a The Quadratic Formula 30

31 Example 1: Solve x 4x 5 = 0 Solution: Here a =, b = -4 and c = -5 By the quadratic formula: x SS 14, 14 x 14 or x 14 31

32 Example : Solve 4x 1x + 9 = 0 Solution: Here a = 4, b = -1 and c = 9 By the quadratic formula: x x 3 3 SS 3

33 Example 3: Solve x 4x + 5 = 0 Solution: Here a =, b = -4 and c = 5 By the quadratic formula: x is not a realnumber. Hence, No real Solution. SS 1 i 6 33

34 b In the quadratic formula, the radicand b 4ac is called the discriminant of the quadratic equation. The discriminant gives the nature of the solutions of a quadratic equation: 0 two real solutions 4 ac 0 one real solution 0 no real solution (i.e., two complex solutions) 34

35 In each of the following, determine the nature of the solutions of the given quadratic equation: 1. x 4x x 3x w 8w

36 Note that the solutions of the quadratic equation ax + bx + c = 0, a 0, is given by b b 4ac b b 4ac r 1 or r a a Taking the sum of these solutions give: r 1 b b 4ac b b 4ac r a a b a b a 36

37 Also, the product of the two solutions is given by: r 1 b b 4ac b b 4ac r a a b b 4ac a a b b 4ac c 4a a 37

38 In each of the following, find the sum and product of the solutions of the given quadratic equation: 1. x 4x x 3x w 8w

39 In this, you have learned that 1. the solution of the linear equation ax + b = 0 is x = -b/a;. the quadratic equation ax + bx + c = 0 can be solve by factoring, completing the squares and by the quadratic formula. 39

40 3. the quadratic formula is given by: x b b 4ac a 4. the sum and product of the solutions of a quadratic equation is given by b r1 r and r1 r a c a 40

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