Strategic Intervention Material In Mathematics Grade 8 Pdf

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My first Strategic Intervention Material. The topic is Solving Quadratic Equation by Factoring

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• 1. + Strategic Intervention Material Mathematics IX Solving Quadratic Equation by Factoring Prepared by: Brian M. Mary T-I
• 2. LEAST MASTERED SKILLS Solving Quadratic Equation by Factoring Sub Tasks + Identifying quadratic equations Rewriting quadratic equations to its standard form Factor trinomials in the form x2 + bx + c Determine roots of quadratic equation ax2 + bx + c = 0, by factoring
• 3. + Overview A quadratic equation in one variable is a mathematical sentence of degree 2 that can be written in the following form ax2 + bx + c = 0, where a, b, and c are real numbers and a 0. How are quadratic equations used in solving real life problems and in making decisions? Many formulas used in the physical world are quadratic in nature since they become second-degree equations when solving for one of the variables. Likewise, many word problems require the use of the quadratic equation. At the enrichment card, we will consider some of the common uses of the quadratic equations.
• 4. + Activity # 1 Quadratic or Not Quadratic? Direction. Identify which of the following equations are quadratic and which are not. Write QE if the equations are quadratic and NQE if not quadratic equation. __________ 1. 3m + 8 = 15 __________ 2. x2 5x 10 = 0 __________ 3. 2t2 7t = 12 __________ 4. 12 4x = 0 __________ 5. 25 r2 = 4r
• 5. Activity # 2 + Set Me to Your Standard! Direction. Write each quadratic equation in standard form, ax2 + bx + c = 0. 1. 3x 2x2 = 7 ____________________ 2. 5 2r2 = 6r ____________________ 3. 2x(x 3) = 15 ____________________ 4. (x + 3)(x + 4)= 0 ____________________ 5. (x + 4)2 + 8 = 0
• 6. + Activity # 3 What Made Me? We learned how to multiply two binomials as follows: factors terms M u l t i p l y i n g (x+2)(x+6) = x2 + 6x + 2x + 12 = x2 + 8x + 12. In factoring, we reverse the operation F a c t o r i n g terms factors x2 + 8x + 12 = (x + 2)(x + 6) The following will enable us to see how a trinomial factors. 12 = 2 (6) x2 + 8x + 12 = (x + 2)(x + 6) 8 = 2 + 6 Product Sum
• 7. In general, the trinomial x2 + bx + c will factor only if there are two integers, which will we call m and n, such that m + n = b and m(n) = c. + Sum Product m + n m(n) x2 + bx + c = (x + m)(x + n) 1. a2 + 11a + 18 m + n = 11 m(n) = 18 2 + 9 = 11 2(9) = 18 The m and n values are 2 and 9. the factorization is, a2 + 11a + 18 = (x + 2) (x + 9) 2. b2 2b 15 m + n = - 2 m(n) = - 15 3 + (-5) = - 2 3(-5) = - 15 The m and n values are 3 and - 5. the factorization is, b2 2b 15 = (x + 3) (x 5)
• 8. Factor the following trinomial in the form x2 + bx + c. x2 + bx + c m + n m(n) (x + m)(x + n) x2 + 4x 12 6 + (-2) 6(-2) (x + 6)(x 2) w2 8w + 12 x2 + 5x - 24 c2 + 6c + 5 + r2 + 5r 14 x2 + 5x + 7 After learning how to factor trinomial in the form x2 + bx + c, we will now determine roots of a quadratic equation using factoring.
• 9. Activity # 4 + Factor then Solve! Some quadratic equations can be solved easily by factoring. To solve each equations, the following procedures can be followed. 1. Transform the quadratic equation into standard form if necessary. 2. Factor the quadratic expression. 3. Set each factor of the quadratic expression equal to 0. 4. Solve each resulting equation. Example. Find the solution of x2 + 9x = -8 by factoring. a. Transform the equation into standard form x2 + 9x = -8 x2 + 9x + 8 = 0 b. Factor the quadratic expression x2 + 9x + 8 = 0 (x + 1)(x +8) = 0 c. Set each factor equal to 0. (x + 1)(x + 8) = 0 x + 1 = 0; x + 8 = 0 d. Solve each resulting equation. x + 1 = 0 x + 1 1 = 0 - 1 x = - 1 x + 8 = 0 x + 8 8 = 0 - 8 x = - 8
• 10. Direction. Determine the roots of the following quadratic equations using factoring. 1. x2 + 8x + 16 = 0 _____________________________________ + _____________________________________ _____________________________________ 2. x2 9x 14 = 0 _____________________________________ _____________________________________ _____________________________________ 3. y2 + 9y + 20 = 0 _____________________________________ _____________________________________ _____________________________________ 4. b2 10b + 21 = 0 _____________________________________ _____________________________________ _____________________________________
• 11. Number Theory * The product of two consecutive even numbers is 168. What are the integers? Solution: Let x = the lesser even integer. Then, x + 2 = the next consecutive even integer. Note: Consecutive even or odd integers are given by x, x+ 2, x + 4, product of two consecutive even integers is 168 x2 + 2x = 168 original equation x2 + 2x 168 = 0 write in standard form + x ( x + 2 ) = 168 ( x + 14 ) ( x 12 ) = 0 factor the left member x + 14 = 0 or x 12 = 0 set each factor equal to zero x = - 14 x = 12 solve each equation when x = - 14, then x + 2 = - 14 + 2 = - 12 and when x = 12, then x + 2 = 12 + 2 = 14 since (-14)(-12) = 168 and (12)(14) = 168, and both solutions are consecutive even integers, the conditions of the problem are met. Therefore, the two integers are 14 and 12 or 12 and 14. Mastery Points! Can you Determine two integers whose product is one number and whose sum is another number? Recognize when the trinomial x2 + bx + c will factor and when it will not? Factor trinomial of the form x2 + bx +c ? Determine roots of a quadratic function in the form ax2 + bx + c?
• 12. Learners Material Mathematics IX, First Edition pp. 27 - 34 Holiday, Berchie. et. al. ALGEBRA 2. USA. The McGraw Hill Companies, C2008. pp. 253 256 Wesner, et. al. ELEMENTARY ALGEBRA with APPLICATIONS. Bernard + J. Klein Publishing, 2006 pp. 152 156
• 13. Activity # 1 Quadratic or Not Quadratic? + 1. NQE 2. QE 3. QE 4. NQE 5. QE Activity # 2 Set Me to Your Standard 1. - 2x2 + 3x 7 = 0/2x2 3x + 7 = 0 2. - 2r2 6r + 5 = 0/2r2 + 6r 5 = 0 3. 2x2 6x 15 = 0 4. x2 + 7x + 12 = 0 5. x2 + 8x + 24 = 0 Activity # 3 What Made Me? x2 + bx + c m + n m(n) (x + m) (x + n) w2 8w + 12 - 6 + 2 -6(2) (w 6)(w + 2) x2 + 5x 24 8 + (-3) 8(-3) (x + 8)(x 3) c2 + 6c + 5 5 + 1 5(1) (c + 5)(c + 1) r2 + 5r 14 7 + (-2) 7(-2) (r + 7)(r 2) x2 + 9x + 20 5 + 4 5(4) (x + 5)(x + 4) Assessment 1. x2 + 8x + 16 = 0 (x + 4)(x + 4) = 0 x + 4 = 0 x + 4 4 = 0 4 x = - 4 2. x2 5x 14 = 0 (x 7)(x 2) = 0 x 7 = 0 x 2 = 0 x 7 + 7 = 0 + 7 x 2 + 2 = 0 + 2 x = 7 x = 2 3. y2 + 9y + 20 = 0 (y + 5)(y + 4) = 0 y + 5 = 0 y + 4 = 0 y + 5 5 = 0 5 y + 4 4 = 0 4 y = - 5 y = - 4 4. b2 10b + 21 = 0 (b 7)(b 3) = 0 b 7 = 0 b 3 = 0 b 7 + 7 = 0 + 7 b 3 + 3 = 0 + 3 b = 7 b = 3
• 14. +

Strategic Intervention Material In Mathematics Grade 8 Pdf

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