Grade 9 Math Lesson 4 - Sum and Product of Roots (PDF)

Module 1 : Lesson 4 The Sum - and Product of the Roots of Quadratic Equations Let X1 = b # D2-4ac...

Module 1 : Lesson 4 The Sum - and Product of the Roots of Quadratic Equations Let X1 = b # D2-4ac b2 fac X2 - = - b => 2a 2a Sum of the Roots of Quadratic a Equation , + X Xz = b+ be 4ac b2 _ Jac b - - t - = 2a 2a = -b + Fac - b - v-4ac) O - (-b = b + ac 4ac) & = 2b) = X + Xz = - Product of the Roots of a Quadratic Equation X, o X2 = b+ be - 4 ac b2 _ Jac - - b = ⑧ 2a 2a 4ac) pac(zx((b Note = (* ) *= (b) 2 (b = X = = = 22a2x ((2a)) 2 S · # = = X X * I 62 +a) 4a2 * I b2 - b2 #192 42 (x)) = x + = x : XiXz = a = =  Therefore , Bonus : We also the can get roots of the b X1 X1 + xz Xe C equation · - = = T From their sum and where X, and Xz are the roots of the QE product. Find the sum and the product of the roots of the FF QE : - Note : Find the roots as well. - 1) 2x2 + 8x - 10 = 0 2) Ep2 + =p + 5 = 0 of roots > sum = - compare to ax + by + c = 0 - = a = E c= 5 b = 7 = - 4 sum of the roots : =I E Product of roots = > - ~+ = · a - Note. - = = _ What are the roots of QE ?? product a = E of the roots : 5 as C Note that the the b 7 r2 552 = · product of : roots (X1 X2) is 5 5 5 - · = c= S - The -5 10 factors of are : 2 ⑤ · 5 I E -.. I # 5 - b The of the roots (x1 + xz) · sum 4 From the is Factors of -5, - -- The factors are : only - 5 and I has - I as the sum. - 5 +1 = - 4 (2x + 5)(vx + 2) = 0 The roots are : Ex +5 = 0 Ex + 2 = 0 # : the roots = of the equation X -5 = and X = 1 X - = = X = X E - X = Find the sum and the product of the roots of the FF QE : 3) (p+ 1)xz - 2px + (q + 1) = 0 Note : in terms of pandq Use the values of a , b , and c in Finding the roots of the QE : 1)0x2 + 12x - 18 = 0 2) x2 + 7x - 18 = 0  Write equation in the the form ax + bx + c 0 which = the numbers solutions. given are 1) 5 I and 2) 2 and I - -  Module 1 : Lesson 5 Equations Transformable to QE Activity. Perform the indicated operation and then express your answer in simplest Form. 1) 2n + 3n 2)2 = t 2 - 15m2 2 1m t+1 + 2 -t 2 - L t - 2 Find the least common denominator (LCD) : - +I = (t 1)(t e) + - 15m2 = 3. 5. m. m 21m = 3 7. · m Find the LCD : (t + 1) (t 2)- 4 CD = 3. 5 · 7 · m. m 2) (t (t 2) - 2) - - I = 105m2 (t 1)(t 2) + - 2n 3n t 2t & = - 4 - t +2 15m2 2 1m (t + 1)(t 2) - = 14n + 15mm = 2 105m2 2) (t + 1)(( = n(14 + 15m) + +0 - m , - 1 105m2 3) * t 2 Recall. Square of a Binomial X+ 1 - Find = the LCD : x(x+ 2) + 2(x + 1) (x + 1) (x + 2) - (x + 1)(x + 2) (x + 1)(x+1) = x2 + 2x + 2x + 2 (x 3)(X 3) - - (x+ 1)(x + 2) = x2 + 4x + 2 < X = - 1 , - 2 (x + 1)(x + 2) Transform the FF. equations to QE. 1) x(x + 4) = 7 2) (x 4)(x+ 5) - = 3 X 2 + 5x - 4x - 20 - 3 = 0 X2 + x - 23 = 0 3) (3x = 2)(2x - 1) = - 4 6X2 - 3x - 4x + 2 = - 4 6x2 - 7x + 4 = 0 4)(X - 7)2 = 6 X2 - 14x + 49 = 6 X2 - 14x + 43 = 0 5) 2 (x b)- + (x + 4)2 = - 20 (x2 - 12x + 36) + (xz + 8x + 14) = - 20 2x2 - 4x + 36 + 14 + 20 = 0 2 - 2 4X + 72 = 0 a) * = + = 3 LCD : 2x Multiply both sides the. LCD by 2x) = =) (3)2x + = 8 + xz = 6X x2 - xx + 8 = 0 = (X 4)(X 2) - - = 0 The roots are X= 4 and X= 2 7) X - 4 + 2 = - 5 3 X - 2 LCD : 3 (X- 2) Multiply LCD to both sides. Using &E : 2)(X -4 3(X 22) (5)3(x 2) - - + = X = - 91 M7 2 (x 2)(X - - 4) + k = 15(X - 2) X2 - 4x - 2x + 8 + 4 = 15X + 30 *2 - 4x - 2x + 15x + 14 - 30 = 0 X = + 9X - 16 = 0 8) X+ 2 - X - 3 = 1 X - 4 X+ 3 LCD : (x -4)(x + 3) (x 4)(X + 3) - ( X-Y) = 1(x-4) x (x + 3)(x + 2) - (x - 4)(x 3) - = (x 4)(x + 3) - (x2 + 5x + 4) = (x2 - 7x + 12) = (x) - X - 12) * 2 + 5x + 6 - X2 + 7x - 12 - x2 + x + 12 = 0 - x2 + 13x + 6 = 0 - (x2 - 13x - 4) = 0 9) - = = (CD: X(x + 3) x(x+ 3)( - (3) = ()x(x + 2) The roots are : X= 6 4(x+ 3) 5x - x(x+3) - = 4 x + 12 - 5X = X2 + 3X 2 + 3x 4x + 5x 12 0 X = 2 X = - - X2 + 4x = 12 = 0 E (x + 4)(X - 2). 10) Find the roots of the equation : X+3 x +4 + = 3 x +1 X - 2 Multiply both sides by LCD : (* + 1) (X-2) (x + 1)(x - 2) (x+3 + xt) = (3) (x +) (x- (x 2)(x + 3) - + (x + 1)(x + 4) = 3(x + W w (x2 + 3x - 2x - 6) + (x2 + 4x + x + 4) = 3)x2 - 2x + x - 2) v V Y 5X & (X2 + x - 4)(X2 + 3x + 4) = 3x2 - 3x - 6 2x2 + ex - 2 - 3x2 + 3x + 6 = 0 - x2 + ax + 4 = 0  Module 1 : Lesson 6 Problems RE Solving involving Mastery Test. Translate English Phrases to mathematical phrases. 7 + X 9 + 2x = 27 2 + 34 8x = 72 X - 6 = 7 3m 1 y - 7 - y 3)x + 10) d 26. n + e = 33 AGE PROBLEMS Let X be the age of Ruby. 2X be the age of Shield. Hence , X. 2x = &8 to solve , & x2 = 49 (Extracting square roots X = 7 Note: We only need + 7 since we are talking about ages. The age of Ruby is 7 old. years The age of Shield is 2(7) = 14 years old IS 7 14 98 ? YES checking : : = Let X be the age of Kirsten X + 5 be the age of Zeke Hence , (x) (x + 5) = 50 To solve , X2 + 5x = 50 the squar - Using completing e... Using (b) (5)2 25 Factoring , = = T 2 X + 5x 50 0 25 = xz + 5x + - => (x - 5)(x + 10) = 50 + 2 X = 5 and X = -10 => (x + E)2 = 20025 4 225 = E)z = X+5 = 5+ 5 = 10 (x + I The age of Kirsten is 5 years old, => x+ 2 =2 while the Zeke is age of => 15-and- X = (5) + 5 = 10 years old = X 1 and 20 - = checking : IS 5. 10 = 50 ? YES. Since we are only talking about ages, we only need positives. X = 5 NUMBER PROBLEMS Example 1: The product of the two numbers is 48 and their sure Is 16. What are the two numbers ? Let X be the First number 16- X be the second number => X + 1 y = = 16 x y - = I Hence & X (16 x) - = 48 16X - x2 - 48 = 0 = x2 + 16X - 48 = 0 4) 8 * (x + 12)) - x + 4) = 0 = - X + 12 0 = = - x + 4 = 0 E X= 12 = X = 4 If X = 12 , then 16-X = 16-12 = 4 X = 1, 16-X 16-4 12 If then = = checking : Is 4 + 12 = 16 ? YES. Hence , the numbers 4 12 48 ? YES Is · =. are. 4 and 12 Example 2. The two consecutive integers is 288.. product of even What are the numbers ? Let X be the First even integer 16 , then X +z = 16 + 2 = 18 If X = X+2 be the next even integer Hence , x (x + 2) = 288 If X = -18 , then X + 2 = -18 + z = - 16 X 2+ 2x = 288 The numbers can be : Using completing the e. square 1) and 18 or -18 and =16 (8)2 (2) 2 = = 1 = checking : X2 + 2x + 1 289 16 18 288 ? YES = Is : = = (x + 1)2 = 289 Is - 16 - 18 = 288 ? YES => X+ 1 = [17 = X= 16 and X = -18  GEOMETRY PROBLEMS & Recall. Represent the in QE : FF. Example. 1 Let w be the length of the s width. V = ewh V = (3 w) (w) (3) V = 18m3 18 = qw2 n = 3m 9= - 18 = 0 Example 2. Let X be the width of the window. I A 315 = = Iw (x+ u)(x) A = 315 cm2 315 = x2 + 6X l = X+1 X 2 + 6x - 315 = 0 w = Y Word Problem : Problem 1 : A rectangular table has an area of 27ft" and a perimeter 24 Ft What are the dimensions of the table ? of. A= 27 F + z A = l w = 27 = ew (Eg 1) P 21+ 2w = 2w(Eg2] = 24 = 21 + p= 24 Ft Let the width w be of the table. 27 table T be the length of the. solution 1: solution 2 : (From teacher P = 21 + zw 2(2) 24 = 2w + Rew 24 = sw = R LCD : W 12 = l+w 2w2 l = 12 - w 24w = 54 + 2w2 - 24w + 54 = 0 substitute to EG1 A = (+ 2 w)(w) - 2 (w= - 12w + 27) = 0 12w 02 27 - = 2(w w2 12w + 27 = 0 - 9)(w 3) - = 0 - Hence 9 (w -9)(v - 3) , w = z or w = w = 9 w = 3 Note that the width : length is larger than. IF W S l= 27 = , then the width is is Hence , eft and the length aft. checking : Is 9 X3 = 27 Ft ? YES Is 2(a) + 2(3) = 24 ? YES , 18 + 4 = 24 Problem 2 :  1) - = T Let b be the base of the triangle · ↓ be the height of the A. height A ↓ 2 A = 52 in (2) 52 = (22 3)2() - ↓ base- b = 2h 3 - Z 104 = 22- 3h 2h 2 - 32 - 104 = 0 a S - - 208 13 8 16 - - = E I > - (2h + 13) (n z) - = 0 22= 13 - n h 8 1 = - = substitute h= 8 to b = 2h 3 - is b = 2(8) - 3 The height 8 inches - the base is 13. inches b = 16 - 3 - b 13 = 2 Let w be the widte l= ? of the T. A = 16 m2 W Let I be the length of the T , l A = ew = 2n + 4 16 = (22 + 4) 2 16 = 202 + &W 0 + 4016 = - - 2 2)(w2 + 2w z) - = 0 w2 + 2w - 8 = 0 (w + 4)(w 2) - = 0 w = - 4 w = 2 2 zw + 4 substitute w to 1 = = 2 = 2(2) + 4 r = = the length of the tarpaulin is 8. meters Unit Test Simulation : O * * O g = ( (x +8)(x - 7)  X = - 5 and X= 3 (x + 5)(X 3) - = 0 => x2 - 3x + 5x - 15 = 0 => xz + 2x - 15 = 8  Module 1: Lesson 7 Quadratic Inequalities Inequality Graphing solution Set Note : 7 [ 1) greater less hollow circle paren the sis * also uses than than parenthesis / & ⑧ [] than less than solid circle greater brackets to or equal to or equal Exampleis : 1) = 24 xxxb - 2 xQ4 Set Notation: 1-2 , 4) 2)0xXx5 * 0 x5 ⑨ ⑳ set Notation : [0 , 5]  X2 X 7 3) 3 - 27xy 3 - ⑳ w Set Notation : [-3 , 2] 4) 2 < X0 ** 2 XX0 < => 7 set Notation ( : - x , 0)U (2 , + / 5) = 17, X7, 4 * X - 1 X34 [ ⑧ ⑳ 7 Set Notation : (-0 , -1)0[4 , + +)  Solving Inequalities Graph and Find the solution set of the FF. Quadratic Equations. 1) x2 + 6x7 - 5 - 2)x2 - 2x - 87 Step. 1 Rewrite # = 2x = z = 0 the in the equation (x + 2)(x 4) 0 - = Form of ax2 + bx + c = 0 X = - 2X = 4 0 x2 + 6x + 5 = & X (x + 1((x + 5) = 0 I f 2 I I 4 ↳ 1) O 2 4 6 z 6 - - - X = -1 and X = - 5 Step. 2 IF X = - 3 Graph in the number line and ( 3)2 - - 2( 3) - - 870 use test points per region. 9 + 6 - 878 15 = z 70 7 > O ? TRUE - p 8 If X = [ 1111111 ↓ > (0) 2 - 2 (0) 8 70 - - 8 - 4 - 5 - 4 - 2 - 10 2 0-870 - 8 >O ? FALSE Test Point substitute Result Region Xz + 4x\ 5 - IF X= 5 A X= - 4 ( - 6)2 + 6) 4)4 - - 5 (592 - 2(5) - 870 FALSE 36 + ( - 34)4 - 5 25 - 10 - 878 0 < - 5 ? 25 - 1870 770 ? TRUE B X = - 2 (- 2)2 + 6) - 2)) - 5 4 - 12 - 5 TRUE - 84 - 5 ? Hence , the solution set is C X = 0 (0) 2 + 4(0)< - 5 1 - 0 , - 2)u(4 , + b) 0 + 0( - 5 0 < - 5 ? FALSE Hence , the solution set is D-5 , -1).  4) 4x2 + 5x = 67, 0 3) 2x3 - x > - 6 => 2x2 - x - 6 = 0 - 12 = - I = (2x + 3)(X 2) - = 0 => 3 X = - and X = 2 or x = - 17 < I 1111117 ⑧ ⑨ < I 11 I I I I > X IF = - 3 - 2 - 10 1 2 3 IF X = - 2 2( - 2)2 - ( 2)x6 - 2(4) + 244 10 ~ >, 90000 d 200 Problem Solving 1 - 2) - &