3.5 Quadratic Equations OBJ:To solve a quadratic equation by factoring

DEF:  Standard form of a quadratic equation ax 2 + bx + c = 0 NOTE:  Each equation contains a polynomial of the second degree.

DEF:  Zero – product property If mn = 0, then m = 0 or n = 0 or both = 0 NOTE:  Solve some quadratic equations by: Writing equation in standard form Factoring Setting each factor equal to 0

P 68 EX 1:  3 c 2 – 10c – 8 = (3c + 2)(c – 4) = 0 c = - 2/3, 4

P 68 EX 2:  5x = 6 – x 2 x 2 + 5x – 6 = 0 (x + 6)(x – 1) = 0 x = - 6, 1

P 69 EX 3:  – 7 x 2 = 21x 7 x x = 0 7x (x + 3) = 0 x = 0, - 3

P 69 EX 3:  25 = 9 n 2 9 n 2 – 25 = 0 (3n – 5)(3n + 5) = 0 n = ± 5/3

EX 5:  5 c 2 + 7c – 6 = (5c – 3)(c + 2) = 0 c = 3/5, -2

EX 6:  7t = 20 – 3 t 2 3 t 2 + 7t – 20 = (3t – 5)(t + 4) = 0 t = 5/3, -4

EX 7:  36 = 25 x 2 25 x 2 – 36 = 0 (5x – 6)(5x + 6) = 0 x = ± 6/5

EX 8:  –2 x 2 = 5x 2 x 2 + 5x = 0 2x(x + 5) = 0 x = 0, - 5

P69 EX 5:  3 n 2 – 15n + 18 = 0 3 (n 2 – 5n + 6) = 0 3 (n – 3)(n – 2) = 0 n = 3, 2

EX 10:  7 n n – 56 = 0 7 (n 2 + 2n – 8) = 0 7 (n + 4)(n – 2) = 0 n = - 4, 2

P69 EX 4:  x 4 – 13 x = 0 (x 2 – 9)(x 2 – 4) = 0 (x – 3)(x + 3)(x – 2)(x + 2) =0 x = ± 3, ± 2

EX 12:  y 4 – 5 y = 0 (y 2 – 4)(y 2 – 1) = 0 (y – 2)(y + 2)(y – 1)(y + 1) = 0 Y = ± 2, ± 1

EX 13:  y 4 – 10 y = 0 (y 2 – 9)(y 2 – 1) = 0 (y – 3)(y + 3)(y – 1)(y + 1) = 0 Y = ± 3, ± 1

EX 14:  y 4 = 20 – y 2 y 4 + y 2 – 20 = 0 (y 2 + 5)(y 2 – 4) = 0 (y 2 + 5)(y – 2)(y + 2) = 0 Y = ± i√ 5, ± 2

EX 15:  y 4 = 12 + y 2 y 4 – y 2 – 12 = 0 (y 2 – 4)(y 2 + 3) = 0 (y – 2)(y + 2)(y 2 + 3) = 0 Y = ± 2, ± i√ 3

6.1 Square Roots OBJ:  To solve a quadratic equation by using the definition of square root DEF:  Square root If x 2 = k, then x = ±√k, for k ≥ 0

P 139 EX 1:  x = 15 x 2 = 10 x = ± √10

P 139 EX 1:  3 y 2 = 75 y 2 = 25 y = ± 5

EX 3:  6 y 2 – 20 = 8 – y 2 7y 2 = 28 y 2 = 4 y = ± 2

EX 4:  3 n = 7 n 2 – = 4n 2 11 = n 2 ±√11 = n

7.3 The Quadratic Formula OBJ:  To solve a quadratic equation by using the quadratic formula DEF:  The quadratic formula x = -b ± √b 2 – 4ac 2a

P169 EX 1:  3 x 2 + 5x – 4 = 0 x = -5 ± √5 2 – 4(3)(-4) 2(3) = -5 ± √ x = -5 ± √73 6

P170 EX 2 :  4 x 2 = x 4 x 2 – 4 x – 11 = 0 x = -(-4)±√ (-4) 2 – 4(4)(-11 ) 2(4) x = 4 ± √ = 4 ± √192 8 = 4 ± 8√3 8 = 4(1 ± 2√ = 1 ± 2√3 2

P 170 EX 3:  5 x 2 – 9x = 0 x(5x – 9) = 0 x = 0, 9/5

P 170 EX 3:  y 2 – 150 = 0 y 2 = 150 y = ± √150 = ± 5√6

EX 5:  4 x 2 – 7x + 2 = 0 x = -(-7) ± √(-7) 2 – 4(4)(2) 2(4) = 7 ± √49 – 32 8 = 7 ± √17 8

EX 6:  9 x 2 = 12x – 1 9 x 2 – 12x + 1 = 0 x = -(-12)±√ (-12) 2 – 4(9)(1) 2(9) = 12 ± √144 – = 12 ± √ = 12 ± 6√3 18 = 12 ± 6√3 18 = 6(2 ± √3) 18 3 = 2 ± √3 3

EX 7:  6 x 2 + 5x = 0 x(6x + 5) = 0 x = 0, -5/6

EX 8:  72 – x 2 = 0 x 2 = 72 x = ± 6√2

8.3 Equations With Imaginary Number Solutions OBJ: To solve an equation whose solutions are imaginary

P 193 EX 1:  3 x = 4x 3 x 2 – 4x + 2 = 0 x = -(-4)±√(-4) 2 – 4(3)(2) 2(3) = 4 ± √16 – 24 6 = 4 ± √-8 6 = 4 ± 2i√2 6 = 2(2 ± i√2 6 3 = 2 ± i√2 3

P193 EX 2:  2 x x 2 – 20 = 0 (2 x 4 – 5 )(x 2 + 4) = 0 x 2 = 5/2 or -4 x = ±√10/2 or ± 2i

EX 3:  2 x = 6x 2 x 2 – 6x + 7 = 0 x = -(-6)±√(-6) 2 – 4(2)(7) 2(2) = 6 ± √36 – 56 4 = 6 ± √-20 4 x = 6 ± 2i√5 4 = 2(3 ± i√5) = 3 ± i√5 2

EX 4:  27 – 6 y 2 = y 4 y y 2 – 27 = 0 (y 2 + 9)(y 2 – 3) = 0 y = ± 3i, ± √3