Download presentation

Presentation is loading. Please wait.

Published byLaurence Golden Modified over 4 years ago

1
3.5 Quadratic Equations OBJ:To solve a quadratic equation by factoring

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

3
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

4
P 68 EX 1: 3 c 2 – 10c – 8 = 0 3 2 3 4 1 4 1 2 +2 3 2 1414 -12 (3c + 2)(c – 4) = 0 c = - 2/3, 4

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

6
P 69 EX 3: – 7 x 2 = 21x 7 x 2 + 21x = 0 7x (x + 3) = 0 x = 0, - 3

7
P 69 EX 3: 25 = 9 n 2 9 n 2 – 25 = 0 (3n – 5)(3n + 5) = 0 n = ± 5/3

8
EX 5: 5 c 2 + 7c – 6 = 0 53 5 2 12 1 3 -3 5353 1212 +10 (5c – 3)(c + 2) = 0 c = 3/5, -2

9
EX 6: 7t = 20 – 3 t 2 3 t 2 + 7t – 20 = 0 34 3 5 15 1 4 -5 3535 1414 +12 (3t – 5)(t + 4) = 0 t = 5/3, -4

10
EX 7: 36 = 25 x 2 25 x 2 – 36 = 0 (5x – 6)(5x + 6) = 0 x = ± 6/5

11
EX 8: –2 x 2 = 5x 2 x 2 + 5x = 0 2x(x + 5) = 0 x = 0, - 5

12
P69 EX 5: 3 n 2 – 15n + 18 = 0 3 (n 2 – 5n + 6) = 0 3 (n – 3)(n – 2) = 0 n = 3, 2

13
EX 10: 7 n 2 + 14n – 56 = 0 7 (n 2 + 2n – 8) = 0 7 (n + 4)(n – 2) = 0 n = - 4, 2

14
P69 EX 4: x 4 – 13 x 2 + 36 = 0 (x 2 – 9)(x 2 – 4) = 0 (x – 3)(x + 3)(x – 2)(x + 2) =0 x = ± 3, ± 2

15
EX 12: y 4 – 5 y 2 + 4 = 0 (y 2 – 4)(y 2 – 1) = 0 (y – 2)(y + 2)(y – 1)(y + 1) = 0 Y = ± 2, ± 1

16
EX 13: y 4 – 10 y 2 + 9 = 0 (y 2 – 9)(y 2 – 1) = 0 (y – 3)(y + 3)(y – 1)(y + 1) = 0 Y = ± 3, ± 1

17
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

18
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

19
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

20
P 139 EX 1: x 2 + 5 = 15 x 2 = 10 x = ± √10

21
P 139 EX 1: 3 y 2 = 75 y 2 = 25 y = ± 5

22
EX 3: 6 y 2 – 20 = 8 – y 2 7y 2 = 28 y 2 = 4 y = ± 2

23
EX 4: 3 n 2 + 9 = 7 n 2 – 35 44 = 4n 2 11 = n 2 ±√11 = n

24
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

25
P169 EX 1: 3 x 2 + 5x – 4 = 0 x = -5 ± √5 2 – 4(3)(-4) 2(3) = -5 ± √25 + 48 6 x = -5 ± √73 6

26
P170 EX 2 : 4 x 2 = 11 + 4x 4 x 2 – 4 x – 11 = 0 x = -(-4)±√ (-4) 2 – 4(4)(-11 ) 2(4) x = 4 ± √16 + 176 8 = 4 ± √192 8 = 4 ± 8√3 8 = 4(1 ± 2√3 8 8 2 = 1 ± 2√3 2

27
P 170 EX 3: 5 x 2 – 9x = 0 x(5x – 9) = 0 x = 0, 9/5

28
P 170 EX 3: y 2 – 150 = 0 y 2 = 150 y = ± √150 = ± 5√6

29
EX 5: 4 x 2 – 7x + 2 = 0 x = -(-7) ± √(-7) 2 – 4(4)(2) 2(4) = 7 ± √49 – 32 8 = 7 ± √17 8

30
EX 6: 9 x 2 = 12x – 1 9 x 2 – 12x + 1 = 0 x = -(-12)±√ (-12) 2 – 4(9)(1) 2(9) = 12 ± √144 – 36 18 = 12 ± √108 18 = 12 ± 6√3 18 = 12 ± 6√3 18 = 6(2 ± √3) 18 3 = 2 ± √3 3

31
EX 7: 6 x 2 + 5x = 0 x(6x + 5) = 0 x = 0, -5/6

32
EX 8: 72 – x 2 = 0 x 2 = 72 x = ± 6√2

33
8.3 Equations With Imaginary Number Solutions OBJ: To solve an equation whose solutions are imaginary

34
P 193 EX 1: 3 x 2 + 2 = 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

35
P193 EX 2: 2 x 4 + 3 x 2 – 20 = 0 (2 x 4 – 5 )(x 2 + 4) = 0 x 2 = 5/2 or -4 x = ±√10/2 or ± 2i

36
EX 3: 2 x 2 + 7 = 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) 4 4 2 = 3 ± i√5 2

37
EX 4: 27 – 6 y 2 = y 4 y 4 + 6 y 2 – 27 = 0 (y 2 + 9)(y 2 – 3) = 0 y = ± 3i, ± √3

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google