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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 - PowerPoint PPT Presentation
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3.5 Quadratic Equations
OBJ: To solve a quadratic equation by factoring
DEF: Standard form of a quadratic equation
ax2 + 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 c2 – 10c – 8 = 0
3 2 3 4 1 4 1 2 +23 21 4 -12(3c + 2)(c – 4) = 0c = - 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 2 + 21x = 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 = 0
5 3 5 2
6 2 1 3
-3
5 3
6 2
+10
(5c – 3)(c + 2) = 0
c = 3/5, -2
EX 6: 7t = 20 – 3 t 2
3 t 2 + 7t – 20 = 03 4 3 54 5 1 4 -53 54 4 +12 (3t – 5)(t + 4) = 0t = 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 2 + 14n – 56 = 0
7 (n 2 + 2n – 8) = 0
7 (n + 4)(n – 2) = 0
n = - 4, 2
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
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
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
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 2 + 5 = 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 2 + 9 = 7 n 2 – 35
44 = 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 ± √b2 – 4ac
2a
P169 EX 1: 3 x 2 + 5x – 4 = 0
x = -5 ± √52 – 4(3)(-4)
2(3)
= -5 ± √25 + 48
6
x = -5 ± √73
6
P170 EX 2 : 4 x 2 = 11 + 4x
4 x 2 – 4 x – 11 = 0x = -(-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= 4(1 ± 2√3 8 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 = 0x = -(-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
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 2 + 2 = 4x
3 x 2 – 4x + 2 = 0x = -(-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 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
EX 3: 2 x 2 + 7 = 6x
2 x 2 – 6x + 7 = 0x = -(-6)±√(-6)2 – 4(2)(7) 2(2) = 6 ± √36 – 56 4 = 6 ± √-20 4x = 6 ± 2i√5 4
= 2(3 ± i√5) 4 = 2(3 ± i√5) 4 2 = 3 ± i√5 2
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
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