3.5 quadratic equations obj:to solve a quadratic equation by factoring
TRANSCRIPT
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
EX: 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: 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: 36 = 25 x 2
25 x 2 – 36 = 0
(5x – 6)(5x + 6) = 0
x = ± 6/5
EX: –2 x 2 = 5x
2 x 2 + 5x = 0
x(2x + 5) = 0
x = 0, - 5/2
EX: 7 n 2 + 14n – 56 = 0
7 (n 2 + 2n – 8) = 0
7 (n + 4)(n – 2) = 0
n = - 4, 2
EX: 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: 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: 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: 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
EX: 6 y 2 – 20 = 8 – y 2
7y 2 = 28
y 2 = 4
y = ± 2
EX: 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
EX: 4 x 2 – 7x + 2 = 0
x = -(-7) ± √(-7)2 – 4(4)(2)
2(4)
= 7 ± √49 – 32
8
= 7 ± √17
8
EX: 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: 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
EX: 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 4 = 6 ± 2i√5 4
= 2(3 ± i√5) 4 = 2(3 ± i√5) 4 2 = 3 ± i√5 2
EX: 27 – 6 y 2 = y 4
y 4 + 6 y 2 – 27 = 0
(y 2 + 9)(y 2 – 3) = 0
y = ± 3i, ± √3