 # OBJ: To solve a quadratic equation by factoring

## Presentation on theme: "OBJ: To solve a quadratic equation by factoring"— Presentation transcript:

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 3 5 2 2 1 3 -3 3 2 +10 (5c – 3)(c + 2) = 0
-3 3 2 +10 (5c – 3)(c + 2) = 0 c = 3/5, -2

EX:  7t = 20 – 3 t 2 3 t 2 + 7t – 20 = 0 -5 5 4 +12 (3t – 5)(t + 4) = 0 t = 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

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

EX:  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 = 12 ± 6√3 = 12 ± 6√3 18 = 6(2 ± √3) 3 = 2 ± √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 = 0 x = -(-6)±√(-6)2 – 4(2)(7)
2(2) = 6 ± √36 – 56 4 = 6 ± √-20 = 6 ± 2i√5 = 2(3 ± i√5) 4 2 = 3 ± i√5

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