[[[20點]]一元二次方程

1.
(x-3)(x+4)=(x-3)(2x-1)
(x-3)(x+4)-(x-3)(2x-1)=0
(x-3)[(x+4)-(2x-1)]=0
(x-3)(-x+5)=0
x=3 or 5

2.
3x^2-2x+11=0

x=-(-2) +/- √[(-2)^2-4(3)(11)]/2(3)

x=[2+/- √(-128)]/6

x=(2+/- √-128)/6

Since √-128 is no real integers,so there is no real roots.

3.
3x-2x^2-4=0
-2x^2+3x-4=0
2x^2-3x+4=0

x=-(-3) +/- √[(-3)^2-4(2)(4)]/2(2)

x=[3+/- √(-23)]/4

x=(3+/- √-23)/4

Since √-23 is no real integers,so there is no real roots.

1.x^2-x-12=2x^2-7x+3
-x^2+6x-15=0
x^2-6x+15=0
x^2-6x+9-9=-15
(x-3)^2=-6
(x-3)=-6
x=-3


2.3x^2-2x+11=0
no 解


3.3x-2x^2-4=0


ｎｏ解

1) (x-3)(x+4)=(x-3)(2x-1)
(x-3)(x+4) - (x-3)(2x-1) = 0
(x-3)[(x+4) - (2x-1)] = 0 <--- Taking common factor
(x-3)(-x+5) = 0
(x-3)(x-5) = 0
∴ x = 3 or 5
2) 3x^2-2x+11=0
∵△=(-2)^2 - 4(3)(11) = -128 < 0
∴There is no real solution.
3) 3x-2x^2-4=0
-2x^2 + 3x - 4 = 0
∵△=(3)^2 - 4(-2)(-4) = -23 < 0
∴There is no real solution.

1.
(x-3)(x+4)=(x-3)(2x-1)
(x-3)(x+4-2x+1)=0
(x-3)(5-x) =0
x=3 or -5

2.
3x^2-2x+11=0
3x-2x +11 =0
x={ -(-2) √[(-2)-4(3)(11)]} (1/6)
∵[(-2)-4(3)(11)] <0
∴quadratic equation3x^2-2x+11=0 have no real roots
3.
3x-2x^2-4=0
∵ [(-2) -4(3)(-4)] <0
∴quadratic equation3x-2x^2-4=0 have no real roots

1. (x-3)(x+4)=(x-3)(2x-1)
x^2 + x - 12 = 2x^2 - 7x + 3
x^2 - 8x + 15 = 0
(x-3)(x-5) = 0
x=3 or x = 5

2. 3x^2-2x+11=0
&Delta; = b^2 - 4ac
= 4 - 4‧3‧11
&lt; 0

there is no solution

3. 3x-2x^2-4=0
&Delta; = b^2 - 4ac
= 9 - 4‧(-2) ‧(-4)
= 9 - 32
&lt; 0

there is no solution

2009-11-01 13:06:55 補充：
Delta 就是那個三角形的符號/discriminant
< 就是那個less than 的符號 (怎麼顯示不了﹖)

1.x=3 or x=5
2.no solution
3.no solution