10分maths

原理要用配方法得出
ax^2+bx+c=0
x^2+(b/a)x+(c/a)=0
x^2+(b/a)x+(b/2a)^2=(-c/a)+(b/2a)^2
(x+b/2a)^2=(-c/a)+(b^2/4a^2)
(x+b/2a)^2=(-4ac/4a^2)+(b^2/4a^2)
(x+b/2a)^2=(b^2-4ac)/(4a^2)
(x+b/2a)=(+-開方b^2-4ac)/2a
x=-b+-開方b^2-4ac)/2a
其實這公式不必像配方法和因式分解法要做許多步驟
example
x^2-6x+5=0(因式分解法)
(x-1)(x-5)=0
x=1 or x=5
x^2-6x+5=0(配方法)
x^2-6x=-5
x^2-6x+(6/2)^2=-5+(6/2)^2
(x-3)^2=4
x-3=+-2
x=5 or x=1
x^2-6x+5=0(二次公式)
a=1, b=-6, c=5
x=-(-6)+-開方((-6)^2-4(1)(5))/(2(1))
x=6+-開方(36-20)/2
x=6+-開方16/2
x=(6+-4)/2
x=5 or 1
由此，可見用二次公式較易掌握。

可以用來找epuation ax^2+bx+c 入面x的值
(a,b,c係常數)
而點推出來呢..........
例如有一條epuation:
x^2-2x-120=0
x^2-2x=120
x^2-2x+1=120+1
(x-1)^2=121
x-1=正負開方121
x-1=正負11
x=1+正負11
x=12 or -10

如果有一條係:
ax^2+bx+c=0
ax^2+bx=-c
x^2+(b/a)x=-(c/a)
x^2+(b/a)x+(b/2a)^2=-(c/a)+(b/2a)^2
[x+(b/2a)]^2=-(c/a)+(b/2a)^2
[x+(b/2a)]^2=-(c/a)+(b^2/4a^2)
[x+(b/2a)]^2=-(4ac/4a^2)+(b^2/4a^2)
[x+(b/2a)]^2=[(b^2-4ac)/4a^2]
x+(b/2a)=(正負開方b^2-4ac)/2a
x=[-b正負開方b^2-4ac]/2a(即是你問的那條)
(/代表分線)

希望幫到你!!!!
^_^

ax^2 + bx + c = 0
a, b, c 係實數, 解方程中既未知數 x
---------------------------------------
ax^2 + bx + c = 0
a[x^2 + (b/a)x] + c = 0
a[x^2 + (b/a)x + (b/2a)^2] - a (b/2a)^2 + c = 0
a[x + (b/2a)]^2 - a (b/2a)^2 + c = 0
a[x + (b/2a)]^2 - b^2/4a + c = 0
a[x + (b/2a)]^2 = b^2/4a - c
a[x + (b/2a)]^2 = (b^2 - 4ac)/(4a)
[x + (b/2a)]^2 = (b^2 - 4ac)/(4a^2)
x + (b/2a) = sqrt[ (b^2 - 4ac)/(4a^2) ] or x + (b/2a) = -sqrt[ (b^2 - 4ac)/(4a^2) ]
x + (b/2a) = sqrt (b^2 - 4ac)/(2a) or x + (b/2a) = -sqrt(b^2 - 4ac)/(2a)
x = [-b + sqrt (b^2 - 4ac)]/(2a) or x = [-b - sqrt (b^2 - 4ac)]/(2a)
有兩個值


2007-10-03 18:53:33 補充：
慢左, 不過請持支我啊,投我一票

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