a)若二次方程px^2+(5p)x+4=0有二重根,求b的值
b)由a小題的結果,解方程px^2+(5p)x+4=0
知道a的答案係16/25
b答案係-5/2
但唔知點計出黍.....
二次方程px^2+(5p)x+4=0有二重根
2007-11-30 6:49 am
回答 (4)
2007-11-30 6:58 am
✔ 最佳答案
a) 若二次方程px^2+(5p)x+4=0有二重根, Delta = 0,(5p)^2 - 4 ( p )( 4 ) = 0
25p^2 - 16p = 0
p ( 25 p - 16 ) = 0
p = 0 ( 捨去) 或 16 / 25
b) ( 16 / 25 )x^2 + 5 ( 16 / 25 )x + 4 = 0
16x^2 + 80x + 100 = 0
( 4x + 10 )^2 = 0
x = -5 / 2
2007-11-29 23:06:47 補充：
a)當p=0時,px^2+(5p)x+4=0會變成4=0, 而這是不成立的, 所以這答案要捨去。
參考: My Maths Knowledge
2007-11-30 10:09 am
a) Delta = 0
Delta = (5p)^2 - 4 ( p )( 4 ) = 0
25p^2 - 16p = 0
p ( 25 p - 16 ) = 0
p = 0 ( rejected) or 16 / 25
b) ( 16 / 25 )x^2 + 5 ( 16 / 25 )x + 4 = 0
16x^2 + 80x + 100 = 0
( 4x + 10 )^2 = 0
x = - 5 / 2
So easy!!
Delta = (5p)^2 - 4 ( p )( 4 ) = 0
25p^2 - 16p = 0
p ( 25 p - 16 ) = 0
p = 0 ( rejected) or 16 / 25
b) ( 16 / 25 )x^2 + 5 ( 16 / 25 )x + 4 = 0
16x^2 + 80x + 100 = 0
( 4x + 10 )^2 = 0
x = - 5 / 2
So easy!!
參考: bb
2007-11-30 7:02 am
二重根 係唔係double root??
b^2-4ac=0
(5p)^2-4px4=0
p=0or16/25
p=0 rejected
16/25x^2+16/5x+4=0
x=-5/2
b^2-4ac=0
(5p)^2-4px4=0
p=0or16/25
p=0 rejected
16/25x^2+16/5x+4=0
x=-5/2
2007-11-30 7:02 am
二重根=有2個一樣的solutions
delta=0
(5p)^2-4(p)(4)=0
25p^2-16p=0
(25p-16)p=0
p=16/25 or 0 (rejected)
16/25x^2+5(16/25)x+4=0
4x^2+20x+25=0
(2x+5)^2=0
x= - 5/2
delta=0
(5p)^2-4(p)(4)=0
25p^2-16p=0
(25p-16)p=0
p=16/25 or 0 (rejected)
16/25x^2+5(16/25)x+4=0
4x^2+20x+25=0
(2x+5)^2=0
x= - 5/2
參考: 自己
收錄日期: 2021-04-24 00:14:38
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