a and b are the real roots of the equation
x2 + ( k-2 )x – (k-1) = 0
if | a | = | b |, find k..
Plz help!!
{F4 A.Math} Absolute functions ~
2006-12-04 4:44 am
回答 (5)
2006-12-04 4:51 am
✔ 最佳答案
a+b=-k+2| a | = | b |
a=b or a=-b
a-b=0 or a+b=0
-k+2=0
k=2
2006-12-04 6:41 am
|a|=|b|
a=b or a=-b
when a=b
delta=0
(k-2)^2+4(k-1)=0
k^2-4k+4+4k-4=0
k^2=0
k=0
when a=-b
a+b=0
-(k-2)=0
-k+2=0
k=2
Hence, k=0 or 2
a=b or a=-b
when a=b
delta=0
(k-2)^2+4(k-1)=0
k^2-4k+4+4k-4=0
k^2=0
k=0
when a=-b
a+b=0
-(k-2)=0
-k+2=0
k=2
Hence, k=0 or 2
參考: me
2006-12-04 6:15 am
a+b = 2-k
ab=1-k
lal=lbl
lal-lbl=0
a^2-2labl+b^2=0
labl=(a^2+b^2)/2
labl=[(a+b)^2-4ab]/2
labl=[k^2-4k+4-4+4k]/2
k=+/-開方2labl
ab=1-k
lal=lbl
lal-lbl=0
a^2-2labl+b^2=0
labl=(a^2+b^2)/2
labl=[(a+b)^2-4ab]/2
labl=[k^2-4k+4-4+4k]/2
k=+/-開方2labl
2006-12-04 5:01 am
| a | = | b |
a=b or a=-b
when a=b
(k-2)^2+4(k-1)=0
k=0
when a=-b
a+b= 2-k =0
k=2
a=b or a=-b
when a=b
(k-2)^2+4(k-1)=0
k=0
when a=-b
a+b= 2-k =0
k=2
2006-12-04 4:53 am
Case 1 a=b
a+b = -(k-2)
a+b = 2-k.....(1)
2a = 2-k
ab = -(k-1)
ab = 1-k.....(2)
a^2 = 1-k
[(1)/2]^2 - (2)
[(2-k)/2]^2 = 1-k
(4-4k+k^2)/4 = 1-k
4-4k+k^2 = 4-4k
k^2 = 0
k=0
Case 2
-a = b
sum of roots
= a +b = a-a = 0
so -(k-2) = 0
2-k = 0
k = 2
so k = 0 or 2
HKCEE 1992
a+b = -(k-2)
a+b = 2-k.....(1)
2a = 2-k
ab = -(k-1)
ab = 1-k.....(2)
a^2 = 1-k
[(1)/2]^2 - (2)
[(2-k)/2]^2 = 1-k
(4-4k+k^2)/4 = 1-k
4-4k+k^2 = 4-4k
k^2 = 0
k=0
Case 2
-a = b
sum of roots
= a +b = a-a = 0
so -(k-2) = 0
2-k = 0
k = 2
so k = 0 or 2
HKCEE 1992
收錄日期: 2021-04-13 13:52:02
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