標題:
{F4 A.Math} Absolute functions ~
發問:
最佳解答:
a+b=-k+2 | a | = | b | a=b or a=-b a-b=0 or a+b=0 -k+2=0 k=2
其他解答:
|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 = 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|||||| 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|||||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
{F4 A.Math} Absolute functions ~
發問:
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a and b are the real roots of the equation x2 + ( k-2 )x – (k-1) = 0 if | a | = | b |, find k.. Plz help!!最佳解答:
a+b=-k+2 | a | = | b | a=b or a=-b a-b=0 or a+b=0 -k+2=0 k=2
其他解答:
|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 = 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|||||| 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|||||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
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