Question

已知a，b为常数，若f（x）=x平方+4x+3，f（ax+b）=x平方+10x+24，则5a+b=多少

Answer 1

f(ax+b)=x²+10x+24
令x=5
则f(5a+b)=25+50+24=99

令f(x)=x²+4x+3=99
x²+4x-96=(x+12)(x-8)=0
x=-12,x=8
所以5a+b=-12或8

Answer 2

f(x)= x^2+4x+3
f(ax+b) = x^2+10x+24
(ax+b)^2 +4(ax+b) +3 =x^2+10x+24
a^2. x^2 + (2ab+4a)x + b^2+4b+3 =x^2+10x+24
=>
a^2=1 (1)
2ab+4a =10 (2)
b^2+4b+3=24 (3)

from (3)
b^2+4b+3=24
b^2+4b-21=0
(b+7)(b-3)=0
b=3 or -7

from (1)
a^2 =1
a=1 or -1

case 1: a=1 , b=3
from (2)
2ab+4a
=2(1)(3) +4(1)
=10
=RS
5a+b = 5(1)+3 =8

case 2: a=1 , b=-7
from (2)
2ab+4a
=2(1)(-7) +4(1)
=-10
≠RS

case 3: a=-1 , b=3
from (2)
2ab+4a
=2(-1)(3) +4(1)
=-2
≠RS

case 4: a=-1 , b=-7
from (2)
2ab+4a
=2(-1)(-7) +4(-1)
=10
=RS
=>
5a+b = 5(-1) -7 = -12

=>
5a+b = 8 or -12

Answer 3

f（ax+b）＝（ax＋b）2+4（ax＋b）＋3
＝a2x2 ＋（2ab＋4a）＋b2+4b＋3

其中常数项
b2+4b＋3 ＝ 24
解出 b＝3 或 b＝－7
a2=1
则a ＝1 或 a ＝ －1
2ab ＋4a ＝ 10
得出a＝1时，b＝3
a＝－1，b＝－7
所以，当a＝1时
5a＋b＝5+3=8
当a＝－1
5a＋b＝－5－7=－12

Answer 4

f(ax+b)=(ax+b)2+4(ax+b)+3=x2+10x+24整理得a2x2+(2ab+4a)x+b2+4b+3=x2+10x+24由左右未知数前系数相等得a2=1,2ab+4a=10,b2+4b+3=10解得a=1，b=3或a=-1，b=-7所以5a-b=2