Question

已知a.b为常数.若f(x)=X2+4x+3,f(ax+b)=X2+10x+24.求5a-b的值.

Answer 1

解:令x=-5
f(-5a+b)=(-5)^2+10x(-5)+24=-1
f(x)=x^2+4x+3=-1
∴f=-2
∴f(-5a+b)=f(-2)=-1
∴-5a+b=-2
∴-(5a-b)=-2
∴5a-b=2

Answer 2

f(ax+b)=(ax+b)^2+4(ax+b)+3=a^2x^2+(2ab+4a)x+(b^2+4b+3)=x^2+10x+24
a^2=1
2ab+4a=10
b^2+4b+3=24
a=1,b=3,5a-b=5-3=2
a=-1,b=-7,5a-b=-5+7=2

Answer 3

设ax+b=t, x=1/a(t-b),
f(t)=1/a(t-b)2+10*1/a(t-b)+24=1/a(t2+b2-2tb)+10/a*t-10/a*b+24
和f(x)=X2+4x+3对照 x的平方系数相等，一次系数也相等，常熟也相等
就把a和b求出来了