Question

已知a,b为常数,若f(x)=x2+4x+3,f(ax+b)=x2+10x=24,则5a-b

Answer 1

∵f(x)=x^2+4x+3

∴f(ax+b)=（ax+b）^2+4(ax+b)+3

=a^2x^2+(4a+2ab)x+b^2+4b+3

=x^2+10x+24

两个多项式相等,那么对应系数相等

∴a^2=1,4a+2ab=10

解得a=1,b=3

∴ 5a -b = 5-3 = 2