当X=1时,ax^2+2bx=5,求当x=-1时,4ax^2-8bx的值!

题目

当X=1时,ax^2+2bx=5,求当x=-1时,4ax^2-8bx的值!

答案

当x=1时,ax^2+2bx=5
x=1带入ax^2+2bx=5
a*1^2+2b*1=5
a+2b=5
x=-1带入4ax^2-8bx
4a*(-1)^2-8b*(-1)
=4a*1+8b
=4(a+2b)
=4*5
=20