已知:a2+4a+1=0,且a4+ma2+12a3+ma2+2a=3,求m的值.
问题描述:
已知:a2+4a+1=0,且
=3,求m的值.
a4+ma2+1 2a3+ma2+2a
答
∵a2+4a+1=0,∴a2+1=-4a,∴(a2+1)2=16a2,∴a4+2a2+1=16a2,即a4+1=14a2,∵a4+ma2+12a3+ma2+2a=3,∴14a2+ma22a(a2+1)+ma2=3,整理得14a2+ma2=-24a2+3ma2,∴(38-2m)a2=0,∵a≠0,∴38-2m=0,∴m=19....