已知方程2x^2+mx-2m+1=0的两实数根的平方和是29/4,求m的值

x1x2=(1-2m)/2
x1+x2=-m/2
x1²+x2²=29/4
=(x1+x2)²-2x1x2
=m²/4-(1-2m)
=m²/4+2m-1
所以
m²+8m-4=29
m²+8m-33=0
(m-3)(m+11)=0
m=3或m=-11
又因为是两实根,所以
Δ=m²-4×2×(1-2m)>=0
m²+16m-8>=0
代入验证,发现m=3