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因为4x^2+mx+1/2m-4=0两个实数根为x1,x2,
所以
4x1^2+mx1+1/2m-4=0
4x2^2+mx2+1/2m-4=0,
x1+x2=-m/4,x1x2=(1/2m-4)/4
从而
6x1²+mx1+1/2m+2x2²-8
=4x1²+mx1+1/2m-4+2x1²+2x2²-4
=2x1²+2x2²-4
=2(x1+x2)²-4x1x2-4
=2(-m/4)²-4*(1/2m-4)/4-4
=0,
即
m²/8-m/2=0
m(m-4)=0
m=0或m=4
1年前
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