Question

已知关于x的一元二次方程4x2+mx+1/2m-4=0,

若方程的两个实数根为x1,x2，且满足6x1²+mx1+1/2m+2x2²-8=0,求m

Answer 1

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

Answer 2

— —用韦达定理 X1乘以X2=负的a分之b X1+X2等于a分之c

Answer 3

因为4x^2+mx+1/2m-4=0两个实数根为x1,x2，,
△=M²-4[4*(1/2 M-4)]≥ 0
M²-8M+64≥0
(M-4)²+48≥0 恒成立

4x1^2+mx1+1/2m-4=0
x1+x2=-m/4,x1x2=(1/2m-4)/4
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
m²/8-m/2=0
m(m-4)=0
m=0或m=4