Question

已知sinα，cosα是方程8x^2+6mx+1=0的两个实数根，求实数m的值

Answer 1

sinα+cosα=-6m/8 sinα*cosα=1/8 (sinα+cosα)^2=36m^2/64 1+1/4=36m^2/64 m^2=20/9
m=±2√5/3 判别式=36m^2-32>=0 m^2>=8/9
实数m的值±2√5/3

Answer 2

根据根与系数的关系可知：
sinα+cosα=-6m/8=-3m/4
sinα*cosα=1/8
(sinα+cosα)^2=1+2sinα*cosα)
9m^2/16=1+1/4=5/4
m^2=20/9
m=±2√5/3

Answer 3

sin²a+cos²a=1
sin²a+cos²a=（sina+cosa）²-2sinacosa
sina+cosa= -b/a
sinacosa=c/a 韦达定理
代入得
9m²/16 -1/4=1
9m²/16=5/4
m²=20/9
m=±2√5/3

然后要注意讨论一下方程有实数根 △≥0
△= 36m²-32≥0
m²≥8/9
m≤-2√2/3 或m≥2√2/3

综上
m=±2√5/3

Answer 4

sinα +cosα=-6m/8 =-3m/4 ;sinα cosα=1/8
=>(sinα +cosα)^2=1-2sinαcosα =>9m^2/4 =1- 1/4 => 9m^2=3 =>m^2= 1/3
=> m=(√3)/3 or (-√3)/3

Answer 5

根与系数的关系
sina+cosa=-(3/4)m
sina*cosa=1/8
将第一个式子平方，得m的值，注意验证

Answer 6

sin²α+cos²α=1
因此两根平方和为1
根据韦达定理,X1+X2=-6m/8=-3m/4,x1x2=1/8
X1²+X2²
=(X1+X2)²-2X1X2
=9m²/16-2/8
=1
9m²/16=5/4
9m²=20
m²=20/9
m=±2√5/3