Question

若sina,cosa是方程4xx+2mx+m=0的两根，则m的值为？

Answer 1

韦达定理
sina+cosa=-m/2
sinacosa=m/4
因为sin²a+cos²a=1
则(sina+coa)²-2sinacosa=1
m²/4-m/2=1
m²-2m-4=0
m=1±√5

判别式大于等于0
4m²-16m>=0
m<=0,m>=4
所以m=1-√5

Answer 2

解：
方程有实根，判别式≥0
(2m)^2-16m≥0
m(m-4)≥0
m≥4或m≤0
由韦达定理，得
sina+cosa=-m/2
sinacosa=m/4
(sina)^2+(cosa)^2=1
(sina+cosa)^2-2sinacosa=1
m^2/4-m/2=1
整理，得
m^2-2m-4=0
(m-1)^2=5
m=1+√5(<4,舍去)或m=1-√5

m=1-√5

Answer 3

sina,cosa是方程4xx+2mx+m=0的两根
sina+cosa=-m/4
sina*cosa=m/4
(sina+cosa)^2=sina^2+cosa^2+2sina*cosa=1+m/2=-m/4
m=-4/3

Answer 4

(sina)^2 + (cosa)^2 = 1

x1^2 + x2^2 = 1

x1^2 + x2^2 = (x1 + x2)^2 - 2x1x2 = (-m/2)^2 - 2(m/4) = m^2/4 - m/2 = 1

m1 = 1+√5 (舍去)
m2 = 1- √5

Answer 5

sina+cosa=-m/2,(sina+cosa)^2=m2/2,1+2sinacosa=m2/2
sinacosa=m/4
所以1+m/4=m2/2
m=√33/16+1/46或-√33/16+1/4