1.若关於x的一元2次方程x^2sin(a)+2x(sin(a)+2)+sina+12=0有实数根,试确定锐角a的范围2.解方程x^4-7x^3+5x^2+7x-6=0小弟不才,希望尽可能的详细.

问题描述:

1.若关於x的一元2次方程x^2sin(a)+2x(sin(a)+2)+sina+12=0有实数根,试确定锐角a的范围
2.解方程x^4-7x^3+5x^2+7x-6=0
小弟不才,希望尽可能的详细.

1.b^2-4ac
=4(sina+2)^2-4sina(sina+12)
=4(sina^2+4sina+4)-4sina^2-48sina
=-32sina+16>0
即:sina 所以:0°
2.解方程x^4-7x^3+5x^2+7x-6=0
x^2(x^2-7x+6)-(x^2-7x+6)=0
(x^2-1)(x-1)(x-6)=0
(x-1)^2(x+1)(x-6)=0
x1=1
x2=-1
x3=6

1、∵方程有实数根
∴判别式△≥0,即4(sina+2)²-4sina(sina+12)≥0
解得sina≤1/2,∵a是锐角
∴0<a≤π/6
2、x^4-7x^3+5x^2+7x-6=0
x^4-7x^3+6x^2-x^2+7x-6=0
x^2(x^2-7x+6)-(x^2-7x+6)=0
(x^2-1)(x-1)(x-6)=0
(x-1)^2(x+1)(x-6)=0
解得x1=1,x2=-1,x3=6

1 因为有实数根所以[2(sina+2)]^2-4sina*(sina+12)>0或=0
解这个不等式有 sina

1、判别式:4(sina+2)^2-4sina(sina+12)>0
4(sin^2a+4sina+4)-4sin^2a-48sina
=14sina+16-48sina=16-44sina>=0
sina02、x^4-7x^3+6x^2-x^2+7x-6
=x^2(x^2-7x+6)-(x^2-7x+6)
=(x^2-1)(x-1)(x-6)
=(x-1)^2(x+1)(x-6)=0
x1=1,x2=-1,x3=6