几道简单的一元二次方程(1)x^2-9x+8=0x1=8x2=1(2)x^2+6x-27=0x1=3x2=-9(3)x^2-2x-80=0x1=-8x2=10(4)x^2+10x-200=0x1=-20x2=10(5)x^2-20x+96=0x1=12x2=8(6)x^2+23x+76=0x1=-19x2=-4(7)x^2-25x+154=0x1=14x2=11(8)x^2-12x-108=0x1=-6x2

Question

几道简单的一元二次方程
(1)x^2-9x+8=0 x1=8 x2=1
(2)x^2+6x-27=0 x1=3 x2=-9
(3)x^2-2x-80=0 x1=-8 x2=10
(4)x^2+10x-200=0 x1=-20 x2=10
(5)x^2-20x+96=0 x1=12 x2=8
(6)x^2+23x+76=0 x1=-19 x2=-4
(7)x^2-25x+154=0 x1=14 x2=11
(8)x^2-12x-108=0 x1=-6 x2=18
(9)x^2+4x-252=0 x1=14 x2=-18
(10)x^2-11x-102=0 x1=17 x2=-6

Answer 1

十字交叉法来做
如
(1)x^2-9x+8=0 答案：x1=8 x2=1
把x^2前面的系数取出是1,把常数项也取出,为8
找两个数,之积为1.设这两个数为A,B
再找两个数,之积为8,设这两个数为C,D
使得A*C+B*D=-9(也就是x前面的系数),如果成立则为(Ax-D)*(Bx-C)=0
x1=D/A x2=C/B