Question

设x1、x2是方程是2x^2+4x-3=0的两个根,利用根与系数的关系,求下列各式的值:

设x1、x2是方程是2x^2+4x-3=0的两个根,利用根与系数的关系,求下列各式的值:
1、(x1+1)(x2+1) 2、x1^2x2+x1x2^2 3、x2/x1+x1/x2 4、(x1-x2)^2
要有过程，清楚点锕~~~，速度

Answer 1

(x1+1)(x2+1)

= x1x2 + x1 + x2 + 1

= (-3/2) + (-4/2) + 1

= -5/2

x1^2x2+x1x2^2

= (x1x2)(x1+x2)

= (-3/2)(-4/2)

= 3

x2/x1+x1/x2

= (x1^2+x2^2)/(x1x2)

= (x1^2+2x1x2+x2^2-2x1x2)/(-3/2)

= ((-4/2)^2-2(-3/2))/(-3/2)

= -14/3

(x1-x2)^2

= (x1+x2)^2 - 4x1x2

= (-4/2)^2 - 4(-3/2)

= 4 + 6

= 10

Answer 2

x1+x2＝-2
x1*x2＝-3/2

1、(x1+1)(x2+1)
＝x1*x2+(x1+x2)+1
＝-3/2-2+1
＝-5/2

2、x1^2x2+x1x2^2
＝x1*x2（x1+x2）
＝(-3/2)*(-2)
＝3

3、x2/x1+x1/x2
＝(x2^2+x1^2)/(x1*x2)
＝[(x1+x2)^2-2x1x2]/(x1*x2)
＝[(-2)^2+3]/(-3/2)
＝-14/3

4、（x1-x2)^2
＝(X1+X2)^2-4x1x2
＝4+6
＝10

Answer 3

x1+x2=-4/2=-2
x1*x2=-3/2
1、(x1+1)(x2+1)=x1*x2+(x1+x2)+1=-2+(-3/2)+1=-5/2
2、x1^2x2+x1x2^2=x1*x2*(x1+x2)=-2*(-3/2)=3
3、x2/x1+x1/x2
=(x1^2+x2^2)/(x1*x2)
=[(x1+x2)^2-2x1*x2]/(x1*x2)
=(x1+x2)^2/(x1*x2) -2
=(-2)^2/(-3/2)-2
=-14/3
4、(x1-x2)^2=(x1+x2)^2-4x1*x2=(-2)^2-4*(-3/2)=10

Answer 4

x1+x2=-2 x1x2=-3/2
1)(x1+1)(x2+1)=x1x2+x1+x2+1=-2.5
2)=x1x2(x1+x2)=3
3)=「(x1+x2)²-2x1x20」/x1x2=-3.5
4)=(x1+x2)²-4x1x2=10