Question

方程4x^2-7x-3=0两根x1,x2,则(1)1/x1+1/x2=__

方程4x^2-7x-3=0两根x1,x2,则(1)1/x1+1/x2=__,(2) |x1-x2|=__,(3) (x1-3)(x2-3)=__,(4) x1x2^2+x1^2x2=__.
在线等。

Answer 1

x1+x2 =7/4,x1*x2 = -3/4
1)1/x1+1/x2=（x1+x2 ）/x1*x2 =-7/4/3/4=-7/3
2)|x1-x2|=根号[(x1+x2 ）^2-4x1*x2 ]=根号[49/16+4*3/4]=根号97/16=根号97/4
3) (x1-3)(x2-3)=x1*x2 -3(x1+x2 )+9=-3/4-21/4+9=1
4) x1x2^2+x1^2x2=x1*x2 (x1+x2 )=-3/4*7/4=-21/16

Answer 2

x1+x2 = -b/a=-7/4
x1*x2 = c/a=-3/4
1/x1+1/x2=（x1+x2）/x1x2=7/3

Answer 3

x1+x2 = -b/a=-7/4
x1*x2 = c/a=-3/4

1/x1+1/x2=（x1+x2）/x1x2=7/3
|x1-x2|=根号下（49+48）÷4=四分之根号97

(x1-3)(x2-3)=x1*x2-3(x1+x2)+9=13.5

x1x2^2+x1^2x2=x1x2（x1+x2）=21/16