Question

设集合A={x^2+-ax+b=0},B={6x^2+(a+2)x+b=0},且A∩B={二分之一},求A∪B

Answer 1

A∩B={1/2}

for A={x^2-ax+b=0}

1/4-a/2+b =0

a = 1/2 +2b --(1)

and for B={6x^2+(a+2)x+b=0}

3/2+(a+2)/2 +b =0

5 + a +2b =0 ---(2)

from (1) (2)

5+1/2 +2b +2b =0

b = -11/8

a = -9/4

A={x|x^2-ax+b=0}

={x|x^2+9x/4-11/8 = 0}

={x|8x^2+18x-11=0}

={x|(2x-1)(4x+11)=0}

={1/2,-11/4}

B={x|6x^2+(a+2)x+b=0}

={x|6x^2-x/4-11/8=0}

={x|48x^2-2x-11=0}

={x|(2x-1)(24x+11)=0}

={1/2,-11/24}

A∪B = {1/2,-11/4}∪ {1/2,-11/24}

= {1/2,-11/4,-11/24}