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
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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
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}
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}
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