已知A={x|x²+px+q=x},B={x|(x-1)²+p(x-1)+q=x+1},当A={2}时,求集合B.
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已知A={x|x²+px+q=x},B={x|(x-1)²+p(x-1)+q=x+1},当A={2}时,求集合B.
已知A={x|x²+px+q=x},B={x|(x-1)²+p(x-1)+q=x+1},当A={2}时,求集合B.
已知A={x|x²+px+q=x},B={x|(x-1)²+p(x-1)+q=x+1},当A={2}时,求集合B.
∵A={2}
而x²+px+q=x二次项系数为1
∴x²+px+q=x有两个相等的实数根,为2
x²+(p-1)x+q=0
根据韦达定理:
x1+x2=-(p-1)=2+2=4
x1x2=q=2*2=4
解得:p=-3,q=4
则(x-1)²+p(x-1)+q=x+1可化为:
(x-1)²-3(x-1)+4=x+1
x²-2x+1-3x+3+4-x-1=0
x²-6x+7=0
x²-6x+9=2
(x-3)²=2
解得:x=3±√2
∴B={3+√2,3-√2}