已知关于x的一元二次方程x²+ax+b=0有整数根且b=2a-1,则a+b=
人气:487 ℃ 时间:2019-11-23 08:16:14
解答
x^2+ax+b=x^2+ax+2a-1=0
x=[-a±√(a^2-8a+4)]/2
a=0,b=-1
a+b=-1貌似还有一种情况等于23,但不知到怎么来的,还求指教a^2-8a+4=n^2
a^2-n^2=8a-4
(a+n)*(a-n)=(2a-1)*4
a+n=2a-1......(1)
a-n=4......(2)
(1)+(2):
无解
a+n=4......(3)
a-n=2a-1......(4)
(3)+(4):
无解
a^2-8a+4=(a-4)^2-12=n^2
(a-4)^2-n^2=12
(a-4+n)*(a-4-n)=1*12
a-4+n=12......(5)
a-4-n=1......(6)
(5)+(6):
a=10.5,b=20,不是整数
a-4+n=6......(7)
a-4-n=2......(8)
(7)+(8):
a=8,b=15,x=-5,-3,符合已知条件,a+b=23
a-4+n=4......(9)
a-4-n=3......(10)
(9)+(10):
a=7.5,不是整数
确实是有一个答案:a+b=23
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