18. K and P are positive integers. Which of the followings must have a negative solution?
1. X^2+kX+P=0
2. X^2-kx+p=0
3. 记不起来好像是不对。
[讨论]看K and P的具体取值，1和2应该都可能有负解。
我认为1一定有负数解。一元二次方程式的求解公式为：x＝－b+-根号b^2-4ac/2a
1中，x＝－k+-根号k^2-4p/2，由于k,p均为正整数，那么 根号k^2-4p作者: lisaislisa 时间: 2003-1-17 08:18
I am with you
1) => x must be negative as -K-root(K^2-4P)
but 2)=> x can be negative too ,as K-root(K^2-4P)<0.. when root(K^2-4P)>K
^^作者: 曼波 时间: 2003-1-17 13:41
but how can root(k^2-4p)>k, given that k,p are positive integers?作者: qqks 时间: 2003-1-17 19:48
我在费费的宝典上看到了这道题目，完整如下：
K and p are positive constants, 问下面那个方程一定有一个负数解？
I x^2+kx+p=0
II X^2 -kx+p=0
III X^2 +kx-p=0
答案是 III作者: lisaislisa 时间: 2003-1-17 20:45
well i 've thought it twice...
it asks 'must have an negative integer ' not can have one , right ?
then from III X^2 +kx-p=0
=> [-k-root(k^2-4p)]/2 must be negative
^^作者: isenetlab12 时间: 2003-1-18 10:15
According to qqks
我在费费的宝典上看到了这道题目，完整如下：
K and p are positive constants, 问下面那个方程一定有一个负数解？
I x^2+kx+p=0
II X^2 -kx+p=0
III X^2 +kx-p=0
the answer shoulbe III.
since I may have no answer if root of (k^2-4p)<0,
but III must have at least one negative solution since -k-root of (k^2+4p) must be less than 0
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