依條係培正數學比賽以前的題目..
有關方程的問題..
link: http://www.2and2.net/files/4978794970e47.bmp
依條答案係 3 ..
我想知個步驟..
可唔可以講個步驟同埋解釋埋比我聽呀??
thx!
1條F.4培正數學比賽問題.. 有關方程
2009-01-23 5:52 am
回答 (2)
2009-01-23 6:54 am
✔ 最佳答案
原式可寫成X^2 * (X^4-2008X^2-126)=0即X^2=0 或 X^4-2008X^2-126=0
由X^2=0知X的一實根為0
而另一方程可看作是X^2的二次方程,
判別式△=2008^2-4(-126)大於0,表示有兩正實根(X^2的兩個可能值是實數);
由根與系數關係知兩根(X^2的兩個可能值)之積=-126,表示X^2的兩個可能值一正一負,
當X^2是正數時,X有兩實根,
當X^2是負數時,X無實根,
綜上X有3實根。
2009-01-22 22:59:01 補充：
更正:判別式大於0表示有兩實根,不是正實根。
2009-01-23 7:15 am
x^6-2008x^4-126x^2=0
=> (x-0)(x-0)(x^4-2008x^2-126)=0
=>x= o (repeated root) or x^2-2008x^2-126=0
for x^4-2008x^2-126=0
we have △=2008^2-(-126*4)>0
so that there must be two real root (+ve and -ve respectively)
x=+/-sqrt of [x^4-2008x^2-126(positive one)]
so there are 4 real roots!!!
=> (x-0)(x-0)(x^4-2008x^2-126)=0
=>x= o (repeated root) or x^2-2008x^2-126=0
for x^4-2008x^2-126=0
we have △=2008^2-(-126*4)>0
so that there must be two real root (+ve and -ve respectively)
x=+/-sqrt of [x^4-2008x^2-126(positive one)]
so there are 4 real roots!!!
收錄日期: 2021-04-21 22:04:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090122000051KK01743