If a and b are the roots of x^2+px-p^2=0
Express 1/a + 1/b in terms of p
Mathsss :)))?
2015-08-28 11:55 am
回答 (3)
2015-08-28 12:00 pm
Sum of the roots : a + b = ‒p
Product of the roots : ab = p²
(1/a) + (1/b)
= (b/ab) + (a/ab)
= (a + b)/ab
= ‒p/p²
= ‒1/p
Product of the roots : ab = p²
(1/a) + (1/b)
= (b/ab) + (a/ab)
= (a + b)/ab
= ‒p/p²
= ‒1/p
2015-08-29 4:25 am
不保證絕對正確
x^2+px-p^2=0
iff x=[-p±√(p^2+4p^2)]/2=(-1±√5)p/2=a or b, a is different from b
→1/a+1/b=(2/p)[(-2)/(-2)]
=2/p
x^2+px-p^2=0
iff x=[-p±√(p^2+4p^2)]/2=(-1±√5)p/2=a or b, a is different from b
→1/a+1/b=(2/p)[(-2)/(-2)]
=2/p
2015-08-28 2:24 pm
收錄日期: 2021-04-18 00:06:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150828035511AAYHZ7Y