咁a>0~佢就係opening upward
a<0就係opening downward
而△>0就係有real roots
△=0就係有一個real root
△<0就係無real root
應該無錯啦可........
咁而家有條數係
find the range of real values of k if kx^2+(4k-1)x+4k(>=)0 for all real values of k.
咁因為k唔知係+ 定-
咁我咪要分case做既
咁我咁做岩唔岩??
Case1 k>0
△(>=)0 (因為all real values of k)
(4k+1)^2-4(k)(4k)(>=)0
k(<=)1/8
Case2 k<0
△(>=)0 (因為all real values of k)
(4k+1)^2-4(k)(4k)(>=)0
k(<=)1/8
(rejected) (因為k<0)
但係個答案係k(>=)1/8
點解會咁既...我做錯左D咩...=3="
仲有~有時知道左個graph點畫咁又可以點幫我計數??
同埋~例如
有時咪要consider y=ax^2+bx+c=0先計既
咁係幾時要幾時就可以(ax^2+bx+c>0)咁樣直接計??
用中文~詳盡D~最好有解釋~
但唔好太深...因為岩岩先學...有好多都唔明...> <"
最慘係唔明自已唔明D咩...=3="""
唔該~=]
更新1:
但係~ -8k(>=)-1 乘除-數唔係要掉符號嫁咩?? mm...例如~ ax^2+bx+c>0 唔係要consider y=ax^2+bx+c=0 之後先可以用ax^2+bx+c=0黎計嫁咩?? 係咩情況下計要consider 係咩情況下計就直接ax^2+bx+c>0?? 咁講會唔會好D??