Aard van die wortels - bepaal p

Re: Aard van die wortels

by Marna Zuhlsdorff -
Number of replies: 0
Hi Chaunte

Baie dankie vir jou  vraag en bewerking! Jou aard van die wortels was spot-on! Jou fout het by ongelykhede (heeeel aan die einde) ingekom.

Hierdie som bestaan uit drie dele: in die eerste deel herskryf ons die kwadratiese vergelyking in standaard vorm (a x squared plus b x plus c equals 0), in die tweede gedeelte bepaal ons die diskriminant (increment equals b squared minus 4 a c) en in die derde deel pas ons die aard van die wortel (increment greater or equal than 0) toe.

Jy kan hier click om die aard van die wortels videos weer te kyk en hier click as jy die ongelykhede weer wil hersien.

Deel 1 - herskryf in standaard vorm:
table row cell 3 x squared plus p equals 4 x plus p x squared end cell row cell 3 x squared minus p x squared minus 4 x plus p equals 0 end cell row cell x squared open parentheses 3 minus p close parentheses minus 4 x plus p equals 0 end cell end table

Deel 2 - bepaal die diskriminant:
table row cell capital delta equals b squared minus 4 a c end cell row cell capital delta equals open parentheses negative 4 close parentheses squared minus 4 open parentheses 3 minus p close parentheses open parentheses p close parentheses end cell row cell capital delta equals 16 minus 4 p open parentheses 3 minus p close parentheses end cell row cell capital delta equals 16 minus 12 p plus 4 p squared end cell row cell capital delta equals 4 p squared minus 12 p plus 16 end cell row cell capital delta equals 4 open parentheses p squared minus 3 p plus 4 close parentheses end cell end table

Deel 3 - pas die aard van die wortel toe:
table row cell m a a r space capital delta greater or equal than 0 end cell row cell b squared minus 4 a c greater or equal than 0 end cell row cell 4 open parentheses p squared minus 3 p plus 4 close parentheses greater or equal than 0 end cell row cell 4 open parentheses p minus 1 close parentheses open parentheses p minus 4 close parentheses greater or equal than 0 end cell row cell therefore p equals 1 space o f space p equals 4 end cell end table

therefore space p less or equal than 1 space o f space p greater or equal than 4
Laat weet my asb as enige iets onduidelik is.

Marna