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f(1)=a+b+c=0
f(2)=4a+2b+c=3
-------------------------subtitusi------------------------
9a-3b+c=28
c=28-9a+3b
--> a+b+c=0
a+b+(28-9a+3b)=0
-8b+4b=-28...............................(persamaan 1
--> 4a+2b+c=3
4a+2b+(28-9a+3b)=3
-5a+5b=-25.............................(persamaan 2
---------------------Eliminasi -----------------------------
-8a+4b=-28 I x 5..................(persamaan 1
-5a+5b=-25 I x 4 ....................(persamaan 2
------------------------------
-40a+20b=-140
-20a+20b=-100
-------------------------------- (-)
-20a = - 40
a = 2
---------------------- Subtitusi -----------------------
-8a+4b = -28
-8(2) + 4 b = -28
4 b = -12
b = -3
a+b+c = 0
2+(-3) + c =0
-1 +c = 0
c = 1
------------------------------------jadi persamaanya------------------------------
2x²-3x+1=y