6. Jika x1 dan x2 adalah akar-akar persamaan kuadrat x2 - 4x + 3 = 0 , maka persamaan kuadrat yang akar-akarnya 1/x12 dan 1/x22 adalah ...
A. x2 - 10x + 9 = 0
B. x2 + 4x - 3 = 0
C. x2 - 4x + 3 = 0
D. x2 + 9x - 10 = 0
E. 9x2 - 10x + 1 = 0
Jawaban : E
x2 - 4x + 3 = 0a = 1
b = -4
c = 3
x1 + x2 = -b/a = 4/1 = 4
x1 . x2 = c/a = 3/1 = 3
akar-akar yang baru :
1/x12 + 1/x22 = -B/A
x22 + x12
-------------- = -B/A
x12x22
(x1 + x2)2 - 2 x1 . x2
-------------------------= -B/A
(x1x2)2
16 - 2 . 3
----------- = -B/A
9
10/9 = -B/A
1/x12 . 1/x22 = C/A
1/ (x1x2)2 = C/A
1/9 = C/A
A = 9
B = -10
C = 1
9x2 - 10x + 1 = 0
Soal 7
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