Question

Akar akar persamaan kuadrat 2x pangkat 2 + x +3 = 0 adalah x1 dan x2. susunlah persamaan kuadrat baru yang akar akarnya a. 3+1/x1 dan 3+ 1/ x2
b. 1/x1-3 dan 1/x2 -3

Answer 1

X1+x2 = -1/2
x1 x2 =  3/2
.
p+q =  3 + 1/x1 + 3 + 1/x2
p+q = 6 + (x1+x2)/(x1.x2) = 6 -  (1/2)/(3/2) = 6 - 1/3 = 17/3
.
p.q = (3+ 1/x1)(3+1/x2) = 9 + 3(x1+x2) + 1/(x1x2)
P.q = 9 + 3(- 1/2) + 1/(3/2)
p.q = 9 - 3/2 + 2/3
p.q = 49/6
..
PK baru = x² - (p+q) x +(pq)= 0
x² - 17/3 x + 49/6 = 0 kalikan 6
6x² - 34 x + 49 = 0
.
2)
cara invers
p = 1/x1 - 3 --> 1/x1 = p + 3
x1 = 1/(p-3) sub ke PK 2x² + x + 3 = 0
2(1/(p-3)² + (1/(p-3)) + 3= 0 
2/ (p-3)² + 1/ (p-3) ) + 3 = 0
kalikan (p-3)²
2 + (p-3) + 3(p-3)² = 
3(p² -18 p + 27 + 2 + p - 3 = 0
3p² - 54 p + 27 + 2+p-3= 0
3x² - 55p + 26 = 0