1.Tentukan persamaan kuadrat yang akar-akarnya 1/x1 dan 1/x2 dari akar akar persamaan kuadrat 3x^2+6x-8=0 2. Tentukan persamaan kuadrat yang akar-akarnya (2x1+1

1) x1 + x2 = -b/a = -6/3
x1.x2 = c/a = -8/3
1/x1 + 1/x2 = (x1+x2)/x1.x2
= (-6/3)/(-8/3)
= (-6/3) × (-3/8) = 6/8 = 3/4
1/x1 . 1/x2 = 1/(x1.x2) = 1/(-8/3) = -3/8
x² - (x1+x2)x + (x1.x2) = 0
x² - (3/4)x + (-3/8) = 0
x² - (3/4)x - 3/8 = 0
kalikan 8
8x² - 6x - 3 = 0

2) x1 + x2 = -b/a = -(-5)/1 = 5
x1.x2 = c/a = -1/1 = -1
2x1 + 1 + 2x2 + 1 = 2(x1+x2) + 2
= 2(5) + 2
= 12
(2x1 + 1)(2x2 + 1) = 4x1.x2 + 2(x1+x2) + 1
= 4(-1) + 2(5) + 1
= 7
x² - 12x + 7 = 0