Question

Diketahui persamaan kuadrat 1/3x² - 6x - 3 = 0. akar-akarnya x1 dan x2. Maka nilai x1² + x2² adalah...
a. 18
b. 19
c. 242
d. 342

(Tolong jawab pake cara)

Answer 1

1/3x² - 6x - 3 = 0
a = 1/3
b = -6
c = -3

x1 + x2 = -b/a
x1 + x2 = -(-6)/(1/3)
x1 + x2 = 6 × 3/1
x1 + x2 = 18

x1 . x2 = c/a
x1 . x2 = -3/(1/3)
x1 . x2 = -3 × 3/1
x1 . x2 = -9

x1² + x2²
= (x1 + x2)² - 2x1x2
= 18² - 2 . (-9)
= 324 + 18
= 342

Jawaban: D