Odpowiedzi

2009-11-17T18:53:48+01:00
a)
x³ -√2 x² -9x +3√18 = 0
x³ -√2 x² -9x +3 √9*2 = 0
x² -√2 x² -9x + 9 √2 = 0
x*(x² - 9) - √2*(x² -9) = 0
(x - √2)*(x² -9) = 0
(x -√2)*(x - 3)*(x +3) = 0
x1 = -3, x2 = √2, x3 = 3
b)

3x³ + x² + 48 x + 16 = 0
3x*(x² + 16) +(x² + 16) = 0
(3x + 1)*( x² + 16) = 0 <=> 3x +1 = 0 <=> 3x = -1 <=>
x = -1/3, ponieważ
x² + 16 ≥ 16 > 0
Odp. x = -1/3