Question

Fie ecuatia 2x²-3x+6=0. Sa se calculeze : a) 1 supra x1 + 1 supra x2 b) 1 supra x1 +3 + 1 supra x2 +3

Answer 1

$\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2 }{x_1 x_2} = \frac{S_1}{S_2}$

Din relatiile lui Viete
$S_1 = \frac{-b}{a} = \frac{3}{2}$
$S_2 = \frac{c}{a} = 3$
$\frac{1}{x_1} + \frac{1}{x_2} = \frac{3/2}{3} = 1/2$
b)

$\frac{1}{x_1+3} + \frac{1}{x_2+3} = \frac{x_1 + x_2 + 6}{(x_1+3)(x_2+3)} =\frac{x_1 + x_2 + 6}{x_1x_2 + 3(x_1+x_2) + 9} = \frac{S_1+6}{S_2 + 3S_1 + 9} = \frac{3/2 + 6}{ 3 + 3* 3/2 + 9}$

$\frac{3/2 + 6}{ 3 + 3* 3/2 + 9} = \frac{3 + 12}{3 + 9 + 18} = \frac{15}{30} =1/2$