[tex]f(x) = x^{2} \sqrt{2} - x\sqrt{6} - \sqrt{3} \\\\a = \sqrt{2} \\b = -\sqrt{6} \\c = -\sqrt{3}[/tex]
Ze wzorów Viet'a obliczamy sumę odwrotności pierwiastków funkcji kwadratowej:
[tex]\frac{1}{x_{1}} + \frac{1}{x_{2}} = \frac{x_{2}}{x_{1}x_{2}} + \frac{x_{1}}{x_{1}x_{2}} = \frac{x_{1}+x_{2}}{x_{1}x_{2}} = \frac{\frac{-b}{a} }{\frac{c}{a} } = \frac{-b}{a} * \frac{a}{c} = \frac{-b}{c}[/tex]
Podstawiamy współczynniki:
[tex]\frac{-b}{c} = \frac{-(-\sqrt{6} )}{-\sqrt{3} } = \frac{\sqrt{6}}{-\sqrt{3} } = -\frac{\sqrt{6}*\sqrt{3} }{\sqrt{3}*\sqrt{3} } = -\frac{\sqrt{18} }{\sqrt{9} } = -\frac{\sqrt{9*2} }{3} = -\frac{\sqrt{9}*\sqrt{2} }{3}= -\frac{3\sqrt{2} }{3} = -\sqrt{2}[/tex]
Odp: D
