Question

A equação do segundo grau: √(3) · x² - 2 · x - 1 = 0 tem quais raízes?

Então, responda:

a) x, + x,,
b) x, · x,,
c) (1/x,) + (1/x,,)
d) x,² + x,,²

Answer 1

$\sqrt{3} .x^2 - 2x -1=0 \\ \\ a = \sqrt{3} \\ b=-2 \\ c=-1 \\ \\ \Delta = b^2-4ac\\ \Delta = (-2)^2-4. \sqrt{3} .(-1)\\ \Delta = 4 + 4 \sqrt{3} \\ \Delta = 10,9\\ \\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \\ \\ x=\dfrac{2\pm3,3}{2 \sqrt{3} } \\ \\ x=\dfrac{2\pm3,3}{3,46} \\ \\ x'=\dfrac{2+3,3}{3,46} = 1,53\\ \\ x"=\dfrac{2-3,3}{3,46} = -0,38\\ \\ \\ a)\ x'+x"=1,53-0,38=1,15\\ \\ b)\ x' . x" = 1,53.(-0,38)=-0,58\\ \\ c)\ \dfrac{1}{x'} +\dfrac{1}{x"}=\dfrac{1}{1,53} -\dfrac{1}{0,58}=-1,07$
$\\d)\ x'^2+x"^2=1,53^2+(-0,58)^2=2,68$

=)

Answer 2

Resolução:
$\sqrt{3} x^{2} -2x-1=0 \\ \\ a= \sqrt{3} \\ \\ b=-2 \\ \\ c=-1$

$x= \frac{-b+/- \sqrt{ b^{2}-4.a.c } }{2.a} \\ \\ x= \frac{2+/- \sqrt{4.(1+ \sqrt{3} } }{2. \sqrt{3} }$

$\frac{2(1+/- \sqrt{1+ \sqrt{3} } }{2. \sqrt{3} } \\ \\ \frac{1+/- \sqrt{1+ \sqrt{3} } }{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} }$

$\frac{ \sqrt{3}+/- \sqrt{1+ \sqrt{3} } }{3}$

a) $x`+x`` \\ \\ \frac{ \sqrt{3}+ \sqrt{1+ \sqrt{3} } }{3}+ \frac{ \sqrt{3}- \sqrt{1+ \sqrt{3} } }{3}= \frac{2 \sqrt{3} }{3}$

b)$\frac{ \sqrt{3}+ \sqrt{1+ \sqrt{3} } }{3}. \frac{ \sqrt{3}- \sqrt{1+ \sqrt{3} } }{3}= \frac{3-2 \sqrt{2+ \sqrt{3} } }{9}$

c) $\frac{1}{ \frac{ \sqrt{3}+ \sqrt{1+ \sqrt{3} } }{3} }+ \frac{1}{ \frac{ \sqrt{3}- \sqrt{1+3} }{3} } \\ \\ \frac{3}{ \sqrt{3}+ \sqrt{1+ \sqrt{3} } }+ \frac{3}{ \sqrt{3}- \sqrt{1+ \sqrt{3} } }$

$\frac{ 6\sqrt{3} }{3- 2\sqrt{2+ \sqrt{3} } }$