Question

resolva a equação: x²+x-1=0

Answer 1

Usando a fórmula de Báskara

 

 

$<var>\boxed{\Delta = b^2 - 4(a)(c)}</var>$

 

$<var>\Delta = 1^2 - 4(1)(-1)</var>$

 

$<var>\Delta = 1 + 4</var>$

 

$<var>\boxed{\Delta = 5}</var>$

 

$<var>x = \frac{-b + ou - \sqrt{\Delta}}{2a}</var>$

 

$<var>x = \frac{-1 + ou - \sqrt{5}}{2}</var>$

 

$<var>x^1 =1+\frac{\sqrt{5} }{2}</var>$

 

$<var>x^2 =1-\frac{\sqrt{5} }{2}</var>$

 

 

$<var>\boxed{S = {1 +\frac{\sqrt{5} }{2}\ e 1-\frac{\sqrt{5} }{2}}}</var>$

Answer 2

$<var>x^{2}+x-1=0 \\\\ \Delta = b^{2} - 4 \cdot a \cdot c \\ \Delta = 1^{2} - 4\cdot (1)\cdot (-1) \\ \Delta = 1 + 4 \\ \Delta = 5 \\\\\\ x = \frac{-b \pm \sqrt{\Delta}}{2\cdot a} \\\\ x = \frac{-1 \pm \sqrt{5}}{2\cdot 1} \\\\ x = \frac{-1 \pm \sqrt{5}}{2} \\\\\\ \boxed{x' = \frac{-1+\sqrt{5}}{2}} \\\\ \boxed{x'' = \frac{-1 -\sqrt{5}}{2}}</var>$

 

$<var>S=\{\frac{-1+\sqrt{5}}{2}, \frac{-1 -\sqrt{5}}{2}\}</var>$