Question

Dado o sistema de equações:

$x + y = 5 \\ {x}^{2} - y = 3$
1) Resolva-o algebricamente.

2) Resolva-o utilizando o Geogebra. Justifique a sua resposta.​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\begin{cases}\mathsf{x + y = 5}\\\mathsf{x^2 - y = 3}\end{cases}$

$\mathsf{x^2 + x = 8}$

$\mathsf{x^2 + x - 8 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = 1^2 - 4.1.(-8)}$

$\mathsf{\Delta = 1 + 32}$

$\mathsf{\Delta = 33}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-1 \pm \sqrt{33}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{-1 + \sqrt{33}}{2}}\\\\\mathsf{x'' = \dfrac{-1 - \sqrt{33}}{2}}\end{cases}}$

$\mathsf{\dfrac{-1 + \sqrt{33}}{2} + y' = 5}$

$\mathsf{-1 + \sqrt{33} + 2y' = 10}$

$\mathsf{2y' = 11 - \sqrt{33}}$

$\mathsf{y' = \dfrac{11 - \sqrt{33}}{2}}$

$\mathsf{\dfrac{-1 - \sqrt{33}}{2} + y'' = 5}$

$\mathsf{-1 - \sqrt{33} + 2y'' = 10}$

$\mathsf{2y'' = 11 + \sqrt{33}}$

$\mathsf{y'' = \dfrac{11 + \sqrt{33}}{2}}$

$\boxed{\boxed{\mathsf{S = \{\:\large(x' = \dfrac{-1 + \sqrt{33}}{2}\large;y' = \dfrac{11 - \sqrt{33}}{2} );\large(x'' = \dfrac{-1 - \sqrt{33}}{2}\large;y'' = \dfrac{11 + \sqrt{33}}{2} )\:\}}}}$

Answer 2

Explicação passo-a-passo:

x+y=5

x²-y=3

________

x²+x=8

x²+x-8=0

a=1, b=1 e c=-8

delta =b²-4ac

delta =1²-4x1x-8

delta =1+32

delta =33

x'= (-1+V33)/2

x"= (-1+V33)/2