Question

{█(x+y=8@x^2+y^2=50) resolva esse sistema de equação de 2 grau

Answer 1

Olá.

Vamos primeiro encontrar uma equação de 2° grau para trabalharmos.

$\left\{\begin{array}{ccc}\mathsf{x+y}&=&\mathsf{8}\\\mathsf{x^2+y^2}&=&\mathsf{50}\end{array}\right\\\\\\ \mathsf{x+y=8}\\\\ \mathsf{x=8-y}\\\\\\\\ \mathsf{x^2+y^2=50}\\\\ \mathsf{(8-y)^2+y^2=50}\\\\ \mathsf{(64-16y+y^2)+y^2=50}\\\\ \mathsf{y^2+y^2-16y+64-50=0}\\\\ \mathsf{2y^2-16y+14=0}$

Tendo encontrado a equação, vamos agora desenvolver por Bhaskara.

$\mathsf{2y^2-16y+14=0}\\\\\\ \Large\mathsf{y=\dfrac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2a}}\\\\\\ \mathsf{y=\dfrac{-(-16)\pm\sqrt{(-16)^2-4\cdot2\cdot14}}{2\cdot2}}\\\\\\ \mathsf{y=\dfrac{16\pm\sqrt{256-4\cdot28}}{4}}\\\\\\ \mathsf{y=\dfrac{16\pm\sqrt{256-112}}{4}}\\\\\\ \mathsf{y=\dfrac{16\pm\sqrt{144}}{4}}\\\\\\ \mathsf{y=\dfrac{16\pm12}{4}}$

Vamos agora encontrar as duas raízes:
$\mathsf{y=\dfrac{16\pm12}{4}}\\\\\\ \mathsf{y'=\dfrac{16+12}{4}}\\\\\\ \mathsf{y'=\dfrac{28}{4}}\\\\\\ \boxed{\mathsf{y'=7}}\\\\\\ \mathsf{y''=\dfrac{16-12}{4}}\\\\\\ \mathsf{y''=\dfrac{4}{4}}\\\\\\\mathsf{y''=1}}$

Os dois valores encontrados são válidos tanto para x quanto para y. Assim:

$\boxed{\boxed{\mathsf{S=\{(x,~y)\in\mathbb{R}~|~(x,~y)=(1,~7)\}}}}}$


Vamos testar:
$\left\{\begin{array}{ccc}\mathsf{x+y}&=&\mathsf{8}\\\mathsf{x^2+y^2}&=&\mathsf{50}\end{array}\right\\\\\\ \textsf{Usemos x~=~1 e y~=~7.}\\\\ \mathsf{1+7=8}\\\boxed{\mathsf{8=8~\checkmark}}\\\\\\\mathsf{1^2+7^2=50}\\\mathsf{1+49=50}\\\boxed{\mathsf{50=50~\checkmark}}$

Testado e aprovado.

Qualquer dúvida, deixe nos comentários.
Bons estudos.