Question

1- Usando a fórmula de Bhaskara resolva as seguintes equações do 2º grau no conjunto IR: a) x + x2 = 182 b) 9x 2 - 12x = 0

Answer 1

 ➯  Aplicando a fórmula de Bhaskara, as raízes das equações de 2° grau são: (a) -14 e 13; (b) 0 e 4/3.

 ➯  (a) x + x² = 182

$\sf{x^2+x-182=0}\\\\\\\sf{Coeficientes:\;a=1,\;b=1\;e\;c=-182.}\\\\\\\sf{\Delta=b^2-4ac}\\\\\sf{\Delta=1^2-4\cdot1\cdot(-182)}\\\\\sf{\Delta=1+728}\\\\\sf{\Delta=729}\\\\\\\sf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\\sf{x=\dfrac{-1\pm\sqrt{729}}{2\cdot1}}\\\\\\\sf{x=\dfrac{-1\pm27}{2}}\\\\\\\sf{x_1=\dfrac{-1+27}{2}=\dfrac{26}{2}=13}\\\\\\\sf{x_2=\dfrac{-1-27}{2}=\dfrac{-28}{2}=-14}\\\\\\\boxed{\sf{S=\{-14,13\}}}$

 ➯  (b) 9x² - 12x = 0

$\sf{Coeficientes:\;a=9,\;b=-12\;e\;c=0.}\\\\\\\sf{\Delta=b^2-4ac}\\\\\sf{\Delta=(-12)^2-4\cdot9\cdot0}\\\\\sf{\Delta=144+0}\\\\\sf{\Delta=144}\\\\\\\sf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\\sf{x=\dfrac{-(-12)\pm\sqrt{144}}{2\cdot9}}\\\\\\\sf{x=\dfrac{12\pm12}{18}}\\\\\\\sf{x_1=\dfrac{12+12}{18}=\dfrac{24}{18}=\dfrac{4}{3}}\\\\\\\sf{x_2=\dfrac{12-12}{18}=\dfrac{0}{18}=0}\\\\\\\boxed{\sf{S=\{ 0,\dfrac{4}{3}\}}}$

 ➯  Saiba mais em:

https://brainly.com.br/tarefa/31418382

Espero ter ajudado.

Bons estudos! :)