Question

Resolva a equação completa de segundo grau a seguir e determine o conjunto solução, isto é, suas raízes:

Answer 1

$\displaystyle \frac{4}{\text t-2} + \frac{4}{\text t-3} = 3$

fazendo o MMC :

$\displaystyle \frac{4(\text t-3)}{(\text t-2)(\text t-3)} + \frac{4(\text t-2)}{(\text t-3)(\text t-2)} = 3$

$\displaystyle 4(\text t-3)+ 4(\text t-2)= 3(\text t-2)(\text t-3)$

$\displaystyle 4\text t-12+ 4\text t-8= 3(\text t^2-5\text t+6)$

$\displaystyle 8\text t-20= 3\text t^2-15\text t+18$

$\displaystyle 3\text t^2-15\text t - 8\text t +18 + 20 =0$

$\displaystyle 3\text t^2-23\text t +38 = 0$

Usando bhaskara :

$\displaystyle \text x = \frac{-(-23)\pm\sqrt{(-23)^2-4.(3).(38)}}{2.3}$

$\displaystyle \text x = \frac{23\pm\sqrt{529-456}}{6} \to \text x = \frac{23\pm\sqrt{73}}{6}$

Portanto suas raízes são :

$\huge\boxed{\bold{\text t = \frac{23+\sqrt{73}}{6} \ \ \text{ou }\ \ \text t = \frac{23-\sqrt{73}}{6} } } \checkmark$

Answer 2

Resposta:

Resposta letra b) (6; 2,5)

Explicação passo a passo: