Question

1) Considerando que a equação
${x}^{2} - 12x + 11 = 0.$
Tem duas raízes reais x1 e x2 diferentes, então pode-se dizer que o valor para a expressão
$\frac{x1 + x2}{2}$
sendo x1 e x2 as raízes é:


a) 8

b) 6

c) 5

d) 4

e) 3



​

Answer 1

$\large\boxed{\begin{array}{lr}x {}^{2} - 12x + 11 = 0 \\ \\ a = 1 \\ b = - 12 \\ c = 11 \\ \\ ∆ = ( - 12) {}^{2} - 4.1.11 \\ ∆ = 144 - 44 \\ ∆ = 100 \\ \\ x = \frac{ - ( - 12) \pm \sqrt{100} }{2.1} \\ \\ x = \frac{12 \pm10}{2} \\ \\ x_{1} = \frac{12 + 10}{2} = \frac{22}{2} = \boxed{11} \\ \\ x_{2} = \frac{12 - 10}{2} = \frac{2}{2} = \boxed{ 1}\end{array}}$

$\large\boxed{\begin{array}{lr} \dfrac{x_{1} + x_{2}}{2} = \dfrac{11 +1 }{2} = \dfrac{12}{2} = \boxed{ \boxed{ \red{6}}}\end{array}}$

$\large\boxed{\begin{array}{lr} \boxed{ \boxed{ \boxed{\boxed{\boxed{\boxed{\boxed{ \huge\blue{opc_{\!\!,}\tilde{ao}~(B)~6}}}}}}}}\end{array}}$

Answer 2

b) 6

x² - 12x +11 =0

$x = \frac{12 + - \sqrt{144 - 44} }{2.1}$

$x1 = \frac{12 - 10}{2}$

x' = 1

$x2 = \frac{12 + 10}{2}$

x''=11

$\frac{x1 + x2}{2} =$

$\frac{1 + 11}{2} = 6$