Question

Determine a soma dos quadrados das raízes da equação

X³ - X ²- 9x + 9 = 0.​

Answer 1

Podemos resolver de duas formas:

1ª forma : Fatorando

$\text x^3-\text x^2-9\text x+9=0 \\\\\\ \text x^2(\text x-1) -9(\text x-1) = 0 \\\\\\ \text (\text x^2-9)(\text x-1)=0 \\\\\\ (\text x-3)(\text x+3)(\text x-1) =0 \\\\\\ \text{raizes : 3,-3,1 }$

Soma dos quadrados das raízes :

$3^2+(-3)^2+1^2 \\\\ 9+9+1 \\\\\ \huge\boxed{19}\checkmark$

2ª Forma de resolver

Relações de Girard para polinômios de grau 3 :

$\text{A.x}^3+\text{B.x}^2+\text{C.x}+\text D$

sendo :

$\text x_1,\text x_2,\text x_3 =$ raízes

temos :

$\displaystyle \text x_1+\text x_2+\text x_3 = \frac{-\text B}{\text A} \\\\\\ \text x_1.\text x_2+\text x_1.\text x_3 +\text x_2.\text x_3 = \frac{\text C}{\text A} \\\\\\ \text x_1.\text x_2.\text x_3 = \frac{-\text D}{\text A}$

Temos o polinômio :

$\text x^3-\text x^2-9\text x+9=0$

queremos a soma dos quadrados das raízes.

seja a, b e c as raízes do polinômio, então queremos determinar :

$\text a^2+\text b^2+\text c^2 = \ ?$

isso é parte do seguinte produto notável :

$(\text {a+b+c})^2 = \text a^2+\text b^2+\text c^2 +2.(\text{a.b}+\text{a.c}+\text{b.c}) \\\\\\ \text a^2+\text b^2+\text c^2 = (\text{a+b+c})^2-2.(\text{a.b}+\text{a.c}+\text{b.c})$

Vamos usar as relações de girard :

$\displaystyle \text a+\text b+\text c = \frac{-(-1)}{1} \to \boxed{\text {a+b+c}=1} \\\\\\ \text a.\text b+\text a.\text c +\text b.\text c = \frac{-9}{1}$

Substituindo no produto notável :

$\text a^2+\text b^2+\text c^2 = (\text{a+b+c})^2-2.(\text{a.b}+\text{a.c}+\text{b.c}) \\\\\\ \text a^2+\text b^2+\text c^2 = 1^2-2(-9) \\\\\\ \text a^2+\text b^2+\text c^2 = 1+18 \\\\\\ \huge\boxed{\text a^2+\text b^2+\text c^2 = 19}\checkmark$