Se x¹ e x² (com x¹ > x²) são as duas raízes reais da equação x - 12/x, com x ≠ 0, o valor da express.... Pergunta de ideia deCaelJ

December 2019 1 68 Report

Se x¹ e x² (com x¹ > x²) são as duas raízes reais da equação x - 12/x, com x ≠ 0, o valor da expressão ( x¹ - x²)² é:
a) 51
b) 53
c) 59
d) 61
e) 63/8

Para hoje mesmo, por favor

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FibonacciTH

Verified answer

Dada a equação:

$\mathsf{x-\dfrac{12}{x}=0}$

Determinando as suas raízes:

$\mathsf{x=\dfrac{12}{x}}\\\\\mathsf{x\cdot x=12}\\\\\mathsf{x^2=12}\\\\\mathsf{\sqrt{x^2}=\sqrt{12}}\\\\\mathsf{x=\pm \sqrt{12}}\\\\\mathsf{x=\pm \sqrt{4\cdot 3}}\\\\\mathsf{x=\pm 2\sqrt{3}}$

Portanto termos duas raízes:

◾ $\mathsf{x_1=2\sqrt{3}}$
◾ $\mathsf{x_2=-2\sqrt{3}}$

Determinado então $\mathsf{\left(x_1-x_2\right)^2=0}$ :

$\mathsf{\left(x_1-x_2\right)^2}\\\\\mathsf{=\left(2\sqrt{3}-\left(-2\sqrt{3}\right)\right)^2}\\\\\mathsf{=\left(2\sqrt{3}+2\sqrt{3}\right)^2}\\\\\mathsf{=\left(4\sqrt{3}\right)^2}\\\\\mathsf{=4^2\left(\sqrt{3}\right)^2}\\\\\mathsf{=16\cdot 3}\\\\\boxed{\mathsf{=48}}\: \: \checkmark$

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