Question

resolva as equações em (R
${x}^{2} - 2 \sqrt{5x} + 4 = 0$
​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{x^2 - 2\sqrt{5}\:x + 4 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-2\sqrt{5})^2 - 4.1.4}$

$\mathsf{\Delta = 20 - 16}$

$\mathsf{\Delta = 4}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{2\sqrt{5} \pm \sqrt{4}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{2\sqrt{5} + 2}{2} = 1 + \sqrt{5}}\\\\\mathsf{x'' = \dfrac{2\sqrt{5} - 2}{2} = -1 + \sqrt{5}}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \left\{1 + \sqrt{5};\:-1 + \sqrt{5}\right\}}}}$

Answer 2

Vamos là.

x² - 2√5*x + 4 = 0

a = 1, b = -2√5, c = 4

delta

d = 20 - 16 = 4

x1 = (2√5 + 2)/2 = 1 + √5

x2 = (2√5 - 2)/2 = -1 + √5