Question

Resolva a equação
$x^{2} - (1 + \sqrt{3})x + \sqrt{3} = 0$
​

Answer 1

$x_{1,2}=\frac{-b±\sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-(-(1+\sqrt{3}))±\sqrt{(-(1+\sqrt{3}))^2-4(1)(\sqrt{3})}}{2(1)}$

$x_{1,2}=\frac{1+\sqrt{3}±\sqrt{1+2\sqrt{3}+3-4\sqrt{3}}}{2}$

$x_{1,2}=\frac{1+\sqrt{3}±\sqrt{1-2\sqrt{3}+3}}{2}$

$x_{1,2}=\frac{1+\sqrt{3}±\sqrt{(1-\sqrt{3})^2}}{2}$

$x_{1,2}=\frac{1+\sqrt{3}±\left|1-\sqrt{3}\right|}{2}$

• Perceba que $\sqrt{3}\gt1$. Logo $\left|1-\sqrt{3}\right|=\sqrt{3}-1$

$x_{1,2}=\frac{1+\sqrt{3}±(\sqrt{3}-1)}{2}$

$x_1=\frac{1+\sqrt{3}+\sqrt{3}-1}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}$

$x_2=\frac{1+\sqrt{3}-(\sqrt{3}-1)}{2}=\frac{1+\sqrt{3}-\sqrt{3}+1}{2}=\frac{2}{2}=1$

→ $\mathbb{S}=\{\sqrt{3},1\}$

Espero ter ajudado. Continue estudando sempre.(◍•ᴗ•◍)♡