Question

Resolva as equações irracionais

Answer 1

Achei que as contas eram pequenas kkkk

Answer 2

De forma geral, temos:

$\sqrt[n]{a} = b \implies (\sqrt[n]{a})^{n} =b^n$

a)

$\sqrt{3x+6}-2=x \implies \sqrt{3x+6}=x +2 \implies (\sqrt{3x+6})^2=(x+2)^2\implies \\\\\implies3x+6=x^2 +4x +4\implies x^2+x-2=0\\\\\Delta=b^2- 4ac\implies\Delta=1^2-4(1)(-2)\implies\Delta=9\\\\x=\dfrac{-b\pm\sqrt{9} }{2a} =\dfrac{-1\pm3}{2}\\\\x'= \dfrac{-1+3}{2}=1\\\\x''= \dfrac{-1-3}{2}=-2\\\\S=\{-2,\: 1\}$

b)  

$\sqrt[3]{x^2-x-4}=2\implies (\sqrt[3]{x^2-x-4})^3=2^3 \implies x^2-x-4=8\\\\\implies x^2-x-12=0\\\\\Delta=b^2- 4ac\implies\Delta=1^2-4(1)(-12)\implies\Delta=49\\\\x=\dfrac{-b\pm\sqrt{49} }{2a} =\dfrac{1\pm7}{2}\\\\x'= \dfrac{1+7}{2}=4\\\\x''= \dfrac{1-7}{2}=-3\\\\S=\{-3,\: 4\}$