Question

√x+2 = √3x+4 -2 vou botar qualquer coisa só pra dar 20 caracteres ​

Answer 1

A idéia para resolver esta equação irracional é extrair a raiz quadrada dos membros para retirar o radical, e assim isolar x encontrando seu valor:

$\begin{array}{l} \\ \sf \sqrt{x+2}=\sqrt{3x+4}-2\\\\\sf (\sqrt{x+2})^2=(\sqrt{3x+4}-2)^2\\\\\sf x+2=(\sqrt{3x+4})^2-2\cdot\sqrt{3x+4}\cdot2+(2)^2\\\\\sf x+2=3x+4-4\sqrt{3x+4}+4\\\\\sf x+2=3x+8-4\sqrt{3x+4}\\\\\sf 2=3x+8-4\sqrt{3x+4}-x\\\\\sf 2=2x+8-4\sqrt{3x+4}\\\\\sf 0=2x+8-4\sqrt{3x+4}-2\\\\\sf 0=2x+6-4\sqrt{3x+4}\\\\\sf4\sqrt{3x+4}=2x+6\\\\\sf \dfrac{4\sqrt{3x+4}}{2}=\dfrac{2x+6}{2}\\\\\sf 2\sqrt{3x+4}=x+3\\\\\sf (2\sqrt{3x+4})^2=(x+3)^2\\\\\sf 4(3x+4)=(x)^2+2\cdot x\cdot3+(3)^3\\\\\sf 12x+16=x^2+6x+9\\\\\sf x^2+6x+9-12x-16=0\\\\\sf x^2-6x-7=0\end{array}$

Por fatoração:

$\begin{array}{l}\sf x^2+x-7x-7=0\\\\\sf x(x+1)-7(x+1)=0\\\\\sf (x+1)\cdot(x-7)=0\\\\\begin{cases}\sf x+1=0\\\\\sf x-7=0\end{cases}\\\\\begin{cases}\sf x'=-1\\\\\sf x''=7\end{cases}\\\\\end{array}$

Fazendo as verificações:

$\begin{array}{l}\begin{cases}\sf Para~x=-1\\\\\sf Para~x=7\end{cases}\\\\\begin{cases}\sf \sqrt{-1+2}=\sqrt{3(-1)+4}-2\\\\\sf \sqrt{7+2}=\sqrt{3(7)+4}-2\end{cases}\\\\\begin{cases}\sf \sqrt{1}=\sqrt{-3+4}-2\\\\\sf \sqrt{9}=\sqrt{21+4}-2\end{cases} \\ \\ \begin{cases}\sf 1=\sqrt{1}-2\\\\\sf 3=\sqrt{25}-2\end{cases} \\ \\ \begin{cases}\sf1=1-2~\to~1\,\neq\,-1\\\\\sf 3=5-2~\to ~3=3\end{cases} \\ \\ \end{array}$

Portanto, nesta equação irracional x = 7

ㅤ

Att. Nasgovaskov