Question

Como soluciona de uma maneira mais fácil equações irracionais?

Answer 1

Olá Diana,

vou te dar alguns exemplos..

$\sqrt{x} -1=3\\ \sqrt{2x-1}=x-2\\ \sqrt{ \dfrac{1}{2}x }+5= \dfrac{1}{2}x+3$

$\sqrt{x} -1=3\\ \sqrt{x} =4\\ ( \sqrt{x})^2=4^2\\ x=16$

$\sqrt{2x-1}=x-2\\ ( \sqrt{2x-1})^2=(x-2)^2\\ 2x-1=x^2-4x+4\\ x^2-6x+5=0\\\\ x= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}= \dfrac{-(-6)\pm \sqrt{(-6)^2-4\cdot1\cdot5} }{2\cdot1}= \dfrac{6\pm \sqrt{16} }{2}\\\\\\ x= \dfrac{6\pm4}{2}\begin{cases}x'= \dfrac{6-4}{2}\Rightarrow x'= \dfrac{2}{2}\Rightarrow x'=1~~(nao~serve)\\\\ x''= \dfrac{6+4}{2}\Rightarrow x''= \dfrac{10}{2}\Rightarrow x''=5 \end{cases}\\\\\\ \huge\boxed{S=\{5\}}$


$\sqrt{ \dfrac{1}{2}x }+5= \dfrac{1}{2}x+3\\\\ \sqrt{ \dfrac{1}{2}x }= \dfrac{1}{2}x+3-5\\\\ \sqrt{ \dfrac{1}{2}x }= \dfrac{1}{2}x-2\\\\ \left(\sqrt{ \dfrac{1}{2}x }\right)^2=\left( \dfrac{1}{2}x-2\right)^2\\\\ \dfrac{1}{2}x= \dfrac{1}{4}x^2-2x+4\\\\ \dfrac{2}{\not4}x= \dfrac{1}{\not4}x^2- \dfrac{8}{\not4}x+ \dfrac{16}{\not4}\\\\ 2x=x^2-8x+16\\ x^2-8x-2x+16=0\\ x^2-10x+16=0$


$x^2-10x+16=0\\\\ x= \dfrac{-(-10)\pm \sqrt{(-10)^2-4\cdot1\cdot16} }{2\cdot1}\\\\\\ x= \dfrac{10\pm \sqrt{100-64} }{2}= \dfrac{10\pm \sqrt{36} }{2}= \dfrac{10\pm6}{2}\begin{cases}x'=2~~(nao~serve)\\ x''=8\end{cases}\\\\\\ \huge\boxed{S=\{8\}}$


Espero ter ajudado você, e tenha ótimos estudos ;D