The solution to the given equation is x = 2 and x = 3
Given the equation;
- [tex]x = 2 +\sqrt{x-2}[/tex]
Subtract 2 from both sides;
[tex]x = 2 +\sqrt{x-2}\\x - 2= 2 - 2+\sqrt{x-2}\\x - 2 = \sqrt{x-2}\\[/tex]
Take the square of both sides:
[tex](x-2)^2 =(\sqrt{x-2} )^2\\(x-2)^2 = x-2\\(x-2)^2 - x+2 = 0\\(x-2)^2-1(x-2) = 0\\x-2[x-2-1] = 0\\(x-2)(x-3) =0\\x = 2 \ or \ 3[/tex]
Hence the solution to the given equation is x = 2 and x = 3
Learn more on rational equations here: https://brainly.com/question/4419050
