Respuesta :
[tex]\bf \sqrt{x+2}+4=8\implies \sqrt{x+2}=8-4\implies \sqrt{x+2}=4
\\\\\\
x+2=16\implies x=16-2\implies x=14\impliedby
\begin{array}{llll}
doesn't\ look\\
extraneous
\end{array}[/tex]
Consider the equation:
[tex]\sqrt{x+2}+4 = 8[/tex]
Subtracting '4' from both the sides of the equation, we get as
[tex]\sqrt{x+2}+4-4= 8-4[/tex]
[tex]\sqrt{x+2}= 4[/tex]
Squaring on both the sides of the equation, we get
[tex](\sqrt{x+2})^2 = (4)^2[/tex]
[tex]x+2 = 16[/tex]
Subtracting '2' from both the sides of the equation, we get
[tex]x+2-2=16-2[/tex]
x=14
Since, An extraneous solution is a solution that arises from the solving process that is not really a solution at all. But, in this equation x=14 is the solution of the given equation.
Hence, it is not an extraneous solution.
