Chapter 1 - 1.6 - Other Types of Equations - 1.6 Exercises - Page 128: 37

$x=1$

$\sqrt {x+8}=2+x$ Squaring both sides of the equation... $x+8=x^2+4x+4$ $0=x^2+3x-4$ $x^2-x+4x-4=0$ $x(x-1)+4(x-1)=0$ $(x+4)(x-1)=0$ $x=-4$ or $x=1$ Check: For $x=1$ $\sqrt {9}=3, 3=3$ For $x=-4$ $\sqrt {4}=2-4, 2=-2$. Thus, $x=-4$ is an extraneous solution The solution is $x=1$.