[tex]Solution, x-8>\sqrt{2x+1}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>3\left(3+\sqrt{2}\right)\:\\ \:\mathrm{Decimal:}&\:x>13.24264\dots \\ \:\mathrm{Interval\:Notation:}&\:\left(3\left(3+\sqrt{2}\right),\:\infty \:\right)\end{bmatrix}[/tex]
Steps:
[tex]\mathrm{Switch\:sides},\sqrt{2x+1}<x-8[/tex]
[tex]\mathrm{Find\:the\:real\:region\:for\:}\sqrt{2x+1}:\quad x\ge \:-\frac{1}{2}[/tex]
[tex]\mathrm{Find\:the\:values\:for\:}x-8>0:\quad x>8[/tex]
[tex]\mathrm{Simplify\:and\:compute\:}\sqrt{2x+1}<x-8:\quad x<3\left(3-\sqrt{2}\right)\quad \mathrm{or}\quad \:x>3\left(3+\sqrt{2}\right)[/tex]
[tex]\mathrm{Combine\:the\:following\:ranges:}[/tex] [tex]x\ge \:-\frac{1}{2}\quad \mathrm{and}\quad \:x>8\quad \mathrm{and}\quad \left(x<3\left(3-\sqrt{2}\right)\quad \mathrm{or}\quad \:x>3\left(3+\sqrt{2}\right)\right)[/tex]
The correct answer is [tex]x>3\left(3+\sqrt{2}\right)[/tex]
Hope this helps!!!
