Question

X-8 >
`√(2x+1)`

Answer 1

`Solution, x-8>√(2x+1)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>3\left(3+√(2)\right)\:\\ \:\mathrm{Decimal:}&\:x>13.24264\dots \\ \:\mathrm{Interval\:Notation:}&\:\left(3\left(3+√(2)\right),\:\infty \:\right)\end{bmatrix}`

Steps:

`\mathrm{Switch\:sides},√(2x+1)<x-8`

`\mathrm{Find\:the\:real\:region\:for\:}√(2x+1):\quad x\ge \:-(1)/(2)`

`\mathrm{Find\:the\:values\:for\:}x-8>0:\quad x>8`

`\mathrm{Simplify\:and\:compute\:}√(2x+1)<x-8:\quad x<3\left(3-√(2)\right)\quad \mathrm{or}\quad \:x>3\left(3+√(2)\right)`

`\mathrm{Combine\:the\:following\:ranges:}`
`x\ge \:-(1)/(2)\quad \mathrm{and}\quad \:x>8\quad \mathrm{and}\quad \left(x<3\left(3-√(2)\right)\quad \mathrm{or}\quad \:x>3\left(3+√(2)\right)\right)`

The correct answer is
`x>3\left(3+√(2)\right)`

Hope this helps!!!