Question

1 ⩽x ⩽ 5, for √(x³ + 36 )

Answer 1

`\bf ~\hspace{10em}1\le x\le 5~\hspace{5em}√(x^3+36) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ √(x^3+36)\implies \sqrt{x^(2+1)6^2}\implies √(x^2x6^2)\implies √((6x)^2x)\implies 6x√(x) \\\\\\ \stackrel{\textit{x = 4}}{6(4)√(4)}\implies 24√(2^2)\implies 24\cdot 2\implies 48`

that's how I read it.... to get some value between 1 and 5, namely 4, to make the expression a rational, well, 48 can be expressed as 48/1.