[tex]\bf \textit{area of a square}\\\\ A=s^2~~ \begin{cases} s=&sides'\\ &length\\ \cline{1-2} A=&150 \end{cases}\implies 150=s^2\implies \sqrt{150}=s~~ \begin{cases} 150=&2\cdot 3\cdot 5\cdot 5\\ &2\cdot 3\cdot 5^2 \end{cases} \\\\\\ \sqrt{2\cdot 3\cdot 5^2}=s\implies 5\sqrt{2\cdot 3}=s\implies 5\sqrt{2}\cdot \sqrt{3}=s[/tex]
well then, we have a couple of known fellows, √2 and √3.
now, let's bear in mind that 2 and 3 are both prime numbers, a prime number is not divisible by anything but itself or 1, so we will never find two same-values that will give us either 2 or 3, namely, there's no exact root for √2 or √3, which means they're both irrationals, and therefore since they're factors of the answer, the answer is irrational.
