Answer :
[tex]\sqrt{69}\ is\ irrational\ number\\\\69=3\times23\\\\therefore\ \sqrt{69}=\sqrt{3\times23}=\sqrt3\times\sqrt{23}\\\\\sqrt3-not\ exist\ in\ the\ set\ of\ rational\ numbres\\and\\\sqrt{23}-not\ exist\ in\ the\ set\ of\ rational\ numbers\\\\You\ can\ only\ rounding\ this\ number:\sqrt{69}\approx8.307[/tex]
