"How on Earth are you meant to calculate something with three nested square roots without a calculator in less than 20 minutes? I'm having trouble with the practice exam and I have a calculator and wolfram alpha and the solution sheet helping me do that."
unless you need ludicrous precision you basically only take the outer root "properly" and for the inner roots you just only take it only as far as you need without affecting the final answer. if you misjudge you'll probably be a bit off in your last sig fig but like. like if you have something like sqrt(sqrt(sqrt(2)+sqrt(3))+sqrt(sqrt(5)+sqrt(7))), you basically get okay, so sqrt 2 + sqrt 3 ~= pi, sqrt 5 + sqrt 7 ~= 4.9, sqrt pi (~= 1.77) + sqrt 4.9 (= 7/sqrt 10 ~= 2.21) ~= 3.98, sqrt of that ~= 1.995. the actual value is closer to 1.996, but yeah.
... it took me like ten minutes just to do that though so it probably isn't practical under exam conditions anyway, though. ... are you sure there isn't an easier way to do it? or maybe you're not actually expected to get such a precise answer?
unless you need ludicrous precision you basically only take the outer root "properly" and for the inner roots you just only take it only as far as you need without affecting the final answer. if you misjudge you'll probably be a bit off in your last sig fig but like. like if you have something like sqrt(sqrt(sqrt(2)+sqrt(3))+sqrt(sqrt(5)+sqrt(7))), you basically get okay, so sqrt 2 + sqrt 3 ~= pi, sqrt 5 + sqrt 7 ~= 4.9, sqrt pi (~= 1.77) + sqrt 4.9 (= 7/sqrt 10 ~= 2.21) ~= 3.98, sqrt of that ~= 1.995. the actual value is closer to 1.996, but yeah.
... it took me like ten minutes just to do that though so it probably isn't practical under exam conditions anyway, though. ... are you sure there isn't an easier way to do it? or maybe you're not actually expected to get such a precise answer?