Evaluate roots of perfect squares/cubes and simplify basic radicals.
Free, timed SHSAT practice sets on square & cube roots, radicals— every question includes a full worked explanation, and there's no account required.
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A root undoes a power. \(\sqrt{144} = 12\) because \(12^2 = 144\), and \(\sqrt[3]{729} = 9\) because \(9^3 = 729\).
Know these by heart so you can answer fast:
To estimate a root that is not a whole number, find the two perfect squares on each side. For \(\sqrt{137}\): \(11^2 = 121\) and \(12^2 = 144\). 137 is much nearer to 144, so \(\sqrt{137} \approx 12\).
For a fraction, take the root of the top and of the bottom: \(\sqrt[3]{\frac{8}{125}} = \frac{2}{5}\), because \(2^3 = 8\) and \(5^3 = 125\).
Common traps:
Take the square root of the top and of the bottom: \(2^2 = 4\) and \(3^2 = 9\), so \(\sqrt{\frac{4}{9}} = \frac{2}{3}\). Taking the root of the top only gives \(\frac{2}{9}\). Taking the root of the bottom only gives \(1\frac{1}{3}\). Forgetting to take the square root gives \(\frac{4}{9}\).
\(18^2 = 324\), so \(\sqrt{324} = 18\). Dividing 324 by 2 instead of taking the square root gives \(162\). Squaring 324 instead of taking the square root gives \(104976\). Forgetting to take the square root gives \(324\).
\(16^2 = 256\) and \(17^2 = 289\), and 260 falls between them, closer to 256. So \(\sqrt{260} \approx 16\). Rounding to the wrong whole number (\(17^2 = 289\) is farther from 260) gives \(17\). Dividing 260 by 2 instead of taking the square root gives \(130\). Giving the nearest perfect square, 256, instead of its square root gives \(256\).