Radicals

A radical is simply a root, like $\sqrt{72}$ or $\sqrt[3]{54}$. (The name comes from the Latin radix, "root," and the $\sqrt{\ }$ sign is called the radical sign.) All we're really doing here is tidying roots up so they're as small and clean as possible, which makes exact answers easy to compare. The rule $\sqrt{ab} = \sqrt a\sqrt b$ has two pieces: $\sqrt{ab}$ (the root of a product) and $\sqrt a\sqrt b$ (the product of the roots), and it says they're the same thing. That lets you split $\sqrt{72}$ into a part you can take the root of exactly and a part you can't. That's most of the lesson.

Where the rules come from. Write the root as an exponent, like last lesson: $\sqrt{ab} = (ab)^{1/2} = a^{1/2}b^{1/2} = \sqrt a\sqrt b$. That's just the rule $(ab)^n = a^nb^n$ wearing different clothes, and division works the same way.

Simplest form means three things: no perfect $n$th-power factors left under the root, no fractions under it, and no roots in a denominator. The point is that two people simplifying the same number end up writing it the same way, so you can compare answers at a glance.

Pulling out perfect squares. Look for the biggest perfect square hiding inside: $\sqrt{72} = \sqrt{36\cdot2} = \sqrt{36}\sqrt2 = 6\sqrt2$.

Rationalizing denominators. To get rid of $\sqrt a$ on the bottom, multiply top and bottom by $\frac{\sqrt a}{\sqrt a}$, which is just $1$. When the bottom is a sum like $3 - \sqrt2$, that doesn't work, but the conjugate $3 + \sqrt2$ does, because $(3 - \sqrt2)(3 + \sqrt2) = 9 - 2$ and the root disappears:

A common mistake. Since $\sqrt{ab} = \sqrt a\sqrt b$, it's tempting to think $\sqrt{a + b} = \sqrt a + \sqrt b$ too. It isn't. Try $a = 9$ and $b = 16$: $\sqrt{25} = 5$, but $\sqrt9 + \sqrt{16} = 7$. Roots behave like exponents, and exponents don't spread over addition.

Distances (the Pythagorean theorem), standard deviations and the period of a pendulum $2\pi\sqrt{L/g}$ are radicals; simplified forms make exact comparisons easy.