Radicals

You know that when you have a fraction with a square root in the denominator, you have to rationalize the denominator (essentially, multiply the fraction by a fraction, equal to one, that will cancel the radical in the original fraction).  Like so:

$\dfrac{2}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$

Although sometimes, when we do a math problem, we end up with a radical in the denominator a fraction, we're not allowed to leave that radical in the denominator.

Why not?

Technically, fractions are supposed to have integers in the numerator and denominator. But mostly, it's customary to rationalize the denominator of fractions, in other words, in order for most math teacher to count an answer as "correct" there cannot be a radical in the denominator (although, interestingly, a radical in the numerator is allowed). 

An imaginary number is the square root of a negative number. How is that imaginary?  It's something that can happen when you're doing math. For instance, what if you have the equation:

$x^2=-16$

You'll square root both sides.

$\eqalign{x^2&=-16\\\sqrt{x^2}&=\sqrt{-16}\\x&=\sqrt{-16}}$

$x$ equals the square root of -16 is an answer that you could get.  But, it's not a mathematically possible number.  Why not?  What number, times itself, is equal to a negative number?

When you multiply a square root by itself, the square root disappears.

$\sqrt{7} \times \sqrt{7}=7$

This leads to a pattern when working with $i$ (which equals $\sqrt{-1}$):

$i^1=\sqrt{-1}$

$i^2=\sqrt{-1} \times \sqrt{-1} =-1$

$i^3=\sqrt{-1} \times \sqrt{-1} \times \sqrt{-1} =-\sqrt{-1}  $

$i^4=\sqrt{-1} \times \sqrt{-1} \times \sqrt{-1} \times \sqrt{-1}= -1 \times -1 = 1$

In math, you can divide just about everything, including radicals!

But, let's remember, we don't put fractions or decimals under radicals (in final answers), so we only divide radicals when the number that will end up under the radical is a whole number.

When radicals are divided by other radicals, you can divide the numbers and keep the product under the radical. If you have a hard time remembering this rule, you can always test the rule with square roots that you know. 

In math, you can multiply just about everything, including radicals!

Let's review what a radical is: a radical is a root.  If there's no superscript number in front of the radical, it's a square root (the 2 in front of the radical is assumed if there is no other number), which asks "what number, times itself, equals this number?"

Because a radical is a square root, a radical squared is the number under the radical:

$\sqrt{12}^2=(\sqrt{12})(\sqrt{12})=12$

You already know that when a number is under a radical (e.g., $\sqrt{49}$), you're supposed to find the square root of the number (the number that, when multiplied by itself, equals the number under the radical).

You also know that most numbers are not perfect squares, so it's nearly impossible for the human mind to calcluate most square roots.

However, we can simplify many square roots -- even if we can't calculate them precisely.

Finding the square root of a perfect square is easy.  What number, times itself, equals your number?

$$\eqalign{\sqrt{81}&=9\\\sqrt{400}&=20}$$

But, what do you do when you need to find the square root of a non-perfect square.

If you have a calculator, it's simple, you use the $\sqrt{\text{  }}$ button.

If you don't have a calculator, you need to estimate.  You won't get a perfect answer by estimating but you can come close. 

It's helpful to think of square roots as the opposites of squares.

The square of a number is what you get when you multiply a number by itself.  The square of $6 = 6^2 = 36$

The square root of a number is the number that you have to multiply by itself to get another number.  So, the square root of $36 = \sqrt{36} = 6$

$6 \times 6 = 36$ so the square root of $36 = 6$.