Question

Note: This is NOT a question. This is a tutorial.

How to take the square root of a number? See comment below to see how!

P.S. This is an algebra tutorial NOT a calculus tutorial, therefore linear approximation or any other method other than the simple method shall be discussed.

Answer 1

The first step in taking the square root of a number is to factor out the number into *prime* factors (i.e. 2, 3, 5, 7, 11, etc). For example, we are given \(\large \sqrt{48}\). The first step would be to factor into *prime* factors: \[\large \sqrt{48} = \sqrt{2 \times 2 \times 2 \times 2 \times 3}\] Now what? Now we take the numbers that repeated twice. In this case, there are two 2s so we take \(\mathbb{one}\) 2 out of the square root. That gives us: \[2\sqrt {\cancel{2 \times 2} \times 2 \times 2 \times 3}\] Now, if we look at what's left in the square root, you'll notice that there are two 2s again. so pull another 2 out of the square root and get: \[2(2) \sqrt{\cancel{2 \times 2 \times 2 \times 2} \times 3}\] Now, there are no longer any repeating numbers so our final answer becomes \(4 \sqrt 3\). But note that there has to be *two* of a kind to be able to pull a number out of the square root. For example we are given \[\sqrt{54}\] If we factor this out, we'll get \[\sqrt{54} = \sqrt{3 \times 3 \times 3 \times 2}\] There are *two* 3's in the root so we pull out *one* 3 out of the root. This gives us: \[3\sqrt{\cancel{3 \times 3} \times 3 \times 2}\] Note that there are no longer *any* two of a kind in the root so our final answer is just \(3\sqrt 6\)

Answer 2

What was the point of that?

Answer 3

lol that @LagrangeSon678 trolled me =_= hahaha

Answer 4

Or, there is always this way :D

Answer 5

lol :))) good one

Answer 6

i like joshes way better

Answer 7

how 'bout showing the algorithm for finding square roots by hand...

Answer 8

Slide rule!

Answer 9

Are you kidding me - now you're teaching square roots? Heights!

Answer 10

lol :P

Answer 11

@dpaInc good one - that would be good :D

Answer 12

2 more basic rules of square roots
\[\Huge \sqrt{a} \times \sqrt{b}=\sqrt{ab}\]
\[\Huge \sqrt{a}\times \sqrt{a}=a\]

Answer 13

guys could you please take a look at my question; i need your help