Thread: Math help - square roots?

1. Math help - square roots?

Hi!

I'm having some trouble with square roots -
I understand the square root of numbers like 49 (7), 9 (3) and all, though I'm very confused with finding the square root of numbers like 18, or 32.

Here is the example the teacher had:

http://i47.tinypic.com/23h0tbk.jpg

Could anyone possibly explain this in some way to me? What I don't understand is how you would even start a problem - like what actual math steps do you take to get those results, and how do you know what numbers to choose, and/or start with?
I'm very confused by this, if anyone could explain or show a better example - Like are there certain steps?
Any help is appreciated, as Math is not my strong subject! : /

2. Maybe it's easier to see it as:

$\displaystyle \sqrt{18}=\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3 \times\sqrt{2}=3\sqrt{2}$

and,

$\displaystyle \sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2} =6\sqrt{2}$

With practice you'll become better at recognising square numbers (like 9, 25, 36, and so on) and the multiples of these.

Larger square numbers can be harder to recognise than smaller ones, so you may sometimes find it easier to do it in steps:

$\displaystyle \sqrt{72}=\sqrt{4\times18}=2\sqrt{9\times2}=2\time s3\times\sqrt{2}=6\sqrt{2}$

3. Hi
Remember !
$\displaystyle \sqrt{a^2}=\left | a \right |$
let's take $\displaystyle \sqrt{18}$
decompose 18 :
$\displaystyle 18=2\times3\times3$.
therefore,
$\displaystyle \sqrt{18}=$$\displaystyle \sqrt{2\times3\times3}=\sqrt{2}\times \sqrt{3^2}=3\sqrt{2}.$

4. There are some very important formulas you need to learn by heart in order to understand square roots. They are :

$\displaystyle \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

$\displaystyle \sqrt{a \times b^2} = b \times \sqrt{a}$

Note that the second one can be proved from the first one :

$\displaystyle \sqrt{a \times b^2} = \sqrt{a} \times \sqrt{b^2} = \sqrt{a} \times b = b \times \sqrt{a}$

This way, if the expression under the square root can be broken into factors, and that some factors can be written as squares, then they can be "extracted" from the square root.

--------------------------------------

Let's take, $\displaystyle \sqrt{48}$ for the example. You want to simplify it.
First, break $\displaystyle 48$ into its prime factors, to make it easier to spot squares that you can remove from the square root.

Note that $\displaystyle 48 = 24 \times 2 = 2 \times 12 \times 2$ $\displaystyle = 2 \times 2 \times 6 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 3 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$

So, $\displaystyle \sqrt{48} = \sqrt{2^4 \times 3}$.

Now, remember $\displaystyle 2^4 = (2^2)^2$, so :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2 \times 3}$

And you see that there is a square. You can remove it :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2} \sqrt{3}$

And the $\displaystyle (2^2)^2$ square root is easily simplified :

$\displaystyle \sqrt{48} = \sqrt{16} \sqrt{3}$

$\displaystyle \sqrt{48} = 4 \sqrt{3}$

Do you understand better now ?

5. Originally Posted by Bacterius
There are some very important formulas you need to learn by heart in order to understand square roots. They are :

$\displaystyle \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

$\displaystyle \sqrt{a \times b^2} = b \times \sqrt{a}$

Note that the second one can be proved from the first one :

$\displaystyle \sqrt{a \times b^2} = \sqrt{a} \times \sqrt{b^2} = \sqrt{a} \times b = b \times \sqrt{a}$

This way, if the expression under the square root can be broken into factors, and that some factors can be written as squares, then they can be "extracted" from the square root.

--------------------------------------

Let's take, $\displaystyle \sqrt{48}$ for the example. You want to simplify it.
First, break $\displaystyle 48$ into its prime factors, to make it easier to spot squares that you can remove from the square root.

Note that $\displaystyle 48 = 24 \times 2 = 2 \times 12 \times 2$ $\displaystyle = 2 \times 2 \times 6 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 3 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$

So, $\displaystyle \sqrt{48} = \sqrt{2^4 \times 3}$.

Now, remember $\displaystyle 2^4 = (2^2)^2$, so :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2 \times 3}$

And you see that there is a square. You can remove it :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2} \sqrt{3}$

And the $\displaystyle (2^2)^2$ square root is easily simplified :

$\displaystyle \sqrt{48} = \sqrt{16} \sqrt{3}$

$\displaystyle \sqrt{48} = 4 \sqrt{3}$

Do you understand better now ?