• May 2nd 2008, 04:38 PM
EmmGee
i need to add 5 square root (7) and 3 square root (28). i can't figure out how to do it. please help me!!!!!
• May 2nd 2008, 04:56 PM
blair_alane
Quote:

Originally Posted by EmmGee
i need to add 5 square root (7) and 3 square root (28). i can't figure out how to do it. please help me!!!!!

First of all, the square root of 28= sq.root(7*4).....so it is also equal to 2 square root 7....so... 3 root 28 = 6 root 7.

so, 5 square root 7 + 6 square root 7 = 11 square root 7.

I dont know if my math is correct, but you just have to simplify your square roots.

Sorry for the sloppy math....
• May 2nd 2008, 05:01 PM
Jhevon
Quote:

Originally Posted by blair_alane
First of all, the square root of 28= sq.root(7*4).....so it is also equal to 2 square root 7....so... 3 root 28 = 6 root 7.

so, 5 square root 7 + 6 square root 7 = 11 square root 7.

I dont know if my math is correct, but you just have to simplify your square roots.

Sorry for the sloppy math....

It is correct (Clapping)
• May 2nd 2008, 05:02 PM
Mathnasium
In general, you can break square roots apart - for example, $\displaystyle \sqrt{44} = \sqrt{4}\times\sqrt{11}$. Of course, the square root of 4 is 2, so you get $\displaystyle 2\sqrt{11}$.

When you look to break apart larger numbers under a square root sign, you want to try to break them up into two factors where one of the factors is a perfect square (4, 9, 16, ...).
• May 2nd 2008, 05:12 PM
blair_alane
yes! the first one I got right! :D here...I'll write it again...

$\displaystyle 28=\sqrt(7*4)$.....so it is also equal to $\displaystyle 2\sqrt7$....so...
$\displaystyle 3\sqrt28=3*2\sqrt7=6\sqrt7$

so, $\displaystyle 5\sqrt7+6\sqrt7=11\sqrt7$

is that somewhat less confusing? (i found out how to do latex! teehee) I probably just made it worse! sorry! I dont know how to lay it out....

hope that helps! :)
• May 2nd 2008, 05:15 PM
blair_alane
Quote:

Originally Posted by Mathnasium
In general, you can break square roots apart - for example, $\displaystyle \sqrt{44} = \sqrt{4}\times\sqrt{11}$. Of course, the square root of 4 is 2, so you get $\displaystyle 2\sqrt{11}$.

When you look to break apart larger numbers under a square root sign, you want to try to break them up into two factors where one of the factors is a perfect square (4, 9, 16, ...).

you are totally right! :)
• May 2nd 2008, 05:19 PM
Jhevon
Quote:

Originally Posted by blair_alane
yes! the first one I got right! :D here...I'll write it again...

$\displaystyle 28=\sqrt(7*4)$.....so it is also equal to $\displaystyle 2\sqrt7$....so...
$\displaystyle 3\sqrt28=3*2\sqrt7=6\sqrt7$

so, $\displaystyle 5\sqrt7+6\sqrt7=11\sqrt7$

is that somewhat less confusing? (i found out how to do latex! teehee) I probably just made it worse! sorry! I dont know how to lay it out....

hope that helps! :)

type $$\sqrt{Anything}$$ to get $\displaystyle \sqrt{Anything}$.

so it would look nicer if you had $\displaystyle \sqrt{7 * 4}$ or better yet $\displaystyle \sqrt{7 \cdot 4}$ ..........use \cdot to get the dot. so the last one we got by typing $$\sqrt{7 \cdot 4}$$
• May 3rd 2008, 08:16 AM
blair_alane
Quote:

Originally Posted by Jhevon
type $$\sqrt{Anything}$$ to get $\displaystyle \sqrt{Anything}$.

so it would look nicer if you had $\displaystyle \sqrt{7 * 4}$ or better yet $\displaystyle \sqrt{7 \cdot 4}$ ..........use \cdot to get the dot. so the last one we got by typing $$\sqrt{7 \cdot 4}$$

teeheehee! Thank you, Jhevy! I shall remember this! :D

Anyway, I hoped that helped EmmGee! :)