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!!!!!

2. 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....

3. 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

4. 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, ...).

5. yes! the first one I got right! 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!

6. 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!

7. Originally Posted by blair_alane
yes! the first one I got right! 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 $$\sqrt{7 \cdot 4}$$

8. 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!

Anyway, I hoped that helped EmmGee!