Simplify Square Roots

Hockeyman

Junior Member
Joined
Dec 8, 2005
Messages
79
Hello everybody,

Can someone help to explain this to me? My teacher gave me this example and I'm not sure how he got the answer.

square root of 72

=square root of 9 multiplied by square root of 8

=3 outside square root and 8 inside which

= 3 outside square root and 4 inside multiplied by the square root of 2

= 6 outside square root and 2 inside

Can someone explain how he got it?

P.S. How do you write the square root signs? :D
 
Do you know the radical properties?.

\(\displaystyle \sqrt{ab}=\sqrt{a}\sqrt{b}\)

\(\displaystyle \sqrt{72}=\sqrt{(9)(8)}=\sqrt{9}\sqrt{8}=3\sqrt{8}\)

Also, \(\displaystyle \sqrt{72}=\sqrt{(36)(2)}=\sqrt{36}\sqrt{2}=6\sqrt{2}\)
 
No I'm afraid I don't, can you explain how you got from square root of 9 multiplied by square root of 8 to get to your answer?
 
Hockeyman said:
square root of 72

=square root of 9 multiplied by square root of 8

=3 outside square root and 8 inside which

= 3 outside square root and 4 inside multiplied by the square root of 2

= 6 outside square root and 2 inside

Can someone explain how he got it?

P.S. How do you write the square root signs? :D

OK, so you're looking for the squareroot of 72. I will go back through these steps and explain what your teacher has done.
. . .
\(\displaystyle \sqrt{72}\)
. . .
So, we start with this. Now, where do we go? Well, when you have an odd number like this that you aren't able to take a regular square root from, you have to break the number down. 72 is the same thing as 9 times 8, correct? So, in essence. . .
. . .
\(\displaystyle \sqrt{72} = \sqrt{(9)(8)} = (\sqrt{9})(\sqrt{8})\)
. . .
Now, you have two numbers under square roots. So, we can see that 9 has a simple square root.
. . .
\(\displaystyle \sqrt{9} = 3\)
. . .
See what I did? Ok, so we took the square root of 9, now where does the 3 go? Well, since the squareroot of 9 was originally multiplied by the squareroot of 8, and then we took the squareroot of 9, we simply multiply that by the squareroot of 8. . . like so:
. . .
\(\displaystyle 3\sqrt{8}\)
. . .
So, that's what your teacher did first. Now, how do we get to the next step? Well, you can't take the squareroot of 8, but you can break 8 down like we did with 72. So, 8 is the same thing as 4 times 2, correct?
. . .
\(\displaystyle \sqrt{8} = \sqrt{(4)(2)} = (\sqrt{4})(\sqrt{2})\)
. . .
Now, we can see that you can easily take the square root of 4, and that's 2. So, what do we do with that 2? Well, we take the two and multiply it by the three that is already on the outside. Like this:
. . .
\(\displaystyle (3)(2)(\sqrt{2}) = 6\sqrt{2}\)
. . .
Do you understand this now? If not, please just write back and tell me what you're having problems with still. I hope this helps!
. . .
. . .
BTW, you can find out how to do any kind of math symbold on here by checking out the LaTex guide. Just scroll your mouse over the button that says "Forum Help" at the top of the screen, then click on a LaTex guide and read through it to find what you need.
 
Your name is awesome, guess you like hockey and maths... just like me :lol:
 
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