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Simplifying Expressions with Radicals

Date: 05/26/98 at 18:51:02
From: Mallorie
Subject: multiplying, dividing, and simplifying radicals.

I am very confused. We're doing a chapter on square roots right now, 
and we have a chapter test Friday. Can you help me? I don't 
understand how to simplify a problem like 14 times the square root of 
320 divided by 2 times the square root of five. Can you explain how I 
go about doing a problem like this? Thank you so much!

Date: 05/27/98 at 13:21:03
From: Doctor Peterson
Subject: Re: multiplying, dividing, and simplifying radicals.

Hi, Mallorie. I'm going to assume that you have the general idea of 
simplifying radicals, but aren't quite confident enough to do these 
big ones. Let's try working through a problem like yours and see if we 
can break it down into smaller steps. I'll simplify:


(As you've probably figured out, "*" means multiply, and "sqrt" means 
square root.)

The first step is to factor the numbers inside the radicals and see 
what you can pull out:

   640 = 64 * 10 = 2^6 * 2 * 5 = 2^7 * 5

(Here "^" means "exponent".)

Look for paired factors -- anything with an even exponent -- and pull 
them aside, since any even power is a square. Here, I take six 2's (an 
even power, so that 2*2*2*2*2*2 = (2*2*2)*(2*2*2) = (2*2*2)^2 is a 
square) and leave the extra one separate:

   640 = 2^7 * 5 = 2^6 * 2 * 5

Now I can apply the rule:

   sqrt(a * b) = sqrt(a) * sqrt(b)

this way:

   sqrt(640) = sqrt(2^6 * 2 * 5) = sqrt(2^6) * sqrt(2) * sqrt(5) 
             = 2^3 * sqrt(2) * sqrt(5)

That last step used the fact that:

   sqrt(a^(2n)) = a^n

Now let's put it all together:

   12 * sqrt(640)   12 * 2^3 * sqrt(2) * sqrt(5)
   -------------- = ----------------------------
    3 * sqrt(5)             3 * sqrt(5)

Notice that you can cancel the sqrt(5)'s, and divide the 12 by 3, and 
we've eliminated the whole denominator. Thus:

   12 * sqrt(640)
   -------------- = 4 * 2^3 * sqrt(2)
    3 * sqrt(5)
                  = 4 * 8 * sqrt(2)
                  = 32 sqrt(2)

If you look back at what we've done, you'll see that we first 
simplified each radical by itself, breaking it apart into a product of 
prime sqrt's, then cancelled out what we could, and finally put it 
back together into the simplest form possible. Take it a step at a 
time, and it's not too bad.

By the way, do you know why we bother simplifying things like this? It 
seems like a lot of work, but it often saves a lot of work later. For 
one thing, now we don't have to figure out the square root of 640. You 
probably even know the square root of 2 already and can calculate this 
on paper. For another, it helps us recognize an expression better -- 
sort of like a mug shot at the police station, where we get it to look 
straight into the camera so its next victim can identify it. You'd 
never have guessed that big ugly expression was just our old friend 
sqrt(2) in disguise, would you?

-Doctor Peterson, The Math Forum
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Associated Topics:
High School Exponents

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