Simplifying Expressions with RadicalsDate: 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: 12*sqrt(640) ------------ 3*sqrt(5) (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 Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/