Dividing RadicalsDate: 02/15/99 at 17:05:38 From: holly Subject: Dividing Radicals What is the process for doing solving problems like 7 sqrt32 -------- 5 sqrt63 ? Date: 02/15/99 at 17:34:33 From: Doctor Pat Subject: Re: Dividing Radicals To divide by a simple radical like the one in your problem, simply multiply both the numerator and denominator by the radical. This is called "rationalizing the denominator" and will leave a whole number in the denominator. Here is what it looks like: 7 sqrt32 7 sqrt(32)*sqrt(63) 7 sqrt(32*63) -------- = ------------------- = -------------- = 5 sqrt63 5 sqrt(63)*sqrt(63) 5 * 63 Now all you have to do is simplify. The 7 in the numerator will divide into the 63 in the denominator, and the sqrt (32*63) can be simplified also (HINT: look for perfect square factors). - Doctor Pat, The Math Forum http://mathforum.org/dr.math/ Date: 02/17/99 at 10:50:19 From: Doctor Bonnie Subject: Re: Dividing Radicals Let us first break down all the stuff that we are taking the square root of into primes: 7 sqrt(2*2*2*2*2) ----------------- 5 sqrt (3*3*7) So now we can do some actual work. Group the 2's in the numerator into groups and the 3's in the denominator: 7 sqrt{(2*2)*(2*2)*2} --------------------- 5 sqrt{(3*3)*7} Because all the stuff under the sqrt sign is multiplied, we can take the sqrt of them separately: 7 sqrt(2*2) sqrt(2*2) sqrt(2) ----------------------------- 5 sqrt(3*3) sqrt(7) So now we can take some sqrts: 7*2*2 sqrt 2 ------------ 5*3 sqrt 7 All we can do now is rationalize the denominator. This is not necessary, but it is sometimes fun to do, and it might lead to a bit simpler answer: (7*2*2 sqrt2)*(sqrt7) --------------------- (5*3 sqrt7)*(sqrt7) So we get: 7*2*2 sqrt 14 ------------- 5*3*7 And doing the multiplication and canceling: 4 sqrt 14 --------- 15 So I think this is about as pretty as you are going to get it. How about you try some? 3 sqrt 8 ---------- 11 sqrt 44 and: 2 sqrt 18 --------- 3 sqrt 12 - Doctor Bonnie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/