Dividing and Multiplying RadicalsDate: 03/07/99 at 09:31:50 From: Marta Subject: Radical Multiplication/Division I do not understand how to divide and multiply radicals. (e.g. square root of 3/4 times square root of 4/5). Another problem is dividing whole numbers and radicals (e.g. 3/(5-square root of 2) and (3 times the square root of 7)/(-1 - square root of 27)) Date: 03/07/99 at 14:35:30 From: Doctor Ezra Subject: Re: Radical Multiplication/Division First of all, the radicals you are asking about are just numbers that involve square roots of other numbers. Let us write sqrt to stand for 'the square-root of'. Rule 1. If you have two positive numbers x and y, then sqrt(x*y) = sqrt(x)*sqrt(y), and sqrt(x/y) = sqrt(x)/sqrt(y). That is, the square root of the product is the product of the square roots, and the square root of the quotient is the quotient of the square roots. That means that to figure out the product of the square root of a fraction with the square root of another fraction, just multiply the fractions and then take their square root. Here is an example that is a bit different from yours: sqrt(5/7)*sqrt(7/11) = sqrt((5/7)*(7/11)) = sqrt((5*7)/(7*11)) (you can cancel the 7's) = sqrt(5/11), which is as far as you can go. If you want to rationalize the denominator, just multiply top and bottom by sqrt(11): sqrt(5/11) = sqrt(5/11)*sqrt(11/11) = sqrt((5*11)/(11*11)) (note sqrt(11*11) = 11) = sqrt(55) / 11 For dealing with fractions with numbers such as 8 - sqrt(3), you have to rationalize the denominator first. If the denominator looks like x + sqrt(y), then multiply top and bottom by x - sqrt(y). The reason this works is that (x + sqrt(y)*(x - sqrt(y)) = x*x - sqrt(y)*sqrt(y) = x^2 - y, which does not have any radicals. So, for example, to figure out what this fraction is: (5 + sqrt(3))/(-7 + sqrt(11)) first multiply top and bottom by (-7 - sqrt(11)). That will give you ((5 + sqrt(3))*(-7 - sqrt(11)))/((-7)^2 -11) = (-35 - 5*sqrt(11) - 7*sqrt(3) - sqrt(33))/(49-11) = (-35 - 5*sqrt(11) - 7*sqrt(3) - sqrt(33))/38. Now you can apply these rules to your problems and get the answers. - Doctor Ezra, The Math Forum http://mathforum.org/dr.math/ |
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