In today's class, we somehow fit two lessons in one day simply because I had to remind her that we didn't have math class tomorrow. Urgh... **groan** The mistake is mine, but the bad is not mine. Anyhow...in the first part of class, we learned about multiplying and dividing radicals. When multiplying radicals, here is one rule, or example to follow.
That's easy enough to do though isn't it? Well of course, that example was just a review of what we've been doing in the previous days. It gets a little more complicated than that. Today, we also learned how to multiply radical binomials. Now this is pretty similar to the things we did in the first unit we did this year on polynomials and factoring them. In the following I will explain, [or try to...] the concept in a step-by-step process in the following example:
Simplify: 6√2 (√6 – 2√4)
1.) In order to simplify this, we must distribute the both the radical and the....oh...the other number, I don't quite know what the exact term is, but in this case it's the 6 right before the square root of 2. You multiply that with the same type of number of both sides of the binomial and then multiplying the radicals with the radicals resulting in...
2.) After getting that answer, simplify the expression even more throughly, breaking it up bit-by-bit into more parts so that we are able to multiply it later on. [SiMpLiFy UnTiL yOu CaN nO lOnGeR sImPlIfY!!!]
But when you are stuck in the situation where you have a radical binomial in the denominator, then it goes something like...
It is okay to have a radical in the numerator, but only then are numbers in their simplest form is when they do not have both a negative exponent and a radical in the denominator.
However, if there is a binomial in the denominator, the "difference of squares" must be multiplied. [REVIEW of DIFFERENCE OF SQUARES: Ex. (2x-3)(2x+3) = 4x2 - 9]
An example of this might be...
[click on the box again. I don't know why the other diagram is goofing up.]