Okay, so it took me a few videos to understand what exactly we were supposed to be doing(your teacher wasn't exactly a ball of fire oop), but I pretty much get it now.
So basically the goal of simplification is to translate as much of the raticand as possible into whole numbers. Right now, most of the raticands are imperfect squares-- numbers whose square roots aren't whole numbers. If we ran these questions through a calculator, we'd get an answer like 17.88854382, which isn't super easy to manipulate.
Instead, we want to break that up into two parts-- the whole number and a much smaller radical. To do this, we need to find the largest factor of each raticand that's a perfect square, which is a number who has a square root of a whole number(for example 25, whose square root is 5).
For example, lets find the largest perfect square of the first problem: (I'm sure there's a better way to do this, but I just try dividing the number by 2, 3, 4, and so on until I find a perfect square)
[tex]320 \div 2 = 160( \sqrt{160} = 12.65)(imperfect) \\ 320 \div 3 = 106.67( \sqrt{106.67} = 10.32)(imperfect) \\ 320 \div 4 = 80( \sqrt{80} = 8.94)(imperfect) \\ 320 \div 5 = 64( \sqrt{64} = 8)(perfect)[/tex]
So now we know that
[tex] \sqrt{320} [/tex]
is the same as
[tex] \sqrt{64} \times \sqrt{5} [/tex]
Finally, we take the square root of 64(in this case 8) and append it to the front of the other radical, like so:
[tex]8 \sqrt{5} [/tex]
and there we go! We have successfully simplified the radical.
I want you to try this out now, and tell me if it makes sense. If it doesn't, tell Gem and I can try to clarify for you on zoom. Sorry for the delay buck, but I hope this is what you needed!
