The majority of the components I learnt this past few days were mostly review from grade 10 or what I learnt last week in math. Although, there were a few things I had to review for the test I had on Thursday. Such as, getting a better understanding on how to deal with decimal exponents.
For example, here is one problem involving an exponent that is a decimal. At first it may look intimidating but all you have to do is change the decimal into a fraction.
The decimal is 1.2, so you can change it into a fraction by putting it over 1 and multiplying the top and bottom by 10. This will change the fraction into 12/10 and you can further reduce it by 2 and it will be 6/5.
In this way, it will be much easier to solve the question, and you don’t have to work with decimals. After you have done that, you put the problem into a radical so you can solve it easier. You put the 6 on the 32 and the 5 to fifth root the 32^6.
The 5th root of 32 is 2 and 2^6 is 64. You also have to put the negative sign outside of the whole equation because it’s not included with the 32. If it were, then the negative sign would be placed inside the brackets with the 32 beforehand.
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I also learnt how to evaluate radicals that are in fraction form. So, if the fraction has 2 perfect squares than it can be square rooted and left into a fraction.
Here are 2 examples:
We have square root of 36/144, so since they’re both perfect squares it can be turned into 6/12. This fraction can also be further simplified into 1/2, if you divide top and bottom by 6.
For cube root of 64/125, you can cube root them separately and the answer will be 4/5 because they’re both perfect cubes.
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The final thing that I was able to reinforce was how to implement restrictions for an expression. For example, with every problem with variables there has to be some sort of restrictions depending on if you’re square rooting, cube rooting, etc.
In this problem, the variable “x” can be equal all real numbers because whatever number you square root when the exponent is even will always be positive.
a) (-2)^2 = 4 since (-2)(-2) = 4
b) (2)^2 = 4 since (2)(2) = 4
In this problem, it doesn’t matter if x is positive or negative the answer will always be positive. So, you have to put “x equals all real numbers” into the answer as well.
Although if you were to square root a variable that was to the odd power then it would need to have a restriction. Since the number inside the square root would be negative and you cannot square root, fourth root, sixth root, etc. a negative number. So, you have to put a restriction of “x has to be greater than or equal to 0.”
Here is an example:
Overall, this week in math wasn’t too bad because the majority of the concepts I had to re-learn didn’t take me a while to figure out.