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So I'm going through the sample questions for the COMPASS test to see what I do and don't know so I can study for the test.
I never really got math in school, but I'm mostly caught up. One question is stumping me. I would look up the type of question and figure it out that way, but I'm not entirely clear on what kind it is. The question is:
8. Saying that 4 < x < 9 is equivalent to saying what about x ?
A. 0 < x < 5
B. 0 < x < 65
C. 2 < x < 3
D. 4 < x < 9
E. 16 < x < 81
There's a square root symbol over the x in the problem, but that didn't copy from the PDF.
My first thought is D, but the answer key says E. I could see that if the problem were 4^2 < sq(x) < 9^2, but it's not. Am I missing something, or is it not the sort of question I think it is?
The answer is indeed E, this is because in the question you have
4 < sqrt(x) < 9
as your question. so you need to do something to the equation to make it look like the equations in the answers. In this case, to remove a square root, you need to square everything.
So you square the first term and get 4^2 = 16
square the middle term and get sqrt(x)^2 = x
square the last term and get 9^2 = 81.
now if you place your new terms into your equation you get
I'm doing a review of squares and such since it looks like I missed some stuff, though that does make sense.
I'm not going to tag this thread (solved) just yet in case I have more questions.
Thanks.
For further notes - this isn't so much squares as it is the rules of less than / greater than signs, which are basically that you can add / subtract / etc anything** to one side as long as you do the same thing to the other side (** - if you multiply or divide by a negative number, then the sign flips, so 4 > 2, and 4 * 1 > 2 * 1, but 4 * -1 < 2 * -1; otherwise the sign will stay in the same direction for pretty much anything else that you'd be doing if you're at this level of math)
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to remove the square root you square each of them
y^2 < x < z^2
Since squaring the square root of x negates each other out.
I think I'm basing this off math I haven't touched in years.
4 < sqrt(x) < 9
as your question. so you need to do something to the equation to make it look like the equations in the answers. In this case, to remove a square root, you need to square everything.
So you square the first term and get 4^2 = 16
square the middle term and get sqrt(x)^2 = x
square the last term and get 9^2 = 81.
now if you place your new terms into your equation you get
16 < x < 81.
Which happens to be the answer E.
I hope this makes sense...
I'm not going to tag this thread (solved) just yet in case I have more questions.
Thanks.
For further notes - this isn't so much squares as it is the rules of less than / greater than signs, which are basically that you can add / subtract / etc anything** to one side as long as you do the same thing to the other side (** - if you multiply or divide by a negative number, then the sign flips, so 4 > 2, and 4 * 1 > 2 * 1, but 4 * -1 < 2 * -1; otherwise the sign will stay in the same direction for pretty much anything else that you'd be doing if you're at this level of math)