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Algebra II, Rational Expressions & Square Roots problems

  1. Nov 23, 2016 #1
    Problem 1
    Simplify/solve:
    2*81/2-7*181/2+5*721/2-50

    Attempt at solution:
    a1/2=√a
    ⇒ 2*√8 - 7*√18 + 5*√72 - 50
    = 2√8 - 7√18 + 5√72 - 50 = ?

    Do not know how to proceed beyond this point. Have experimented with little luck.


    Problem 2
    Simplify/solve:
    a-1(1+1/a2)-1/2 * (1+a2)1/2

    Attempt at solution:
    a-1 = 1/a, 1-1/2 = √-1, (1/a2)-1/2 = √-1/a2
    ⇒ 1/a * √-1 + √-1 / a (the a is now simplified, the 1 is not) * √1 + a (the a is simplified, 1 is not)
    = ?


    Do not know how to proceed beyond this point. Have given up before experimenting.

    These are the problems I cannot solve. In fact, I cannot solve any problems of this type. I have no clue what I'm doing, just following certain rules I believe to be correct.
    These rules are described in my attempts at a solution at the top of the problem.
     
  2. jcsd
  3. Nov 23, 2016 #2

    fresh_42

    Staff: Mentor

    Why don't you try to factorize the numbers under the roots and express everything in terms of ##\sqrt{2}##?
    E.g. ##\sqrt{6}=\sqrt{2} \cdot \sqrt{3}##
     
  4. Nov 23, 2016 #3

    BvU

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    For problem 1 you could try ##\sqrt{ab} = \sqrt a \sqrt b## for a and b both positive.
     
  5. Nov 23, 2016 #4

    Merlin3189

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    Well, you know all the technical stuff you needed and it's easy now.
    You need to evaluate all the sq roots. You could use a calculator, but given the nature of the question (you could have put the original into a calculator!) you might look at the roots and see if you can simplify them. You seem to know a lot about powers and roots: do you know √(a x b) = √a x √b ?
    So factors might help.
    Here you go astray: 1-1/2 ≠ √-1 and I'm not sure where you got this from.
    So this went wrong as well.

    With fractional and negative indices, how about dealing with one thing at a time, just as you did in the first question?

    First, negative index means reciprocal ##a^{-n} = \frac{1}{a^n} ## or usefully ## \left(\frac a b \right)^{-n} = \left(\frac b a \right)^n ##
    See if you can make some progress using the rules a step at a time.
    Your stating the rules you used was very helpful in understanding what you were doing, thanks.
    "Just following rules" may be fine, but it sounds as if you may not have much confidence in them! If you need some explanation of these rules , which are perfectly sensible, not arbitrary, just ask next time.

    To me, these seem slightly complex problems for someone who is not already familiar and confident with powers and surds. Maybe you could practice on some simpler ones and build up to these?

    Anyhow, keep posting any questions on here. There are plenty of people happy to help.

    Edit: sorry, two other posts appeared while I was struggling with the Latex!
     
  6. Nov 23, 2016 #5

    epenguin

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    I think that most of the rules that you need to apply here boil down to

    a(n + m) = anab ... (1)
    (ab)n = anbn ... (2)
    and
    (a)nm = anm ...(3)

    If these are not sufficiently obvious, revise them in an elementary textbook - we cannot write the textbooks for you. You may have to use these either forward or backwards.
    You use these to drive towards the simplification you want.
    For example where you have (an1/n), this by (3) equals a1 = a.

    So using this on your 8½, look for a perfect square factor because this is going to simplify to 1 when you multiply the index 2 by ½:

    8½ = (4*2)½ (I picked out the factor that is a square)
    = (22*2)½ = (22)½*2½ by (2)
    = 1* 2½ by (3)
    = 2½
    And you have a couple more quite like this in that problem.
     
  7. Nov 24, 2016 #6

    epenguin

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    For the second problem, again there are two aspects - knowing and understanding the rules, but then also seeing how they can be applied to simplify your expression.

    Frequently we have a sum of two or more fractions Inside a bracket - example
    (1 + 1/a2). Usually there is not much we can do leaving it as a sum of fractions. What we have to do in such cases is express the thing in brackets as a single fraction - that is as one numerator divided by one denominator. The procedure is explained in algebra books, But it is exactly the same as you do in ordinary arithmetic e.g. in an expression like
    (1 + 1/7) - how do you express that as a single fraction? Do the same with the expression above, and you will find that is part of the entire expression it suggests something.
     
  8. Nov 24, 2016 #7

    BvU

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  9. Nov 24, 2016 #8

    epenguin

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    I had thought to mention that your #3 is an example of my eq. (2) but then I thought that was overloading.
     
  10. Nov 24, 2016 #9
    Thank you all for your assistance! I found it very useful!
    I discovered that the textbook I was using was actually meant for 4th years revising for their GCSE-type exams (it's a bit different in my country, but about equivalent in difficulty), since the exam covers just about everything from high school. In fact, I never even took rational exponents yet (and didn't even realize). I was experimenting with them to see if I could figure them out. Seems like I partly did, but I'll have to wait to actually study them rigorously.

    I messed up yesterday's test quite badly, but so did the rest of my class. I got a C-, which is better than I expected. However, when our teacher had us go over the problems again, I found that I solved most of them effortlessly, writing down the rules used in the process (so that I have a reference for later). So I suppose that means two things:

    a) I was doing problems way more advanced than the ones made for me.
    b) Experimenting with these problems strengthened my ability to solve the simpler alternatives.

    I'm feeling way more confident in my rules now, both thanks to today's lesson and your replies. Thank you all again!
     
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