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Confusing answers.

  1. Apr 25, 2004 #1
    I am currently reviewing my whole algebra textbook for a CAT6 test on tuesday. Apparently, it is very difficult, so I'm taking every minute to review.
    Anywho, I ran into 3 problems that are currently driving me crazy. :tongue:

    One concerns roots.
    Here is the problem the book presents:
    (6x2x)x1/2

    My answer was (36x3)1/2. But the book says the answer is (6x)5/2

    BTW, how do you write root signs using the latex program?

    another root problem:

    y3/y1/4

    My answer was x3/4. The book says x11/4 is the answer.

    Now, here is a factoring problem:
    -4(x+16)4 + 9(x+16)2 + x + 16

    On this problem, the part that confuses me is the answer:
    [-4(x+16)3 + 9x + 145) (x+16)

    I arrived at this answer on my own, except I went one step further:
    when I got to [-4(x+16)3 + 9x + 145) (x+16) , I went ahead and rewrote (x+16)3 as (x+16)(x+16)2 so that I could distribute the -4. I then arrived at (x+16)3 (5x+81). Apparently, my final step was incorrect. I would like to know why.

    Also, I did most of these operations in my head (as it is easier for me to do so), so if you would like me to explain anything that I did, I should point out that I may sound unmathematical in a sense.

    Thank you for your time.
     
  2. jcsd
  3. Apr 25, 2004 #2
    The book is wrong on the first problem, and so are you. ;) Remember the rule that says [tex]a^xa^y = a^{x + y}[/tex]. This rule works even when x or y are fractions.

    [tex](6x^2x)x^{1/2} = (6x^3)x^{1/2} = 6x^3x^{1/2} = 6x^{3 + 1/2} = 6x^{7/2}[/tex]

    For the second problem, remember [tex]\frac{a^x}{a^y} = a^{x - y}[/tex], even when x or y are fractions.

    [tex]\frac{y^3}{y^{1/4}} = y^{3 - 1/4} = y^{11/4}[/tex]
     
  4. Apr 25, 2004 #3

    matt grime

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    And the last part on factorization. It's just wrong. You cannot pull a factor of x+16 out of the expression again after you've simplified it once. What do you mean to do in "distributing the -4"? You have done something that is against the laws of arithmetic and algebra, that's all.
     
  5. Apr 25, 2004 #4
    Muzza: Thank you for your kind explanation. It seems to me that from now on, it is far more efficient to calculate roots exponentially; without using a radical.

    Mattgrime: When I said "distribute the -4", I meant applying the distributive property.

    But I see the mistake now. They weren't like terms? (I am referring to the factoring problem)
     
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