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Need help real bad math

  1. Apr 7, 2003 #1
    the following two questions are in math..and i have no clue how to even start..
    i got them in a test..
    i would appreciate it if u could tell me what i should do , or what concept is involved...

    1) if a,b,c belong to the set of all real no.s and a!=b(not equal to)
    then

    a)a^2+b^2+ab=3c^2 for some c in (a,b)
    b)a^2+b^2+ab=3c^2 for not even one c in (a,b)
    c)a^2+b^2-ab=3c^2 for some c in (a,b)


    2)the no. of non-negative integral solutions for the eqn.
    2x+y+z=20 are
    a)132
    b)121
    c)144
    d)none of the above

    thanks in advance
     
  2. jcsd
  3. Apr 8, 2003 #2
    1) Take a=0,b=1
    a^2+b^2+ab=1;3*c^2=1 => c is in (a,b);
    Correct answer : a)

    2) x can be 0,1,2,3,...10

    y+z can be 20,18,16,.....4,2,0;
    y can take 21,19,17,.....5,3,1 values for every y+z;
    So 21+19+17+...+3+1=121...
    Correct answer : b)


    [?]
     
  4. Apr 8, 2003 #3

    HallsofIvy

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    You have to be a little more careful, Bogdan. Showing that there exists such a c for a=0 b= 1 doesn't show that there is such a c for any a,b.

    (a+b)2= a2+ 2ab+ b2 so

    a2+ b2+ ab= (a+b)2- ab

    If (a+b)2> ab (a=1, b=0 for example then the right side is positive and there exist c such that 3c2 is equal to it.

    If (a+b)2< ab (example, a= 1, b= -1) then the right hand side is negative and there is no such c (3c2 can't be negative).

    The correct answer is b.
     
  5. Apr 10, 2003 #4
    Sorry but .... how did u reach this conclusion ?
    I haven't reached any answer yet, but i have an approach.
    x => 0
    y => 0
    z => 0
    Let's first look at z.
    Rearrange the original equation.
    2x+y+z=20
    z=20-2x-y
    Therefore as x and y get smaller, z will get bigger.
    So a the lower bound of x and y, you will have the upper bound of z.
    The lower bound of x is 0, the lower bound of y is 0 too.
    So the upper bound of z is 20-(2*0)-(0) = 20
    So now we have a smaller range of z.
    0 <= z <= 20
    Do the same for x and y, you will reah the following conclusions.
    0 <= x <= 10
    0 <= y <= 20
    And ... i am still thinking of remaining (here is a clue, for each value of x, a certain value of y+z will be determined, just find the number of posibilities for y+z on each value of x (of course not by trial and error :wink: )).
    I will work on it and reply ASAP.
     
  6. Apr 10, 2003 #5
    121

    i get 121 for the last question.
     
  7. Apr 11, 2003 #6
    Maybe I'm wrong...
    But b) can't be (busta rhymes...) the correct answer...
    I demonstrated that for a=0, b=1,c=sqrt(1/3)...so there is a c in (a,b)...there is an example...mine...
    I think the problem is if the answer is a or c...because a=0 b=1 makes a and c correct...take a=2,b=3...and you'll probably find out that a is the correct answer...
    Correct me if I'm wrong...
    mercury, do you have the correct answers ?
     
  8. Apr 12, 2003 #7
    I was finally able to continue my method (which i started in the last post) to find the answer of the question (of the second question).
    The answer is 121.
    Let me show you how ...
    From the last post, i showed that
    0 <= x <= 10
    now suppose you wanted to solve the question by trial (try each value of x, and find the possible values of y and z at each value of x), you will get something like this.
    (please note that if you are given a value for x, and a value of y, there will be only a single value for z that satisfies the equation, therefore it is not needed to list the values of z)
    > x=10 (y+z=0)
    >> only one posibility, which is that y = 0
    > x=9 (y+z=2)
    >> posibility one : y=0
    >> posibility two : y=1
    >> posibility three : y=2
    > x=8 (y+z=4)
    >> posibility one : y=0
    >> posibility two : y=1
    >> posibility three : y=2
    >> posibility four : y=3
    >> posibility five : y=4
    ... so on ...
    so far you have the following answers
    (x,y,z)

    (10,0,0)

    (9,0,2)
    (9,1,1)
    (9,2,0)

    (8,0,4)
    (8,1,3)
    (8,2,2)
    (8,3,1)
    (8,4,0)

    ... so on ...
    Now you may see that the number of possibilities of y and z for each value of x can be expressed as
    P=21-2x
    (this is very logical **)
    So the number of answers is
    [sum](from x=0 to x=10)(21-2x)
    =21*(10+1)[sum](from x=0 to x=10)(-2x)
    =21*11 -2 * [sum](from x=0 to x=10) (x)
    =21*11 -2*(10+1)*(10/2)
    =231-2*(11)*(5)
    =231-110
    =121

    Any comments ?
     
  9. Apr 16, 2003 #8

    Hurkyl

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    Gold Member

    If (a+b)2 < ab (example, a= 1, b= -1)

    Slight problem with this statement... the left hand side of this inequality is 0 while the right hand side is negative. It doesn't hold!


    a) is correct for problem 1:
    (without loss of generality, let a < b)

    case 1: 0 <= a < b

    notice that 3a^2 < a^2 + b^2 + ab < 3b^2, because:

    a^2 = a^2 < b^2
    a^2 < b^2 = b^2
    a^2 < ab < b^2

    Since 3c^2 is a continuous function, the intermediate value theorem says that there is a c such that:

    3c^2 = a^2 + b^2 + ab

    case 2: a < b <= 0

    This time, 3b^2 < a^2 + b^2 + ab < 3a^2, and the same argument holds.

    case 3: a < 0 < b

    3a: |a| < |b|

    3*0^2 < a^2 + b^2 + ab < 3b^2

    Therefore there is a c in (0, b) for which equality is satisfied, and thus c is in (a, b)

    3b: |b| < |a|

    3*0^2 < a^2 + b^2 + ab < 3a^2

    Therefore there is a c in (a, 0), and thus c is in (a, b)

    3c: |a| = |b|

    a^2 + b^2 + ab = b^2 + b^2 - b^2 = b^2

    Setting c = b / sqrt(3) yields equality, and c is in (a, b)

    Hurkyl
     
  10. Apr 16, 2003 #9
    Thank you Hurkyl for the time spent proving rigorously the first problem...as I said...it was a)
     
  11. Apr 17, 2003 #10
    hey people...thanks loads for answering my question and taking a lot of pains to explain it in detail...i have learned a lot of new things..
    i suppose that's what this forum is for ...?

    anyway i appreciate it..

    cheers and peace

    mercury
     
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