Results 1 to 9 of 9

Math Help - Alegbra

  1. #1
    Junior Member
    Joined
    Oct 2005
    Posts
    37

    Smile Alegbra

    well, I've got some problems that I can't solve, I'm doing the work all the day but I just can't solve those... I would appreciate your help even if you can help just in one...
    by the way I need it for wednesday so it's pretty urgent...

    formula: ax^2+bx+c(trinom)
    1.
    6x^3+x^2-x+12x^2+2x-2

    2.
    30x^3-15x^2-14x^2+7x-4x+2


    prove:

    (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9

    (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400

    prove that:

    (n^2+2n)(n^2-1)
    ------------------
    24

    and last one:
    given:
    a+b=5
    ab=3
    How much is:
    a^2+b^2
    a^2b+ab^2
    a^3b+ab^3

    Thanks a lot for everyone who can help, even one of all these...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by dgolverk
    well, I've got some problems that I can't solve, I'm doing the work all the day but I just can't solve those... I would appreciate your help even if you can help just in one...
    by the way I need it for wednesday so it's pretty urgent...

    formula: ax^2+bx+c(trinom)
    1.
    6x^3+x^2-x+12x^2+2x-2

    2.
    30x^3-15x^2-14x^2+7x-4x+2


    prove:

    (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9

    (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400

    prove that:

    (n^2+2n)(n^2-1)
    ------------------
    24

    and last one:
    given:
    a+b=5
    ab=3
    How much is:
    a^2+b^2
    a^2b+ab^2
    a^3b+ab^3

    Thanks a lot for everyone who can help, even one of all these...
    Hello,

    I cann't do all your problems 'cause I haven't much time now.
    to 1. 6x^3+x^2-x+12x^2+2x-2=
    x(6x^2+x-1)+2(6x^2+x-1)
    (x+2)(6x^2+x-1) now you can factorize the 2nd factor easily.

    to 2. 30x^3-15x^2-14x^2+7x-4x+2=
    15x^2(2x-1)-7x(2x-1)-2(2x-1)
    (15x^2-7x-2)(2x-1) now you can factorize the 1rst factor easily.

    (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9
    set x=7a-8 and y=11-7a. Then you get:
    x^2+2xy+y^2=9\  \Longleftrightarrow \ (x+y)^2=3^2 Solve for x and y and then resubstitute to get a and b.

    (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400 Same procedure: x= 11a+5 and y=11a-15 and you'll get:
    x^2-2xy+y^2=400\  \Longleftrightarrow \ (x-y)^2=20^2

    Now I've to leave the forum. I hope this was of some help to you.

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by dgolverk

    prove that:

    (n^2+2n)(n^2-1)
    ------------------
    24
    I assume that you want to prove that is dividable with 24, so:
    \frac{{(n^2  + 2n)(n^2  - 1)}}{{24}} = \frac{{(n - 1)n(n + 1)(n + 2)}}{{24}}

    So, n can be any number 24*x,(x \in Q,x \ne 0) and it can be n=2 and n=-3 because 24=1*2*3*4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    given:
    a+b=5
    ab=3
    How much is:
    a^2+b^2
    a^2b+ab^2
    a^3b+ab^3
    \begin{array}{l}<br />
 (a + b)^2  = 5^2  \\ <br />
 a^2  + 2ab + b^2  = 25 \\ <br />
 a^2  + b^2  = 25 - 2ab \\ <br />
 a^2  + b^2  = 25 - 2ab \\ <br />
 a^2  + b^2  = 25 - 2*3 \\ <br />
 a^2  + b^2  = 19 \\ <br />
 \end{array}<br />

    \begin{array}{l}<br />
 a^2 b + ab^2  = ab(a) + ab(b) \\ <br />
 3a + 3b = 3(a + b) = 3*5 = 15 \\ <br />
 \end{array}<br />

     a^3 b + ab^3  = ab(a^2 ) + ab(b^2 )
     3a^2  + 3b^2  = 3(a^2  + b^2 ) = 3*19 = 57
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let me just explain DenMac's post that
    \frac{n(n+1)(n-1)(n+2)}{24} is an integer.
    Because 24=4\times 3\times 2\times 1.
    That means that one of the integers,
    n,n+1,n-1,n+2 contains a multiple of 4
    because there are 4 of them in increasing order. Similarly one of them is also a multiple of 3 and finally a multiple of 2. Thus,
    n(n+1)(n-1)(n+2) is a multiple of 24.
    -------------------------
    If you want a more advanced explanation (I do not know if you learned this thus I do not want to post this one) is through congruences:
    Let x=n-1
    Then, x+1=n, x+2=n+1, x+3=n+2.
    Thus, (under modulo 24)
    x\equiv r_1
    x+1\equiv r_2
    x+2\equiv r_3
    x+3\equiv r_4
    Now non-of these are congruent to each other for then,
    x+i\equiv x+j
    Thus,
    i\equiv j and 0\leq i,j\leq 3
    Which is impossible.
    Thus, the integers r_1,r_2,r_3,r_4 must be the integers 1,2,3,4 in some order by the pigeonhole principle. Thus, r_1r_2r_3r_4=24.
    Thus, upon multiplication of the congruences we have that
    x(x+1)(x+2)(x+3)\equiv 24\equiv 0
    Q.E.D.
    -----------------------
    Finally, there is a theorem from number theory that the product of n-consecutives integers is always divisible by n! which is the case here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2005
    Posts
    37

    Smile I'm impressed!

    You did great job
    It's all very helped but I not really understood the answers for these two:
    <br />
(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9<br />

    and
    <br />
(11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400<br />

    I mean how can I find the X and the Y?
    Thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by dgolverk
    You did great job
    It's all very helped but I not really understood the answers for these two:
    <br />
(7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9<br />

    and
    <br />
(11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400<br />

    I mean how can I find the X and the Y?
    Thanks
    Notice the squares formula a^2+2ab+b^2=(a+b)^2
    Overhere (in first problem) we have that,
    a=(7a-8)
    b=(11-7a)
    Thus,
    (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2
    Becomes,
    [(7a-8)+(11-7a)]^2=[3]^2=9
    The same idea is used in the second problem.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2005
    Posts
    37

    Oh, now I see...

    Thanks a lot!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Welcome
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Number and alegbra
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 5th 2009, 06:11 AM
  2. Alegbra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 17th 2009, 03:15 PM
  3. simple alegbra question:
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2008, 09:32 AM
  4. alegbra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 20th 2008, 08:04 AM
  5. [SOLVED] Please Help!!Alegbra
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 12th 2007, 03:09 PM

Search Tags


/mathhelpforum @mathhelpforum