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Thread: Alegbra

  1. #1
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    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...
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  2. #2
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    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. $\displaystyle 6x^3+x^2-x+12x^2+2x-2=$
    $\displaystyle x(6x^2+x-1)+2(6x^2+x-1)$
    $\displaystyle (x+2)(6x^2+x-1)$ now you can factorize the 2nd factor easily.

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

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

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

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

    Greetings

    EB
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  3. #3
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    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:
    $\displaystyle \frac{{(n^2 + 2n)(n^2 - 1)}}{{24}} = \frac{{(n - 1)n(n + 1)(n + 2)}}{{24}}$

    So, n can be any number $\displaystyle 24*x,(x \in Q,x \ne 0)$ and it can be n=2 and n=-3 because 24=1*2*3*4.
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  4. #4
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    given:
    a+b=5
    ab=3
    How much is:
    a^2+b^2
    a^2b+ab^2
    a^3b+ab^3
    $\displaystyle \begin{array}{l}
    (a + b)^2 = 5^2 \\
    a^2 + 2ab + b^2 = 25 \\
    a^2 + b^2 = 25 - 2ab \\
    a^2 + b^2 = 25 - 2ab \\
    a^2 + b^2 = 25 - 2*3 \\
    a^2 + b^2 = 19 \\
    \end{array}
    $

    $\displaystyle \begin{array}{l}
    a^2 b + ab^2 = ab(a) + ab(b) \\
    3a + 3b = 3(a + b) = 3*5 = 15 \\
    \end{array}
    $

    $\displaystyle a^3 b + ab^3 = ab(a^2 ) + ab(b^2 ) $
    $\displaystyle 3a^2 + 3b^2 = 3(a^2 + b^2 ) = 3*19 = 57$
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  5. #5
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    Let me just explain DenMac's post that
    $\displaystyle \frac{n(n+1)(n-1)(n+2)}{24}$ is an integer.
    Because $\displaystyle 24=4\times 3\times 2\times 1$.
    That means that one of the integers,
    $\displaystyle n,n+1,n-1,n+2$ 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,
    $\displaystyle n(n+1)(n-1)(n+2)$ 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 $\displaystyle x=n-1$
    Then, $\displaystyle x+1=n$,$\displaystyle x+2=n+1$,$\displaystyle x+3=n+2$.
    Thus, (under modulo 24)
    $\displaystyle x\equiv r_1$
    $\displaystyle x+1\equiv r_2$
    $\displaystyle x+2\equiv r_3$
    $\displaystyle x+3\equiv r_4$
    Now non-of these are congruent to each other for then,
    $\displaystyle x+i\equiv x+j$
    Thus,
    $\displaystyle i\equiv j$ and $\displaystyle 0\leq i,j\leq 3$
    Which is impossible.
    Thus, the integers $\displaystyle r_1,r_2,r_3,r_4$ must be the integers $\displaystyle 1,2,3,4$ in some order by the pigeonhole principle. Thus, $\displaystyle r_1r_2r_3r_4=24$.
    Thus, upon multiplication of the congruences we have that
    $\displaystyle x(x+1)(x+2)(x+3)\equiv 24\equiv 0$
    Q.E.D.
    -----------------------
    Finally, there is a theorem from number theory that the product of n-consecutives integers is always divisible by $\displaystyle n!$ which is the case here.
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  6. #6
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    Smile I'm impressed!

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

    and
    $\displaystyle
    (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400
    $

    I mean how can I find the X and the Y?
    Thanks
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  7. #7
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    Quote Originally Posted by dgolverk
    You did great job
    It's all very helped but I not really understood the answers for these two:
    $\displaystyle
    (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2=9
    $

    and
    $\displaystyle
    (11a+5)^2-2(11a+5)(11a-15)+(11a-15)^2=400
    $

    I mean how can I find the X and the Y?
    Thanks
    Notice the squares formula $\displaystyle a^2+2ab+b^2=(a+b)^2$
    Overhere (in first problem) we have that,
    $\displaystyle a=(7a-8)$
    $\displaystyle b=(11-7a)$
    Thus,
    $\displaystyle (7a-8)^2+2(7a-8)(11-7a)+(11-7a)^2$
    Becomes,
    $\displaystyle [(7a-8)+(11-7a)]^2=[3]^2=9$
    The same idea is used in the second problem.
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  8. #8
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    Oh, now I see...

    Thanks a lot!
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  9. #9
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    Welcome
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