Results 1 to 5 of 5

Math Help - Confusing System of Equations

  1. #1
    Member
    Joined
    Mar 2011
    Posts
    99

    Question Confusing System of Equations

    Hi MHF

    Before I continue I would like to ask you guys for some help.
    I want to know where/how can I practice hard Algebra questions?

    How did you guys get so good? Any tips? Any book? Any website?
    Thanks

    Now, for the question

    We have to find the value of x^4+y^4+z^4 when x , y and z are real numbers.

    Following three equalities:

    x+y+z=3
    x^2+y^2+z^2=9
    xyz =-2

    First, from the first two equations we have (this is actually in the question)
    xy+yz+xz=A

    Next using

    (x^2+y^2+z^2)^2=x^4+y^4+z^4+B[(xy)^2+(yz)^2+(zx)^2]

    We have
    (x^4)+(y^4)+(z^4)=C

    I know NOTHING about this question. It asks to find A,B and C ;all these equations are given.

    So,
    any tips?

    Last time I asked something here I got a really good response and actually got to know that I had missed somethings in school.
    Am I missing out something?

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

  2. #2
    Newbie
    Joined
    May 2011
    Posts
    3
    To find out A, B and C, we have to use the three equalities.

    To find A, expand the first equality. Substitute using the other 2 equalities when necessary.

    To find B, expand (x^2+y^2+z^2)^2. Rearrange the expanded equation and you can get B eventually.

    From the question, we know that x^4+y^4+z^4 = C. So, we can rewrite the equation (x^2+y^2+z^2)^2 = C + B[(xy)^2+(yz)^2+(zx)^2)]. And the tricky part now is to find [(xy)^2+(yz)^2+(zx)^2)] which is by expanding the equation (xy+yz+xz)^2. Substitute when necessary and you will get the answer C.

    Sorry that my explaination might be difficult to understand. Do ask if you have more questions.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2011
    Posts
    99
    Thanks for replying ahyan1234!

    I really thought there was an easy way to do this, like some property from the systems of equations that I had forgotten.
    This requires a lot of work, doesn't it? hahaha

    This is great!!!!! Thank you so much for your reply!
    I really appreciate it :3


    Edit:


    So, I was trying to get A in this question, and it's not that simple.
    Because x is actually 9-3y-3z-3y+y+yz-3z+yz+z (using the FOIL method)
    and not only x=9-y-z (expanding the second equation)

    I got to go in circles, what one equation does the other undoes.

    This so...er... head-aching! I'm sorry, sometimes I really suck in algebra.
    Last edited by Zellator; May 6th 2011 at 05:44 AM. Reason: tried to to the equation, and still no go
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,550
    Thanks
    542
    Hello, Zellator!

    I have a different approach.
    I'll get you started . . .


    \text{Given: }\;\begin{array}{ccccc}x+y+z &=& 3 & [1] \\ x^2+y^2+z^2 &=& 9 & [2] \\ xyz &=& \text{-}2 & [3] \end{array}

    \text{Find: }\;x^4 + y^4 + z^4

    \text{Square [1]: }\qquad\qquad\; (x + y + z)^2 \;=\;3^2

    . . x^2 + 2xy + 2xz + y^2 + 2yz + z^2 \;=\;9

    . . \underbrace{x^2 + y^2 + z^2}_{\text{This is 9}} +\;2(xy + yz + xz) \;=\;9

    . . . . . . . .Hence: . \boxed{xy + yz + xz \;=\;0}


    \text{Cube [1]: }\qquad (x + y + z)^3 \;=\;3^3

    x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3y^zz + 3yz^2 + 3x^2z + 3xz^2 + 6xyz \;=\;27

    x^3 + y^3 + z^3 + 3x^2y + 3xy^2 + 3y^zz + 3yz^2 + 3x^2z + 3xz^2 + {\bf9xyz} \;=\;27 + {\bf3xyz}

    x^3 + y^3 + z^2 + 3(x^2y + xy^2 + xyz) + 3(xyz + y^2z + yz^2) + 3(x^2z + xyz + xz^2)
    . . . . . . =\;27 + 3xyz

    x^3 + y^3 + z^2 + 3xy(x + y + z) + 3yz(x + y + z) + 3xz(x + y + z) \;=\;27 + 3xyz

    x^3 + y^3 + z^3 + 3\underbrace{(x+y+z)}_{\text{This is 3}}\underbrace{(xy + yz + xz)}_{\text{This is 0}} \;=\;27+ 3\underbrace{(xyz)}_{\text{This is -2}}

    . . . . . . . . . . . . . . . x^3 + y^3 + z^3 + 3(3)(0) \;=\;27 - 6

    . . . . . . . . . . . . . . . . . . . . . \boxed{x^3 + y^3 + z^3 \;=\;21}


    Get the idea?

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2011
    Posts
    99
    Absolutely! Wow!

    I'm loss for words!
    I really wish I had your experience in this hahahhaha!

    I'm sorry, but why did you made the Cubic of the expression?
    You probably need the cubic to get the ^4 right?

    This is really amazing! It seem so easy when you guys do it!
    I'll probably dig the forum up for more equations and problems like this.

    Thanks for you time ahyan1234 and Soroban.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. system of equations when x,y and z = 1
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 3rd 2011, 08:33 AM
  2. Replies: 2
    Last Post: April 20th 2010, 03:26 PM
  3. Confusing system of simultaneous equations...
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 1st 2009, 09:14 PM
  4. Confusing System of Equations
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 11th 2009, 09:10 AM
  5. system of equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 26th 2008, 02:38 PM

Search Tags


/mathhelpforum @mathhelpforum