Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - Help - equation

  1. #1
    Junior Member
    Joined
    Dec 2005
    Posts
    58

    Equation help

    Can someone help me to solve this equation?

    x^4-2ax^2+x+a^2-a=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by DenMac21
    Can someone help me to solve this equation?

    x^4-2ax^2+x+a^2-a=0
    Notice that x^4-2ax^2+a^2=(x^2-a)^2
    Thus,
    x^4-2ax^2+x+a^2-a=(x^2-a)^2+(x-a)=0
    Now notice that,
    (x^2-a)^2=(x-a)^2(x+a)^2
    Thus,
    (x^2-a)^2+(x-a)=(x-a)^2(x+a)^2+(x-a)=0
    Further, factor (x-a) thus,
    (x-a)[(x-a)(x+a)^2+1]=0.

    But I did not finish it, I do not know if that helps.
    Thus, one solution is x=a
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by ThePerfectHacker
    Notice that x^4-2ax^2+a^2=(x^2-a)^2
    Thus,
    x^4-2ax^2+x+a^2-a=(x^2-a)^2+(x-a)=0
    Now notice that,
    (x^2-a)^2=(x-a)^2(x+a)^2
    (x^2-a)^2=(x-a)^2(x+a)^2 is not correct, right side isn't equal to the left side.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2005
    Posts
    58

    Help - equation

    Can someone help me to solve this equation?

    x^4-2ax^2+x+a^2-a=0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by DenMac21
    Can someone help me to solve this equation?

    x^4-2ax^2+x+a^2-a=0
    x^4 -2ax^2 +x +a^2 -a = 0
    x^4 -2ax^2 +a^2 = a -x
    (x^2 -a)^2 = a-x
    x^2 -a = +,-sqrt(a-x) ----------(1)

    The RHS +,-sqrt(a-x):
    ----"a" cannot be zero because sqrt(-x) is not real.
    ----"a" cannot be negative because the radicand will be negative.
    So, a > 0 ----------(i)
    Then,
    ----for (a-x) to be positive, x < a -----------(ii)
    ----since "a" cannot be negative and x<a, then x cannot be negative too.

    In the whole (1):
    ---x cannot be zero because (-a) cannot be equal to +,-sqrt(a).

    Hence, x > 0

    Therefore, 0 < x < a, but x cannot be zero. --------answer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    I don't really understand a lot of your answer...

    "----"a" cannot be zero because sqrt(-x) is not real."?

    And what if x is negative? Then surely sqrt(-x) exists, no? It's not specified that x is a positive number, I think it's just a real number...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by TD!
    I don't really understand a lot of your answer...

    "----"a" cannot be zero because sqrt(-x) is not real."?

    And what if x is negative? Then surely sqrt(-x) exists, no? It's not specified that x is a positive number, I think it's just a real number...
    Can you show x can be negative? Let us see it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Ha ha, I missed that, you are right.

    Try this,
    x^4-2ax^2+a^2+x-a=0
    x^4-ax^2-ax^2+a^2+x-a=0
    x^2(x^2-a)-a(x^2-a)+(x-a)=0
    (x^2-a)(x^2-a)+(x-a)=0
    (x-a)[(x+a)(x^2-a)+1]=0
    Thus,
    x-a=0,x=a
    But for the second factor I did not solve it.
    I do not know if this helps.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by TD!
    The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.
    WOW, I did the same mistake again. For some reason I see x^2-a as a difference of two squares.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker
    WOW, I did the same mistake again. For some reason I see x^2-a as a difference of two squares.
    Well it is, but the second square is \sqrt{a}^2 then
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631

    Re TD! comment on my answer.

    Quote Originally Posted by ticbol
    Can you show x can be negative? Let us see it.
    TD!

    I have to go to work now to put some food on the table. I will be back in about 12 hours from now. Maybe by then you could show us why you think x could be negative. Then I would comment on your posting.

    ticbol

    -----------
    Like I said before somewhere in this Forum, I do not comment on somebody's answer whether that answer is correct, wrong, lacking, too long, whatever.

    But, of course, I will comment on any comment re my answers.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    DenMac21 do not make double threads on the same topic please.

    As I am looking on the equation I do not see any basic way of doing it do you TD!? There is definetly a solution to it by the fundamental theorem.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    I haven't really looked into the problem so I haven't been able to come up with a more elegant solution, but here's a go.

    We have a 4th degree polynomial without a 3rd degree term so a possible factorization would be two second degree polynomials with opposite lineair terms.

    \begin{gathered}<br />
  x^4  - 2ax^2  + x + a^2  - a = 0 \hfill \\<br />
  \left( {x^2  + \alpha x + \beta } \right)\left( {x^2  - \alpha x + \gamma } \right) = 0 \hfill \\<br />
  x^4  + \left( {\alpha ^2  - \beta  - \gamma } \right)x^2  + \left( {\gamma  - \beta } \right)x + \beta \gamma  = 0 \hfill \\ <br />
\end{gathered}

    Identifying coefficients gives us a 3x3 system

    <br />
\left\{ \begin{gathered}<br />
  \alpha ^2  - \beta  - \gamma  = 2a \hfill \\<br />
  \alpha \left( {\gamma  - \beta } \right) = 1 \hfill \\<br />
  \beta \gamma  = a^2  - a \hfill \\ <br />
\end{gathered}  \right.<br />

    Solving yields \alpha  = 1 \wedge \beta  =  - a \wedge \gamma  = 1 - a

    So we have the following factorization

    \left( {x^2  + x - a} \right)\left( {x^2  - x - a + 1} \right) = 0

    Then it's just solving two quadratics, solutions are

    x = \frac{{1 \pm \sqrt {4a - 3} }}<br />
{2} \vee x =  - \frac{{1 \pm \sqrt {4a + 1} }}<br />
{2}

    Of course, these solutions for x depend on a, so I don't see why a negative solution would be impossible - for certain values of the parameter a.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    You can always use the quadtric formula!!!
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  2. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM
  3. Replies: 2
    Last Post: May 18th 2009, 12:51 PM
  4. Replies: 2
    Last Post: April 28th 2009, 06:42 AM
  5. Replies: 1
    Last Post: October 23rd 2008, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum