Can someone help me to solve this equation?
$\displaystyle x^4-2ax^2+x+a^2-a=0$
Notice that $\displaystyle x^4-2ax^2+a^2=(x^2-a)^2$Originally Posted by DenMac21
Thus,
$\displaystyle x^4-2ax^2+x+a^2-a=(x^2-a)^2+(x-a)=0$
Now notice that,
$\displaystyle (x^2-a)^2=(x-a)^2(x+a)^2$
Thus,
$\displaystyle (x^2-a)^2+(x-a)=(x-a)^2(x+a)^2+(x-a)=0$
Further, factor $\displaystyle (x-a)$ thus,
$\displaystyle (x-a)[(x-a)(x+a)^2+1]=0$.
But I did not finish it, I do not know if that helps.
Thus, one solution is $\displaystyle x=a$
x^4 -2ax^2 +x +a^2 -a = 0Originally Posted by DenMac21
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.
Ha ha, I missed that, you are right.
Try this,
$\displaystyle x^4-2ax^2+a^2+x-a=0$
$\displaystyle x^4-ax^2-ax^2+a^2+x-a=0$
$\displaystyle x^2(x^2-a)-a(x^2-a)+(x-a)=0$
$\displaystyle (x^2-a)(x^2-a)+(x-a)=0$
$\displaystyle (x-a)[(x+a)(x^2-a)+1]=0$
Thus,
$\displaystyle x-a=0,x=a$
But for the second factor I did not solve it.
I do not know if this helps.
TD!Originally Posted by ticbol
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
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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.
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.
$\displaystyle \begin{gathered}
x^4 - 2ax^2 + x + a^2 - a = 0 \hfill \\
\left( {x^2 + \alpha x + \beta } \right)\left( {x^2 - \alpha x + \gamma } \right) = 0 \hfill \\
x^4 + \left( {\alpha ^2 - \beta - \gamma } \right)x^2 + \left( {\gamma - \beta } \right)x + \beta \gamma = 0 \hfill \\
\end{gathered} $
Identifying coefficients gives us a 3x3 system
$\displaystyle
\left\{ \begin{gathered}
\alpha ^2 - \beta - \gamma = 2a \hfill \\
\alpha \left( {\gamma - \beta } \right) = 1 \hfill \\
\beta \gamma = a^2 - a \hfill \\
\end{gathered} \right.
$
Solving yields $\displaystyle \alpha = 1 \wedge \beta = - a \wedge \gamma = 1 - a$
So we have the following factorization
$\displaystyle \left( {x^2 + x - a} \right)\left( {x^2 - x - a + 1} \right) = 0$
Then it's just solving two quadratics, solutions are
$\displaystyle x = \frac{{1 \pm \sqrt {4a - 3} }}
{2} \vee x = - \frac{{1 \pm \sqrt {4a + 1} }}
{2}$
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.