Can someone help me to solve this equation?

$\displaystyle x^4-2ax^2+x+a^2-a=0$

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- Jan 9th 2006, 05:32 PMDenMac21Equation help
Can someone help me to solve this equation?

$\displaystyle x^4-2ax^2+x+a^2-a=0$ - Jan 9th 2006, 07:08 PMThePerfectHackerQuote:

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 $\displaystyle x=a$ - Jan 9th 2006, 07:43 PMDenMac21Quote:

Originally Posted by**ThePerfectHacker**

- Jan 10th 2006, 08:57 AMDenMac21Help - equation
Can someone help me to solve this equation?

$\displaystyle x^4-2ax^2+x+a^2-a=0$ - Jan 10th 2006, 10:19 AMticbolQuote:

Originally 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. - Jan 10th 2006, 10:24 AMTD!
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... - Jan 10th 2006, 10:27 AMticbolQuote:

Originally Posted by**TD!**

- Jan 10th 2006, 10:41 AMThePerfectHacker
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. - Jan 10th 2006, 10:44 AMTD!
The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.

- Jan 10th 2006, 10:50 AMThePerfectHackerQuote:

Originally Posted by**TD!**

- Jan 10th 2006, 10:52 AMTD!Quote:

Originally Posted by**ThePerfectHacker**

- Jan 10th 2006, 10:55 AMticbolRe TD! comment on my answer.Quote:

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

-----------

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. - Jan 10th 2006, 10:55 AMThePerfectHacker
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. - Jan 10th 2006, 11:02 AMTD!
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. - Jan 10th 2006, 11:17 AMThePerfectHacker
You can always use the quadtric formula!!! :D