Help - equation

**DenMac21** · January 9th 2006, 05:32 PM

Can someone help me to solve this equation?

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

**ThePerfectHacker** · January 9th 2006, 07:08 PM

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 $x=a$

**DenMac21** · January 9th 2006, 07:43 PM

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.

**DenMac21** · January 10th 2006, 08:57 AM

Can someone help me to solve this equation?

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

**ticbol** · January 10th 2006, 10:19 AM

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.

**TD!** · January 10th 2006, 10:24 AM

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...

**ticbol** · January 10th 2006, 10:27 AM

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.

**ThePerfectHacker** · January 10th 2006, 10:41 AM

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.

**TD!** · January 10th 2006, 10:44 AM

The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.

**ThePerfectHacker** · January 10th 2006, 10:50 AM

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.

**TD!** · January 10th 2006, 10:52 AM

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

**ticbol** · January 10th 2006, 10:55 AM

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.

**ThePerfectHacker** · January 10th 2006, 10:55 AM

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.

**TD!** · January 10th 2006, 11:02 AM

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.

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

Identifying coefficients gives us a 3x3 system

$ \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 $\alpha = 1 \wedge \beta = - a \wedge \gamma = 1 - a$

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

$x = \frac{{1 \pm \sqrt {4a - 3} }} {2} \vee x = - \frac{{1 \pm \sqrt {4a + 1} }} {2}$

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.

**ThePerfectHacker** · January 10th 2006, 11:17 AM

You can always use the quadtric formula!!!

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