Autor Tema: find x

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19 Junio, 2011, 07:25 pm
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jacks

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find value of \( x \) in \( (x^2-2x)(x^2-4) = 2 \)

19 Junio, 2011, 07:39 pm
Respuesta #1

javier m

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me sumo a la pregunta

20 Junio, 2011, 04:01 am
Respuesta #2

argentinator

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Espero no equivocarme, pero la forma usual de resolver este tipo de problemas es igualar cada factor a un divisor de 2. En este caso, dado que 2 es primo, entonces se puede expresar como \( 1\times{}2 \) ó \( -1\times{}-2 \). Pero, sin embargo, el ejercicio no me da soluciones enteras, incluso se puede verificar en wolfram alpha. Por lo que te pregunto, está bien copiado el ejercicio?

Translation:

crimeeee says: Maybe I am wrong, but the usual way to solve this kind of probles if by equating each factor to a divisor of 2.
In this case, since 2 is prime, it can be written as \( 1\times{}2 \) or \( -1\times{}-2 \).
But I cannot find integer solutions in this exercise. Moreover, it can be verified in wolfram alpha.
So, I have a question for you: is well copied this exercise?



jacks: I have the same problem that crimeeee, and I also think that probably the exercise is not well copied.


20 Junio, 2011, 08:05 am
Respuesta #3

jacks

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yes my friends crimeeee, you are saying right.

The given equation is not satisfied for \( x=\pm 1,\pm 2 \).So the given expression has quadratic factor.

\( x^4-2x^3-4x^2+8x-2 = \left(x^2+ax+2\right)\left(x^2+bx-1\right) = \left(x^2+ax-2\right)\left(x^2+bx+1\right)  \)

now we will calculate \( a \) and \( b \)

20 Junio, 2011, 08:48 am
Respuesta #4

argentinator

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If

\(  x^4-2x^3-4x^2+8x-2 =  \left(x^2+ax+2\right)\left(x^2+bx-1\right)
=x^4+(a+b)x^3+(1+ab)x^2+(-a+2b)x-2  \)

Then, by equating the polynomial coefficients we obtain the equations

\( \begin{align*}a+b&=-2\\
1+ab &=-4\\
-a+2b &= 8\end{align*}
 \)

From the 1st and 3rd equations we have: \( a=-4, b =2 \).
Replacing this results in 2nd equation we have \(  1 + ab = -7 \), that is not equal to -4.

Mmmm  This is not working.


Tell me if you think I am wrong.


20 Junio, 2011, 09:02 am
Respuesta #5

argentinator

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What about this factorization?:

\(  x^4-2x^3-4x^2+8x-2 =  \left(x^2+ax+2r\right)\left(x^2+bx-1/r\right)
=x^4+(a+b)x^3+(2r-1/r+ab)x^2+(-a/r+2rb)x-2  \)

Now we have to find \( a, b,  \)r, with these equations:

\(
\begin{align*}a+b&=-2\\
2r-1/r+ab &=-4\\
-a/r+2rb &= 8\end{align*}
 \)

From the 1st equation: \( a = -2-b \).

By replacing in the 2nd equation: \( 2r-1/r+(-2-b )b=-4 \).
The 3rd equation would become: \( -2/r-b/r+2rb = 8. \)

Now you have to solve for \( b, r \), in these last two equations.

I think this is perhaps the right way. It seems cumbersome, but at least it works.

20 Junio, 2011, 05:29 pm
Respuesta #6

argentinator

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By multiplying by \( r \) both equations and rearranging terms:

\( 2r^2 -r(b^2+2b+4)-1=0 \)

\( -2-b-8r+2r^2b=0 \)

Solving for \( b \) in the 2nd equation:

\( b=\dfrac{2+8r}{2r^2-1} \)

I put this result in 1st equation.

\( 2r^2-r\left(\left(\dfrac{2+8r}{2r^2-1}\right)^2+2\dfrac{2+8r}{2r^2-1}+4\right)-1=0 \)

and then I try to solve for \( r \).

By I will obtain an equation of degree 4 again...

I think the only feasible way to work out this problem is by using the standard formulas to find roots in an equation of degree 4.




21 Junio, 2011, 02:35 am
Respuesta #7

jacks

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yes argentinator you are saying right.

i have also tried, but could not getting.