Root of the symmetric equation

If that term were there, then the whole point of the exercise would become clear, and you could use the "hint" you were given to solve it.
  • #1
Physicsissuef
908
0

Homework Statement



Solve this equation, and find x.

[tex]6x^5-5x^4-29x^2-5x+6=0[/tex]

Homework Equations



if [tex]x= \alpha[/tex] is root of the symmetric equation, then [tex]x= \frac{1}{\alpha}[/tex], is also root of the symmetric equation

The Attempt at a Solution



I tried first to write like this

[tex]6x^5-5x^4-30x^2+x^2-5x+6=0[/tex]

and then multiplying by five

[tex]30x^5-25x^4-150x^2+5x^2-25x+30=0[/tex]

then trying to divide with [tex]x^3[/tex]

[tex]30x^2-25x-150 \frac{1}{x} +5 \frac{1}{x} -25 \frac{1}{x^2}+30 \frac{1}{x^3}=0[/tex]

then adding [tex]-30x^3+30x^3[/tex]

[tex]-30x^3+30x^3 + 30x^2-25x-150 \frac{1}{x} +5 \frac{1}{x} -25 \frac{1}{x^2}+30 \frac{1}{x^3}=0[/tex]

And I am stuck in here. Please help me. Thank you.
 
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  • #2
Hi Physicsissuef!

Be logical. :cool:

1. How many linear factors (ie, of the form x+a) are there?

2. Using the hint they gaave you, what can you say about all those linear factors? :smile:
 
  • #3
tiny-tim said:
Hi Physicsissuef!

Be logical. :cool:

1. How many linear factors (ie, of the form x+a) are there?

2. Using the hint they gaave you, what can you say about all those linear factors? :smile:

Hmm... I don't know, what you want to say.
 
  • #4
[tex]6x^5-5x^4-29x^2-5x+6[/tex] is fifth-order (in other words, it starts with x^5).

So it has five linear factors, which means it equals:
(x + a)(x + b)(x + c)(x + d)(x + e), for some values of a b c d and e.

So, using the hint they gaave you, what can you say about a b c d and e? :smile:
 
  • #5
tiny-tim said:
[tex]6x^5-5x^4-29x^2-5x+6[/tex] is fifth-order (in other words, it starts with x^5).

So it has five linear factors, which means it equals:
(x + a)(x + b)(x + c)(x + d)(x + e), for some values of a b c d and e.

So, using the hint they gaave you, what can you say about a b c d and e? :smile:

[tex](x^5 + \frac{1}{x^5})(x^4+ \frac{1}{x^4})+(x^3+ \frac{1}{x^3})+(x^2+ \frac{1}{x^2})+(x+ \frac{1}{x})[/tex]
like this? What's next? :smile:
 
  • #6
Physicsissuef said:
[tex](x^5 + \frac{1}{x^5})(x^4+ \frac{1}{x^4})+(x^3+ \frac{1}{x^3})+(x^2+ \frac{1}{x^2})+(x+ \frac{1}{x})[/tex]
like this? What's next? :smile:

No - I meant with a b c d and e constants.

So (x + a) etc are first-order in x (linear).

like this: (x + a)(x + b)(x + c)(x + d)(x + e).

(so a b c d and e are like the [tex]\alpha[/tex] in the hint.)

Try again - what can you say about a b c d and e? :smile:
 
  • #7
but how I will find a,b,c,d,e? This equation obviously doesn't have real roots.
a,b,c,d,e are complex numbers.
 
  • #8
Physicsissuef said:
but how I will find a,b,c,d,e? This equation obviously doesn't have real roots.

(i) That doesn't seem obvious to me. Why do you think that?

(ii) The hint they gave you applies to complex numbers also.

(iii) I'm not asking you to find a b c d and e yet - I'm only asking what can you say about them?

a,b,c,d,e are complex numbers.

Can they all be complex numbers?

And if, say, a and b are complex numbers, what does that tell you about a + b or a - b? :smile:
 
  • #9
tiny-tim said:
(i) That doesn't seem obvious to me. Why do you think that?

(ii) The hint they gave you applies to complex numbers also.

(iii) I'm not asking you to find a b c d and e yet - I'm only asking what can you say about them?



Can they all be complex numbers?

And if, say, a and b are complex numbers, what does that tell you about a + b or a - b? :smile:

I don't know. Can we please start solving this equation?
 
  • #10
You can start any time you like.

But why would you simple assert that "This equation obviously doesn't have real roots."
Any odd degree polynomial equation "obviously" has a least one real root! Do you see why?

It is also true that complex roots of polynomial equations with real coefficients come in pairs. That was tiny-tim's point.
 
  • #11
HallsofIvy said:
You can start any time you like.

But why would you simple assert that "This equation obviously doesn't have real roots."
Any odd degree polynomial equation "obviously" has a least one real root! Do you see why?

It is also true that complex roots of polynomial equations with real coefficients come in pairs. That was tiny-tim's point.

And how can I start? Should I combine the elements somehow?
 
  • #12
Physicsissuef said:
And how can I start? Should I combine the elements somehow?
I would start by expanding the brackets, then you can compare the coefficients.
 
  • #13
If you sow above, I tried with -30x^2+x^2, can you please give me, some hint how to expand the brackets, please?
 
  • #14
(btw, thanks, HallsofIvy!)
Hootenanny said:
I would start by expanding the brackets, then you can compare the coefficients.

No, don't start that way … it's horrible! :frown:

Start with the hint:
If (x + a) is a factor, then so is (x + 1/a).​

Now, what does that tell you about b c d and e? :smile:
 
  • #15
tiny-tim said:
(btw, thanks, HallsofIvy!)


No, don't start that way … it's horrible! :frown:

Start with the hint:
If (x + a) is a factor, then so is (x + 1/a).​

Now, what does that tell you about b c d and e? :smile:
Oops, the hint does make it a little easier... :blushing:

[Hides in the corner]
 
  • #16
tiny-tim can you please start, with solving the equation, just start... I will continue...
 
  • #17
Um...guys? This equation doesn't look too "symmetric" to me. When you plug in [itex]x=\frac{1}{\alpha}[/itex], you get the following.

[tex]\frac{6}{\alpha^5}-\frac{5}{\alpha^4}-\frac{29}{\alpha^2}-\frac{5}{\alpha}+6=0[/tex]

Multiplying both sides by [itex]\alpha^5[/itex] gives the following.

[tex]6-5\alpha-29\alpha^3-5\alpha^4+6\alpha^5=0[/tex].

That's not the same equation.

Physicsissuef, can you please look carefully at the exercise you were given to make sure that there is not supposed to be another term in there, namely [itex]-29x^3[/itex]?
 
  • #18
Tom Mattson said:
Um...guys? This equation doesn't look too "symmetric" to me. When you plug in [itex]x=\frac{1}{\alpha}[/itex], you get the following.

[tex]\frac{6}{\alpha^5}-\frac{5}{\alpha^4}-\frac{29}{\alpha^2}-\frac{5}{\alpha}+6=0[/tex]

Multiplying both sides by [itex]\alpha^5[/itex] gives the following.

[tex]6-5\alpha-29\alpha^3-5\alpha^4+6\alpha^5=0[/tex].

That's not the same equation.

Physicsissuef, can you please look carefully at the exercise you were given to make sure that there is not supposed to be another term in there, namely [itex]-29x^3[/itex]?

No, there isn't. I asked also my professor, but also he told me that there is some problem with -29. I don't know even if this task is possible to solve. I tried the others, and solved them. Also I sow the results in the book, and they wrote [tex]x_1[/tex]=1 which is not correct. So maybe they have typo error.
 
  • #19
Tom Mattson said:
Um...guys? This equation doesn't look too "symmetric" to me.

hmm … never spotted that … :redface:

It should be [tex]6x^5-5x^4-29x^3-29x^2-5x+6=0[/tex].

Take my word for it, this factors very nicely. :smile:

(So nicely, you could actually guess it!)
(But no telling, please, except of course for Physicsissuef!)
 
  • #20
Physicsissuef said:
So maybe they have typo error.

There's no "maybe", if the hint that was given to you is supossed to mean anything, then that term should be there.
 
  • #21
Again, something is not correct.
I came up with this equation.
[tex]6x^4-40x^3-18x^2-11x+6=0[/tex]
 
  • #22
Physicsissuef said:
Again, something is not correct.
I came up with this equation.
[tex]6x^4-40x^3-18x^2-11x+6=0[/tex]

Now you've completely lost me - that's not remotely symmetric. :confused:

Physicsissuef, it's definitely
[tex]6x^5-5x^4-29x^3-29x^2-5x+6=0\,.[/tex]​

(Many thnks to Tom Mattson for pointing that out!)

So stick to that equation, and solve it using the hint they gave you and by asking yourself what you can say about a b c d and e. :smile:
 
  • #23
tiny-tim said:
Now you've completely lost me - that's not remotely symmetric. :confused:

Physicsissuef, it's definitely
[tex]6x^5-5x^4-29x^3-29x^2-5x+6=0\,.[/tex]​

(Many thnks to Tom Mattson for pointing that out!)

So stick to that equation, and solve it using the hint they gave you and by asking yourself what you can say about a b c d and e. :smile:

[tex](x+1)(6x^4-40x^3-18x^2-11x+6)=0[/tex]
One solution is x=-1, and the other, I don't know... Again -11x is problem.
 
  • #24
Excellent so far …

Physicsissuef said:
[tex](x+1)(6x^4-40x^3-18x^2-11x+6)=0[/tex]
One solution is x=-1, and the other, I don't know... Again -11x is problem.

Oh I see!

Yes, excellent :smile::smile:
(except the 40x^3 should be 11x^3, to match the 11x, which is what confused me).

(I take it you got that by seeing that (x + a) was on its own, and so a = 1/a, so a = ±1?)

Right - that's half the battle (well, maybe a third).

Now the problem is to write
[tex]6x^4-11x^3-18x^2-11x+6[/tex]​
in the form (x + b)(x + c)(x + d)(x + e).

So what can you say about b c d and e? Can you write any of them in terms of the others? :smile:
 
  • #25
Look,

[tex]6x^5-5x^4-29x^3-29x^2-5x+6=0[/tex]

[tex]6(x^5+1)-5x(x^3+1)-29x^2(x+1)=0[/tex]

[tex]6(x+1)(x^4-x^3+x^2-x+1)-5x(x+1)(x^2-x+1)-29x^2(x+1)=0[/tex]

[tex](x+1)(6x^4-6x^3+6x^2-6x+6-5x^3+5x^2-5x-29x^3-29x^2)=0[/tex]

so yes, it is -40x^3

[tex](x+1)(6x^4-40x^3-18x^2-11x+6)=0[/tex]
 
Last edited:
  • #26
Nope. Your
[tex](6x^4-6x^3+6x^2-6x+6-5x^3+5x^2-5x-29x^3-29x^2)[/tex]​
should be
[tex](6x^4-6x^3+6x^2-6x+6-5x^3+5x^2-5x -29x^2)[/tex]​

(Personally, I'd just do long division - it's quicker, and you're less likely to make a mistake :blushing:)
 
  • #27
Ok, now I sow my error.

[tex]6x^4-11x^3-18x^2-11x+6=0[/tex]

So,

[tex] 6(x^2+ \frac{1}{x^2})-11(x+ \frac{1}{x})-18=0[/tex]

[tex]6(y^2-2)-11y-18=0[/tex]

[tex]6y^2-12-11y-18=0[/tex]

[tex]6y^2-11y-30=0[/tex]

Ok, I know now. Thanks for the help, guys.
 
Last edited:

Related to Root of the symmetric equation

What is a root of a symmetric equation?

A root of a symmetric equation is a value that, when substituted into the equation, makes the equation true. In a symmetric equation, this value will be the same for both sides of the equation.

How do I find the roots of a symmetric equation?

To find the roots of a symmetric equation, you can use algebraic methods such as factoring or the quadratic formula. You can also graph the equation and find the x-intercepts, which will be the roots.

Can a symmetric equation have more than one root?

Yes, a symmetric equation can have multiple roots. In fact, quadratic equations, which are a type of symmetric equation, can have two distinct roots.

Are the roots of a symmetric equation always real numbers?

No, the roots of a symmetric equation can be real or complex numbers. For example, the equation x^2 + 1 = 0 has no real solutions, but it has two complex solutions: i and -i.

How are the roots of a symmetric equation related to the graph of the equation?

The roots of a symmetric equation are the x-intercepts of the graph of the equation. This means that the points where the graph crosses the x-axis are the values for which the equation is true.

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