Is this form possible?

[nothing]

Is it possible to have an equation
such as quadratic
whose two root solutions are 1 and phi^3?

 

[Romsek]

[MATH] (x-1)(x-\phi^3)=0[/MATH]
what's \(\displaystyle \phi\)?

 

[Dr.Peterson]

What do you mean by "such as"? What feature are you interested in?

Possibly you mean phi to be the golden ratio, and are thinking of polynomials whose roots are irrational numbers in general.

But the answer depends (depending on the generalization you have in mind) on whether you require the coefficients of the polynomial to be real numbers, integers, or from some other set. What have you learned about roots of polynomials?

 

[nothing]

I am interested in polynomials whose roots are both real and irrational.

Take [MATH]ax^2+bx^2 \pm c[/MATH]= 1, 4.23606797...

wherein only by changing the +/- (not the value) of c, 1 becomes phi^3 and vice versa.
I am wondering if such a thing exists.

 

[topsquark]

I am interested in polynomials whose roots are both real and irrational.

Take [MATH]ax^2+bx^2 \pm c[/MATH]= 1, 4.23606797...

wherein only by changing the +/- (not the value) of c, 1 becomes phi^3 and vice versa.
I am wondering if such a thing exists.

I don't see any reason that they couldn't be. As Romsek says, the quadratic would be [math](x - 1)(x - \phi ^3 )[/math]. On the other hand, when you expand this out to the standard [math]ax^2 + bx + c[/math] form none of a, b, and c are going to be integers in general.

-Dan

 

[pka]

I am interested in polynomials whose roots are both real and irrational.

\(\displaystyle x^2-(1+\sqrt 2)x+\sqrt 2=0\) is one such.

 

[nothing]

\(\displaystyle x^2-(1+\sqrt 2)x+\sqrt 2=0\) is one such.

Yes like that but so that the only two valid solutions are 1 and 4.23606797... 0 can not be a valid solution.

 

[nothing]

[MATH] (x-1)(x-\phi^3)=0[/MATH]
what's \(\displaystyle \phi\)?

I don't see any reason that they couldn't be. As Romsek says, the quadratic would be [math](x - 1)(x - \phi ^3 )[/math]. On the other hand, when you expand this out to the standard [math]ax^2 + bx + c[/math] form none of a, b, and c are going to be integers in general.

-Dan

Thanks for the replies,

I see it as [math]\pi+\pi\sqrt5/2\pi[/math] and I should have indicated x≠0, the only valid solutions can be 1 and the exact value of phi^3

 

[pka]

Yes like that but so that the only two valid solutions are 1 and 4.23606797... 0 can not be a valid solution.

Your post did not say anything about zero having to be a root.
\(\displaystyle x^2-\sqrt 2 x=0\) has \(\displaystyle 0~\&~\sqrt 2\) as roots.

 

[nothing]

Your post did not say anything about zero having to be a root.
\(\displaystyle x^2-\sqrt 2 x=0\) has \(\displaystyle 0~\&~\sqrt 2\) as roots.

No I'm saying 0 can neither be a root nor an answer.

{Standard form equation} = 1, 4.23606797...

Nothing else

 

[pka]

No I'm saying 0 can neither be a root nor an answer.
{Standard form equation} = 1, 4.23606797...
Nothing else /QUOTE]
Sorry to tell you this, but you do understand this topic to even ask a clear question.

 

[nothing]

Sorry to tell you this, but you do understand this topic to even ask a clear question.

I never said 0 had to be a root. You got that from literally nowhere.
I said from first post, only two solutions of 1 and 4.23606797...

 

[Dr.Peterson]

@nothing, let's look at what you have said, to clarify what this is about:

Yes like that but so that the only two valid solutions are 1 and 4.23606797... 0 can not be a valid solution.

You are the one who brought up whether 0 could be a solution, when you said this. Nothing said before that mentioned 0 as a solution. So why did you say that?

You said this in answer to the offer of [MATH]x^2-(1+\sqrt 2)x+\sqrt 2=0[/MATH], a polynomial with an integer (1) and an irrational number ([MATH]\sqrt{2}[/MATH]) as its only roots? There is no zero there. (You said real and irrational, forgetting that irrational numbers are real.)

If you want an equation with 1 and [MATH]\phi^3[/MATH] as its only roots, you were given that, too; expanded, it is [MATH]x^2 - (3 + \sqrt{5})x + \sqrt{5} = 0[/MATH].

I see it as [math]\pi+\pi\sqrt5/2\pi[/math] and I should have indicated x≠0, the only valid solutions can be 1 and the exact value of phi^3

No one until then had given an equation with 0 as a root. Why do you have to say that x is not 0?

And what are you saying about [MATH]\pi+\pi\sqrt5/2\pi[/MATH]? Where did that number come from?

No I'm saying 0 can neither be a root nor an answer.

{Standard form equation} = 1, 4.23606797...

Nothing else

No one has said otherwise. You were given [MATH](x-1)(x-\phi^3)=0[/MATH], which does what you want, and I've just expanded that to [MATH]x^2 - (3 + \sqrt{5})x + \sqrt{5} = 0[/MATH] for you.

Please tell us, all at once, exactly what you are asking for, so we don't have to talk past one another.

 

[nothing]

@nothing, let's look at what you have said, to clarify what this is about:

You are the one who brought up whether 0 could be a solution, when you said this. Nothing said before that mentioned 0 as a solution. So why did you say that?

It was brought up because a solution which was not 1, phi^3 was offered, despite the op clearly stating whose two root solutions are 1 and phi^3.
It can not be any other two values, only those two.

You said this in answer to the offer of [MATH]x^2-(1+\sqrt 2)x+\sqrt 2=0[/MATH], a polynomial with an integer (1) and an irrational number ([MATH]\sqrt{2}[/MATH]) as its only roots? There is no zero there. (You said real and irrational, forgetting that irrational numbers are real.)

It needs to be in standard form complete with radicals.

If you want an equation with 1 and [MATH]\phi^3[/MATH] as its only roots, you were given that, too; expanded, it is [MATH]x^2 - (3 + \sqrt{5})x + \sqrt{5} = 0[/MATH].

Nothing can equal 0.
eqn = 1, 4.23606797...
and that is all. No 0 anywhere.
I was given a 0 solution asking for only 1, 4.23606797...

No one until then had given an equation with 0 as a root. Why do you have to say that x is not 0?

The equation is not supposed to have variables in it, just radicals and operators whose arrangement equals 1, 4.23606797...

And what are you saying about [MATH]\pi+\pi\sqrt5/2\pi[/MATH]? Where did that number come from?

I was asked what phi is equal to, so I showed what phi is equal to, in radical form, which is the form the equation must be in.

No one has said otherwise. You were given [MATH](x-1)(x-\phi^3)=0[/MATH], which does what you want, and I've just expanded that to [MATH]x^2 - (3 + \sqrt{5})x + \sqrt{5} = 0[/MATH] for you.

Please tell us, all at once, exactly what you are asking for, so we don't have to talk past one another.

I want one equation which equals only two possible values: 1 and 1, 4.23606797... depending on the +/- root

 

[Dr.Peterson]

Is it possible to have an equation
such as quadratic
whose two root solutions are 1 and phi^3?

The equation is not supposed to have variables in it. Just radicals whose arrangement equals 1, 4.23606797...

I was asked what phi is equal to, so I showed what phi is equal to, in radical form, which is the form the equation must be in.

I want one equation which equals only two possible values: 1 and 1, 4.23606797... depending on the +/- root

I think I'm starting to see the problem. You don't know what an equation is, or what a root or solution is!

We all interpreted your question as asking for a quadratic equation (that is, something like ax^2 + bx + c = 0 with specific numbers in place of a, b, c), whose solutions (also called roots) are 1 and [MATH]\phi^3[/MATH]. That is, when x is replaced with 1 or with [MATH]\phi^3[/MATH], it becomes true. We gave you such an equation, and also another whose solutions are 1 and [MATH]\sqrt{2}[/MATH] because you had said that, more generally, you wanted one whose roots are "real and irrational".

But that's not what you want; you said it all wrong.

If you want something without variables in it, you are evidently asking for a single expression, like your example of [MATH]\pi+\pi\sqrt5/2\pi[/MATH], that has two values, 1 and \phi^3. But what does that even mean? If it is not an equation with variables, then what do you mean by "two root solutions"?

I'm guessing that by "quadratic" you meant "containing radicals", though that is not at all what it means.

Let's get back to the start. It looks like post #4 may be the closest you came to stating what you mean:

I am interested in polynomials whose roots are both real and irrational.

Take [MATH]ax^2+bx^2 \pm c[/MATH] = 1, 4.23606797...

wherein only by changing the +/- (not the value) of c, 1 becomes phi^3 and vice versa.
I am wondering if such a thing exists.

Apparently what you meant by "equation" is "polynomial expression", and you want it to have a plus-or-minus in it so that it yields two values (not roots), one of which is "real" (you may have meant integer) and the other irrational (though "both" suggests you may mean that both "roots" should be real, irrational numbers). None of this is the correct meaning of the OP, but let's try to figure it out.

What's still missing is the role of x. Here you showed a polynomial expression with a variable, but in post #14 you said there should be no variables.

Can you correct my guesses about your meaning?

 

[nothing]

I think I'm starting to see the problem. You don't know what an equation is, or what a root or solution is!

We all interpreted your question as asking for a quadratic equation (that is, something like ax^2 + bx + c = 0 with specific numbers in place of a, b, c), whose solutions (also called roots) are 1 and [MATH]\phi^3[/MATH]. That is, when x is replaced with 1 or with [MATH]\phi^3[/MATH], it becomes true. We gave you such an equation, and also another whose solutions are 1 and [MATH]\sqrt{2}[/MATH] because you had said that, more generally, you wanted one whose roots are "real and irrational".

But that's not what you want; you said it all wrong.

If you want something without variables in it, you are evidently asking for a single expression, like your example of [MATH]\pi+\pi\sqrt5/2\pi[/MATH], that has two values, 1 and \phi^3. But what does that even mean? If it is not an equation with variables, then what do you mean by "two root solutions"?

I'm guessing that by "quadratic" you meant "containing radicals", though that is not at all what it means.

Let's get back to the start. It looks like post #4 may be the closest you came to stating what you mean:

Apparently what you meant by "equation" is "polynomial expression", and you want it to have a plus-or-minus in it so that it yields two values (not roots), one of which is "real" (you may have meant integer) and the other irrational (though "both" suggests you may mean that both "roots" should be real, irrational numbers). None of this is the correct meaning of the OP, but let's try to figure it out.

What's still missing is the role of x. Here you showed a polynomial expression with a variable, but in post #14 you said there should be no variables.

Can you correct my guesses about your meaning?

No - you have it. Regarding the variable x, I meant this to designate whatever the coefficients had to be in order to produce the desired effect. Fundamentally, the "polynomial expression" can only have one variable: shifts from + to - and everything else stays identical.

Please forgive the terrible use of terminology, I am not a mathematician however this problem is important to me because I may have found a real proof of the Reimann Hypothesis being true, in an very bizarre way.

 

[Romsek]

[MATH]\{\text{I am not a mathematician }\} \cap \{\text{I may have found a real proof of the Reimann Hypothesis}\} = \emptyset[/MATH]

 

[nothing]

[MATH]\{\text{I am not a mathematician }\} \cap \{\text{I may have found a real proof of the Reimann Hypothesis}\} = \emptyset[/MATH]

Would solving for the proper radical identity/expression of phi count, such that the derived expression yields 1, 4.23606797... as their solutions?

 

[pka]

Would solving for the proper radical identity/expression of phi count, such that the derived expression yields 1, 4.23606797... as their solutions?

Good grief, A TROLL What a waste of everybody's time .

 

[nothing]

Good grief, A TROLL What a waste of everybody's time .

Please do not scapegoat your own nature onto others - I asked a question. Would a radical expression whose only two solutions are 1 and 4.23606797... count? It's a yes/no question.