Quadratic in x and linear in y problem

[baldbrain]

Homework Statement

If $x$ is a rational function of $y$, such that (ax² + bx + c)y + (a'x² bx' + c') = 0
prove that (ac' - a'c)² = (ab' - a'b)(bc' -b'c)

Homework Equations

The quadratic formula

The Attempt at a Solution

This equation can be rewritten as:
(ay + a')x² (by + b')x (cy +c') = 0
Since x is rational, the discriminant of the above quadratic in $x$ is a perfect square.
(by + b')² - 4(ay + a')(cy + c') is a perfect square.
(b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') is a perfect square.
(b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') = λ²
,where λ is an integer
That's all I reckon. How to proceed?

 

[epenguin]

Are you sure you have given the whole question?- looks incomplete to me, nothing to constrain these constants.
You seem to have forgotten a ' on last c of first line - c' ?

 

[baldbrain]

You seem to have forgotten a ' on last c of first line - c' ?

Yes, sorry.
Fixed

 

[baldbrain]

Are you sure you have given the whole question?- looks incomplete to me, nothing to constrain these constants.
You seem to have forgotten a ' on last c of first line - c' ?

That's it, there's no constraints to the constants. When no constraints are given, we assume them to be real numbers.

 

[baldbrain]

Also, I can notice an analogy. What we must prove is same as the condition for 2 separate quadratics (in x)
a₁x² + b₁x + c₁ = 0
&
a₂x² + b₂x + c₂ = 0
to have a common root, which is:
(c₁a₂ - c₂a₁)² = (a₁b₂ - a₂b₁)(b₁c₂ - b₂c₁)
But I doubt this has anything to do with this problem.

 

[BvU]

Does the problem statement restrict a, b and c to be rational numbers ? (as opposed to real numbers)

 

[baldbrain]

Does the problem statement restrict a, b and c to be rational numbers ? (as opposed to real numbers)

No. What I've typed out is the entire problem.

 

[SammyS]

Homework Statement

If $x$ is a rational function of $y$, such that (ax² + bx + c)y + (a'x² bx' + c') = 0
prove that (ac' - a'c)² = (ab' - a'b)(bc' -b'c)

Homework Equations

The quadratic formula

The Attempt at a Solution

This equation can be rewritten as:
(ay + a')x² (by + b')x (cy +c') = 0
Since x is rational, the discriminant of the above quadratic in $x$ is a perfect square.
(by + b')² - 4(ay + a')(cy + c') is a perfect square.
(b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') is a perfect square.
(b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') = λ²
,where λ is an integer
That's all I reckon. How to proceed?

I suppose you meant to have the prime symbol on b, not on x, in the second quadratic as well as a plus sign leading the b', so it would be: (a'x² + b'x + c')

Here's an idea. I'm not sure it works.

Consider your result the discriminant.

(b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') is a perfect square.

But this is a quadratic in y. For it to be a perfect square, its discriminant must be ZERO.

See if that helps.

Added in Edit: Yup, That seems to do it !

 

[epenguin]

Also, I can notice an analogy. What we must prove is same as the condition for 2 separate quadratics (in x)
a₁x² + b₁x + c₁ = 0
&
a₂x² + b₂x + c₂ = 0
to have a common root, which is:
(c₁a₂ - c₂a₁)² = (a₁b₂ - a₂b₁)(b₁c₂ - b₂c₁)
But I doubt this has anything to do with this problem.

I bet it does. If the question had been find a condition for the equation to represent a pair of straight lines, wouldn't that be the answer? And you could say what they are.
At least I've found a question to your answer.

 

[baldbrain]

I suppose you meant to have the prime symbol on b, not on x, in the second quadratic as well as a plus sign leading the b', so it would be: (a'x2 + b'x + c')

Yes, sorry again

 

[baldbrain]

I suppose you meant to have the prime symbol on b, not on x, in the second quadratic as well as a plus sign leading the b', so it would be: (a'x² + b'x + c')

Here's an idea. I'm not sure it works.

Consider your result the discriminant.But this is a quadratic in y. For it to be a perfect square, its discriminant must be ZERO.

See if that helps.

Added in Edit: Yup, That seems to do it !

Oh right, thanks!
Damn! Why did I not think this...
Thank you to @epenguin and @BvU as well.

 

[baldbrain]

I still have one question.
What if, there is an arbitrary non-zero λ such that (b'² - 4a'c' - λ²,y₁) is a point on the y-quadratic?

 

[SammyS]

I still have one question.
What if, there is an arbitrary non-zero λ such that (b'² - 4a'c' - λ²,y₁) is a point on the y-quadratic?

I'm not sure about that at all.

Let's go back to the problem statement which states that $\ x\$ is a rational function of $\ y \$.

I take the use of the term "rational function" to mean that $\displaystyle \ x = \frac{P(y)}{Q(y)} \$, where both $\ P(y)\$ and $\ Q(y) \$ are polynomial functions of$\ y\$.

This being the case, I do not expect that $\ \lambda \$ would be a constant, let alone an integer. I hadn't given this much thought until you posed this last question.

 

[baldbrain]

This being the case, I do not expect that $\ \lambda \$ would be a constant, let alone an integer. I hadn't given this much thought until you posed this last question.

If λ isn't a constant, then what we must prove isn't necessarily true?

 

[SammyS]

If λ isn't a constant, then what we must prove isn't necessarily true?

No. The arguments about the discriminants still make sense. (Come to think of it, we do need to have the coefficients be integers or at least be rational numbers.)

See, if $\ x\$ is to be a rational function, then the solution for $\ x\$ can't have radical expressions in its makeup.

 

[epenguin]

I admit i misunderstood the question - though I would claim, .understandably.
I had naturally misread that y is a rational function of x. Which it obviously is, in fact a 'proper rational' function, from the equation.
But if, as the question states and I understandably missed, x is also a rational function of y.*, suppose this be the ratio of two polynomials in y of degree m and n. Then substituting this rational function of y for x and clearing of fractions gives us a polynomial in y of degree (2m + 1) or 2n, whichever of these is the higher. Equated to 0, your equation is something which can be true at maximum this last number of real or nonreal complex values.

I cannot see from the statement of the problem, any constraint on what these, or the constants involved, are.

It may be possible to deduce from from the formula it is required to prove, what the question is as we often do here. I tried in #9 and am not going to try further in this case. Unless my understanding is still mistaken.

* However, the OP made so many mistranscripts that I think this needs to be verified.

 

[baldbrain]

Equated to 0, your equation is something which can be true at maximum this last number of real or nonreal complex values.

I get didn't understand that. (Typos maybe)

However, the OP made so many mistranscripts that I think this needs to be verified.

There's no problem there: "$x$ is a rational function of $y$"

 

[baldbrain]

Equated to 0, your equation is something which can be true at maximum this last number of real or nonreal complex values.

Do you mean to say that, assuming b² - 4ac < 0, (b² - 4ac)y² + [2bb' - 4(ac' + a'c)]y + (b'² - 4a'c') = 0 occurs at the maxima of f(y)?

 

[epenguin]

I get didn't understand that. (Typos maybe)

Which can be true at only (2m + 1) or 2n, whichever is the larger number, of values of y.

 

[baldbrain]

Which can be true at only (2m + 1) or 2n, whichever is the larger number, of values of y.

Ok. Thanks again