1. ## Solve this Quadratic Equation

Is it possible to solve (for x and y) this quadratic equation:

x^2 + 7xy - 3 = 0

2. Originally Posted by acevipa
Is it possible to solve (for x and y) this quadratic equation:

x^2 + 7xy - 3 = 0
What sort of solutions are you looking for?

3. ## not sure exactly

Hmmm. I really am not sure on this one, I just tried typing it into my fancy calculator and got a strange result so I don't think you can.

4. Originally Posted by acevipa
Is it possible to solve (for x and y) this quadratic equation:

x^2 + 7xy - 3 = 0

$\displaystyle x^2 + 7xy - 3 = 0$

$\displaystyle 7xy = -x^2 + 3$

$\displaystyle y = \frac{-x^2 + 3}{7x}$

$\displaystyle y = -\frac x7 + \frac 3{7x}$

It should be relatively clear from here that there are an infinite number of solutions.

5. Originally Posted by angel.white

$\displaystyle x^2 + 7xy - 3 = 0$

$\displaystyle 7xy = -x^2 + 3$

$\displaystyle y = \frac{-x^2 + 3}{7x}$

$\displaystyle y = -\frac x7 + \frac 3{7x}$

It should be relatively clear from here that there are an infinite number of solutions.
Alternatively, solving for x using the quadratic formula:

$\displaystyle x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}$.

The interesting question would be to find all integer solutions (x, y).

6. Originally Posted by mr fantastic
Alternatively, solving for x using the quadratic formula:

$\displaystyle x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}$.

The interesting question would be to find all integer solutions (x, y).
I'll take a stab at it.
Premise: x, y are integers

$\displaystyle x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}$

$\displaystyle 2x = -7y \pm \sqrt{49y^2 + 12}$

setup case 1: $\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$

$\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$

$\displaystyle \sqrt{49y^2 + 12} = 2x+7y$

$\displaystyle 49y^2 + 12 = 4x^2+28xy+49y^2$

$\displaystyle 12 = 4x^2+28xy$

$\displaystyle 3 = x^2+7xy$

setup case 2: $\displaystyle -7y - \sqrt{49y^2 + 12} = 2x$

$\displaystyle \sqrt{49y^2 + 12} = -2x-7y$

$\displaystyle 49y^2 + 12 = 4x^2 +28xy+49y^2$

$\displaystyle 12 = 4x^2 +28xy$

$\displaystyle 3 = x^2 +7xy$

Continued from $\displaystyle 3 = x^2 +7xy$
So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of $\displaystyle 3 = x^2 +7xy$

Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.

$\displaystyle \frac 3x =x + 7y$

Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.

So $\displaystyle x \in \{ -3, -1, 1, 3\}$

proof case 1: x = -3
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{-3} =-3 + 7y$

$\displaystyle \frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} -3$

proof case 2: x = -1
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{-1} =-1 + 7y$

$\displaystyle -\frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} -1$

proof case 3: x = 1
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{1} =1 + 7y$

$\displaystyle \frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} 1$

proof case 4: x = 3
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{3} =3 + 7y$

$\displaystyle -\frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} 3$

Therefore there are no integer values of x for which y can be an integer also.

So the solution to the problem "find all integer solutions (x, y)" is the empty set.

7. Originally Posted by angel.white
I'll take a stab at it.
Premise: x, y are integers

$\displaystyle x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}$

$\displaystyle 2x = -7y \pm \sqrt{49y^2 + 12}$

setup case 1: $\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$

$\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$

$\displaystyle \sqrt{49y^2 + 12} = 2x+7y$

$\displaystyle 49y^2 + 12 = 4x^2+28xy+49y^2$

$\displaystyle 12 = 4x^2+28xy$

$\displaystyle 3 = x^2+7xy$

setup case 2: $\displaystyle -7y - \sqrt{49y^2 + 12} = 2x$

$\displaystyle \sqrt{49y^2 + 12} = -2x-7y$

$\displaystyle 49y^2 + 12 = 4x^2 +28xy+49y^2$

$\displaystyle 12 = 4x^2 +28xy$

$\displaystyle 3 = x^2 +7xy$

Continued from $\displaystyle 3 = x^2 +7xy$
So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of $\displaystyle 3 = x^2 +7xy$

Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.

$\displaystyle \frac 3x =x + 7y$

Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.

So $\displaystyle x \in \{ -3, -1, 1, 3\}$

proof case 1: x = -3
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{-3} =-3 + 7y$

$\displaystyle \frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} -3$

proof case 2: x = -1
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{-1} =-1 + 7y$

$\displaystyle -\frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} -1$

proof case 3: x = 1
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{1} =1 + 7y$

$\displaystyle \frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} 1$

proof case 4: x = 3
$\displaystyle \frac 3x =x + 7y$

$\displaystyle \frac 3{3} =3 + 7y$

$\displaystyle -\frac 27 = y$

But this contradicts that y is an integer, so $\displaystyle x \not{=} 3$

Therefore there are no integer values of x for which y can be an integer also.

So the solution to the problem "find all integer solutions (x, y)" is the empty set.
This might have no solutions with $\displaystyle x$ and $\displaystyle y$ in $\displaystyle \mathbb{Z}$ but it does have four such solutions in $\displaystyle \mathbb{Q}$.

(Apply the rational root theorem to get a list of candidates for $\displaystyle x$, then find the (if it exists) $\displaystyle y$ that gives that root. This also gives a neater proof of no solutions in $\displaystyle \mathbb{Z}$)

RonL

(Of course ther is a deliberate mistake here in that there is an implicit assumption that 7y is an integer, correcting the mistake gives an infinite number of rational solutions)

8. Originally Posted by CaptainBlack
This might have no solutions with $\displaystyle x$ and $\displaystyle y$ in $\displaystyle \mathbb{Z}$ but it does have four such solutions in $\displaystyle \mathbb{Q}$.

(Apply the rational root theorem to get a list of candidates for $\displaystyle x$, then find the (if it exists) $\displaystyle y$ that gives that root. This also gives a neater proof of no solutions in $\displaystyle \mathbb{Z}$)

RonL
I think I understand, I looked up the rational root theorem, and got $\displaystyle \pm \frac {1, 3}{1, 3}$

So I could have just jumped straight to the 4 proof cases.

(I assume this is what you were suggesting, because there should be an infinite number of rational solutions to the problem.)