Question

Find all the non-negative integral solutions of \(y^2 + 6xy - 8x =0 \). Does it contain a positive integral solution which is a multiple of 3?

Answer 1

@surjithayer

Answer 2

Dunno, have you tried Quadratic formula?

Answer 3

Yeah, will not work for sure.

Answer 4

DON'T TRY IT - it will make everything complicated.

I think, there should be a trick in there.

Answer 5

I don't see how the Quadratic formula would've worked here.

Answer 6

What if we do this : y^2 = 8x - 6xy

y/x = 8/y - 6

Answer 7

I have a hunch, but I rather not look like an idiot if it backfires.

Answer 8

There, I divided both sides by xy ...

\(\cfrac{y^2}{xy} = \cfrac{8x}{xy} - \cfrac{6xy}{xy} \)

\(\cfrac{y}{x} = \cfrac{8}{y} - 6 \)

@iambatman may want to suggest something too.

Answer 9

Finally someone has got an idea

mathmale yupiieee

Answer 10

rational root theorem?

Answer 11

What is it wio, never heard about it. Sorry

Answer 12

Hello!!

I'd suggest you treat y as a variable and x as a constant. You could then re-write your equation as \[y^2+(6x)y-8x=0,\]...and for the purpose of applying the quadratic formula, we could say that a=1, b=6x and c=-8x.

Applying the quadratic formula, \[y=\frac{ -(6x)\pm \sqrt{(6x)^2}-4(1)(-8x) }{ 2(1) }, ~or\]

Answer 13

Yes sir, that was the method I went for, but this didn't lead to any conclusion for me.

Answer 14

\[y=\frac{ -6x \pm \sqrt{36x^2+32x} }{ 2 }\]

Answer 15

Now ask yourself what happens if the discriminant, 36x^2+32x, happens to be zero. How many roots will you then have? Will they be real, imaginary or complex? Equal or unequal?

Answer 16

Integer or non-integer?

Answer 17

If D = 0 then roots will be real, equal but non-integer.

Answer 18

As D = 0 means

36x^2 + 32x = 0

36x^2 = -32x

x = -32/36

and the roots were : -3x

-3 * (-32/36) = 32 / 12 = 8 /3 = roots.

Answer 19

Next, look at the discriminant, 36x^2+32x, again. Are there any values of x for which the discriminant is both positive and a perfect square? If so, that should enable us to find two real, unequal, integer roots.

If you do that, all you have to do as a last step is to determine whether or not either or both of the resulting real, unequal, integer roots is divisible by 3.

Answer 20

let me verify this: By "integral," do you mean that your roots are to be INTEGERS ONLY, or did you have something else in mind?

Answer 21

By integral, I only mean "integers" .. nothing else sir!

Answer 22

Good, that simplifies matters!

When we started out here, I suggested that we treat y as a variable and x as a constant.

If the discriminant is zero, then there are two real, equal roots whose values depend upon the value of the "constant" x, right? Does x HAVE TO BE an integer?

Answer 23

I don't think x wil be an integer then.

Answer 24

*will.

Answer 25

I have not yet solved this problem. Rather, I'm posing questions that I hope may guide you towards finding a solution of your own.

Look at \[y=\frac{ -6x \pm \sqrt{36x^2+32x} }{ 2 }~again.\]

Answer 26

If it is in fraction, the denominator should be a multiple of 3, right? if it is an integer, anything will work ...

We have -3x

Answer 27

We could do case studies on this to determine whether or not there is/are any solution/s divisible by 3.

Case 1: Assume that x MUST be an integer and proceed accordingly.

Case 2: Drop the assumption that x must be integer and determine whether there could still be one or more real, integer roots to the original equation divisible by 3.

Answer 28

In Case 1, we'd want to know whether there's some integer x which, when substituted into the discriminant 36x^2+32x, results in a positive perfect square. Is there such a beast?
:)

Answer 29

x = -1 ?

Answer 30

Try it. If you substitute x=-1 into the discriminant formula, is the result (+)? And is the result a perfect square?

Answer 31

\[32(-1)^2+32(-1)=?\]

Answer 32

36(1) - 32 = 4

Answer 33

perfect square and positive ...

Answer 34

So both criteria are satisfied!

Thus, with x=-1, your solutions are ... what? There should be 2 such solutions.

We'll ask later whether either or both of these 2 solutions are divisible by 3.

Answer 35

\(\cfrac{-6(-1) \pm 2 }{2} = \cfrac{6 \pm 2}{2} = 4 ~ or ~ 2 \)

Answer 36

Neither of these are divisible by 3.

Answer 37

You do apparently have two solutions to the original equation, even tho' it happens that neither is divisible by 3. So, how would y ou answer this question?

Answer 38

It's important that you check your answer(s). We're saying that x=-1 and that y = either 2 or 4, right? Is (-1,2) a solution of the original equation? Is (-1,4) such a sol'n? They must be, or else we have not found solutions.

Answer 39

o.O this denies a fact actually :

\(y^2 + 6xy - 8x =0 \)
\(y^2 = 8x - 6xy\)
\(y^2 = x(6- 6y)\)
\(x = \cfrac{y^2}{6 - 6y}\)

Answer 40

I see what you're doing, but wonder whether there's any advantage to what appears to be making the problem more complicated than it is.

Answer 41

We have to find non-negative solutions, so, we can't take x = -1 into consideration...

x has to be positive. Am I wrong in my thinking sir?

Answer 42

My question: If:\[y^2 + 6xy - 8x =0,\] is (-1,2) a solution?

Answer 43

sorry, I did mistake there...

4 + 6(-1) ( 2) + 8 = 4 -12 + 8 = 0

its a solution, sir!

Answer 44

I get:\[(2)^2+6(-1)(2)-8(-1),\]This differs from your result. Why?

Answer 45

OK, so you have found that the point (-1,2) satisfies the original equation.

Answer 46

Yes sir. It satisfied.

Answer 47

The wording is just a little confusing and troubling. We are to find "nonnegative integer solutions." Our x=-1 is certainly not nonnegative. However, you'll recall that I arbitrarily dubbed y as the variable and x as a sort of constant. If this be the case, then our solutions are 2 and 4, both of which are nonnegative integers. What do you think?

Answer 48

I don't think that we were supposed to take x as a constant or a sort of constant. As, what we have the solutions of the given equation as : (x,y) i.e. (-1,2) or (-1,4)....

y depends on x and x depends on y. neither of them can be taken as constant.

Answer 49

Though, if we follow some logical thinking, we could have done it in 2 lines. I just got a new but correct (hopefully) way to do this problem.

As each term has a variable, thus, the first solutions are clearly : (0,0)

And the fact is that this is the only solution.

Answer 50

* this is the only non-negative integral solution.

I am trying to prove it though.

Answer 51

Recall that I proposed doing this problem as a set of case studies. Case #1 involved restricting x to integers only. We've apparently found valid solutions.

Case #2 would involve removing the restriction that x be integer. Were we very lucky and were able to find an x value that made the discriminant a positive, perfect square whose square root had a denominator equal to the denominator of 6(-x), then we might have additional integer solutions. Sort of makes my head swim, however.

Answer 52

We can certainly divide problems up into cases as we've done here.

Yes, the wording "not supposed to treat x as a constant" is troublesome. We know that y and x are actually variables. Perhaps one way in which you could look at this problem differently would be to regard y as a function of x. Not certain whether or not this would be appropriate.

The acid test is whether our "solutions" are actually solutions to the original equation. We have already shown that (-1,2) IS a solution, and could probably show that (-1,4) is also a solution.

Answer 53

This is what I came up with. Not sure whether this is a perfect method or not.

We had : \(y^2 + 6xy - 8x = 0\)

solving for x in terms of y :

\(x = \cfrac{y^2}{8 - 6y}\) ; but x is positive.

And thus, \(8 - 6y > 0\) and y should be a non-negative integer. This, there are only 2 possible values of y : 0 and 1.

for y = 1, x is not an integer. thus, we are only left with (0,0) as the solution.

This method answers what the question requires but

Answer 54

I always love the way by which you take some cases and then go further in solving the question. This reduces the chances of mistakes. And with the method we followed (using quadratic formula) , we also come up with the same answer, that :

Only non-negative solution of x,y is (0,0) . Hence the equation has only one non-negative integral solution. So, it has no positive integral solutions which are multiples of 3.

We only missed one condition there, to think of x can be zero too!

Answer 55

I'm happy you're happy with our results. The only area where I remain a bit uncomfortable lies in having to decide for ourselves what "non-negative solution" means in this context.

It seems that "non-negative" would apply just fine to a scalar numerical answer, such as x=5. But here we're discussing a POINT, which has two coordinates instead of one.

Would we need to argue that (-1,2) is negative because one of its two coordinates is negative?

Answer 56

Actually non negative solution refers to the condition when both x and y are more than or equal to zero.

Answer 57

If (and that's a big IF) we accept that as the definition of "non-negative solution," then clearly (-1,2) and (-1,4) do not qualify, but (0,0) does.

Answer 58

Yeah

Answer 59

@mathslover: Does this discussion leave you fully satisfied, or are there still loose ends to address?

Answer 60

This discussion was actually perfect for me! I had some wrong concepts in mind, and those are vanished now. And yes sir, I'm satisfied. Thanks a lot for your help. I appreciate it a lot.


\(\checkmark\)

Answer 61

so pleased to have had another opportunity to work with you! See you again soon!

Answer 62

Same here sir! :-) Thank you again.