# Thread: find all (x;y) real numbers if:

1. Find all if :

2.

3. metalari, we won't do your homework for you.

4. Obviously, its solution is not an equation, but an inequation ... because here is one equation for two unknown numbers ... .

Maybe ... , here is no analytical solution, rather than numerical solution ... .

5. Originally Posted by metalari
Find all if :

Write the equation as

The LHS is so its minimum value is 2. On the other hand, the maximum value of the RHS is 2; this is because the minimum value of is 1 and so the maximum value of is 1. Hence equality occurs when both sides are equal to 2, i.e. at and those values of for which .

6. Could it not be re written as a quadratic type (as shown below) where the "x" in the quadratic formula is , factored and further simplified? When I tried, my result matched that from matematica.

7. Originally Posted by Differential-Cheese
Could it not be re written as a quadratic type (as shown below) where the "x" in the quadratic formula is , factored and further simplified? When I tried, my result matched that from matematica.
How would you factor it? It doesn’t look very factorable to me. I tried completing the square but found that it hardly simplified things.

My method above is basically this: LHS ≥ 2 for all x, y; RHS ≤ 2 for all x, y; therefore LHS = RHS if and only if LHS = 2 = RHS.

8. Excuse me ... .

Originally Posted by metalari
Find all if :

If , then alias ... .

with

Thus, ... .

The RHS in last equation depends only ... .

Note: ... .

9. Then how would you deduce from your method? You realize that the equation has no solution in for don’t you?

10. If , then

,

,

,

,

or

with ... .

11. You have not answered my question. How can you tell from this

Originally Posted by trfrm
that you cannot have ?

12. To jump in here, it follows from the inside square root and the fact that we want real solutions.

13. Originally Posted by Crimson Sunbird
that you cannot have ?
In general, we have to use table of and square root or use calculator ... . Obviously, we can find for some 's ... because ... . However, for a , in general, the solution is not unique ... .

14. Originally Posted by trfrm
Originally Posted by Crimson Sunbird
that you cannot have ?
In general, we have to use table of and square root or use calculator ... . Obviously, we can find for some 's ... because ... . However, for a , in general, the solution is not unique ... .
You are still not answering my question. I did not say anything about x being unique or not. My question was: how can you deduce from your expression relating x and y? In any case river_rat has answered my question on your behalf – so I think I will stop now.

FYI the problem posed in this thread looks very much like a Mathematical Olympiad question. Are you familiar with such questions? Besides the International Mathematical Olympiad (IMO) there are math olympiad contests at national and local levels. If you’ve tried to do some of the problems from these contests, you’ll get an idea of what’s expected for each question – and experience can also tell you how best to go about solving a typical question.

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