How do you solve this?

1. Apr 9, 2016

gede

1. The problem statement, all variables and given/known data

$x + y + 3 = 2 \sqrt{x - 1} + 4 \sqrt{y - 1}$

2. Relevant equations

3. The attempt at a solution

My brain is blank. I give up.

2. Apr 9, 2016

SteamKing

Staff Emeritus
Well, it doesn't require calculus.

You should try to post HW threads in the proper forum.
You have one equation in two unknowns. You won't find a unique solution unless you can provide another equation.

3. Apr 9, 2016

Jingfei

Are they asking you to express y as a function of x?

4. Apr 9, 2016

epenguin

Polya principle - have I seen anything like this before? I'm sure you have.

However, to remind you (and in consideration of the fact it looks like quite a lot of calculation you will have to do) you have square roots there. What else can you do to bring those in relation with anything than square them? Not just square things at random of course, but do algebraically valid operations that involve squaring. In fact you have to square both sides of the equation. Then, as typical with this kind of problem, you will still have a square root in the equation. You have to manipulate it in the appropriate way so that you have square root by itself on one side of the equation. And which point at least the next step should be obvious.

It is not looking very easy, but we shall have to see. However at some point or other it will be useful for you - more useful than solving the problem - to take a look at your textbook, and problems you have done before, to recognise got there is something with something a bit like this, so that there was no need for your brain to be blank.

You might find it slightly helpful to change the variables into different ones by letting (x-1) = X, and (y-1)= Y

5. Apr 10, 2016

haruspex

There's an even handier change which will avoid any complicated squaring.
In the present case, it is only necessary to assume x and y are real.

6. Apr 10, 2016

epenguin

That is good hint towards solving the problem, whatever the problem is. I think you could find some integer values that satisfy the equation, if that were the question.

Which brings to an earlier Polya principle: have you understood the question? As the OP hasn't provided a question, we have given some pretty good help for whatever it is!

7. Apr 10, 2016

micromass

Staff Emeritus
What are you supposed to do? State $x$ in terms of $y$? State $y$ in terms of $x$? Draw a graph? Implicit differentiation?

8. Apr 10, 2016

haruspex

Almost surely, find x and y. See my reply to SteamKing in post #5.

9. Apr 10, 2016

gede

Anyone can solve this problem?

10. Apr 10, 2016

micromass

Staff Emeritus
Not if you don't post a problem, no.

11. Apr 10, 2016

epenguin

Well haruspex I now realise, can even do that, probably.

It is something you probably wouldn't realise unless you work on the problem. For that you have to

1. Start. That means knowing and stating what the problem is.

12. Apr 10, 2016

haruspex

Use epenguin's hint at the end of post #4. Then see if you can get my hint in post #5. If you can't, I'll spell it out a bit more, but try first.

13. Apr 10, 2016

Staff Emeritus
I can.

But I am not going to show you. You need to put some effort in yourself.

14. Apr 11, 2016

ehild

You can find an unique real solution.

15. Apr 11, 2016

ehild

It is necessary to assume both x and y greater than 1.
And I think you can give the OP that handier change of variables. This is a nice problem, the OP would learn a lot from it if he could go ahead.

16. Apr 11, 2016

haruspex

If x and y are real, each square root in the equation must be either real or lie on the positive imaginary axis (by the standard definition of the principal values of the square root function in the complex plane). So the imaginary parts cannot cancel in the sum.

17. Apr 14, 2016

Staff: Mentor

Thread closed for Moderation. We do not allow schoolwork questions to be asked with zero effort shown. It is against the PF rules.