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- Feb 14, 2012

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How many real solutions does the following system have?

$x+y=2$

$xy-z^2=1$

$x+y=2$

$xy-z^2=1$

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

How many real solutions does the following system have?

$x+y=2$

$xy-z^2=1$

$x+y=2$

$xy-z^2=1$

Here is my solution:

Given $x+y = 2$ and $xy-z^2 = 1\Rightarrow xy = 1+z^2 \geq 1$

So $xy\geq 1$ Means either both$x,y$ are positive quantity OR both $x,y$ are negative quantity

But Given $x+y = 2$. So $x,y>0$

Now Using $\bf{A.M\geq G.M}$, we get $\displaystyle \frac{x+y}{2}\geq \sqrt{xy}$

So $xy\leq 1$ and equality hold when $x = y= 1$ but above we have got $xy\geq 1$

So $xy = 1$ means $x=y =1$. So $z=0$

So there is only one solution which is given as $(x,y,z) = (1,1,0)$

So $xy\geq 1$ Means either both$x,y$ are positive quantity OR both $x,y$ are negative quantity

But Given $x+y = 2$. So $x,y>0$

Now Using $\bf{A.M\geq G.M}$, we get $\displaystyle \frac{x+y}{2}\geq \sqrt{xy}$

So $xy\leq 1$ and equality hold when $x = y= 1$ but above we have got $xy\geq 1$

So $xy = 1$ means $x=y =1$. So $z=0$

So there is only one solution which is given as $(x,y,z) = (1,1,0)$

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- #3

- Feb 14, 2012

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Thanks for participating,Here is my solution:

So $xy\geq 1$ Means either both$x,y$ are positive quantity OR both $x,y$ are negative quantity

But Given $x+y = 2$. So $x,y>0$

Now Using $\bf{A.M\geq G.M}$, we get $\displaystyle \frac{x+y}{2}\geq \sqrt{xy}$

So $xy\leq 1$ and equality hold when $x = y= 1$ but above we have got $xy\geq 1$

So $xy = 1$ means $x=y =1$. So $z=0$

So there is only one solution which is given as $(x,y,z) = (1,1,0)$

And thanks also for showing us a nice method to solve the problem!

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- #4

- Feb 7, 2012

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- Mar 31, 2013

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xy - z^2 = 1

or 1-p^2 - z^2 = 1 or p^2 + z^ 2 = 0

p = 0 and z = 0 -> x =y = 1 , z = 0

so only one solution

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