MHB Find the number of real solutions

[anemone]

How many real solutions does the following system have?

$x+y=2$

$xy-z^2=1$

 

[juantheron]

Here is my solution:

Spoiler

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 $(x,y,z) = (1,1,0)$

 

[anemone]

Here is my solution:

Spoiler

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 $(x,y,z) = (1,1,0)$

Thanks for participating, jacks!

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

 

[Opalg]

[sp]Substitute $y = 2-x$ in the second equation: $x(2-x) - z^2 = 1$. Then $-(x-1)^2 - z^2 = 1-1 = 0$, or $(x-1)^2 + z^2 = 0$. So the only solution is $(x,y,z) = (1,1,0).$[/sp]

 

[kaliprasad]

Spoiler

x = 1 - p, y = 1 + p give

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