If $x+y=2$ and $xy-z^2=1,$ what is the sum of all real solutions of $x$ , $y$ and $z$ ?

The answer is 2.

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$2x-x^{2}-z^{2}=1\Rightarrow x^{2}-2x+1+z^{2}=0\Rightarrow (x-1)^{2}+z^{2}=0$ $\therefore x=1, y=1, z=0$

Dieuler Oliveira
- 6 years, 10 months ago

beautiful question

Milan Kashyap
- 6 years, 10 months ago

What is D ?

Brad Cork
- 6 years, 11 months ago

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The discriminant of $x^2-2x+(z^2+1)$ .

mathh mathh
- 6 years, 11 months ago

D is short called for Delta

Pathern Duy Đinh
- 6 years, 11 months ago

how can you calculate D

John Kamel
- 6 years, 11 months ago

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The discriminant of $ax^2+bx+c$ , $a\neq 0$ is $D=b^2-4ac$ . You can read more about it here .

mathh mathh
- 6 years, 11 months ago

I thought the D was a simbol of z in formulae

Alif Nizam
- 6 years, 11 months ago

i think is x= 0 , y = 2 and z=1 . so simple .. SUM 0+2+1 = 3 ..no need to do soo much :)

vinit mahajan
- 6 years, 11 months ago

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No, that does not satisfy the second equation. We get $0\cdot 2-(1)^2=-1\ne 1$ .

Daniel Liu
- 6 years, 11 months ago

Loved the problem (y)

Jayeeta Sarkar
- 6 years, 10 months ago

When you got z = 0, the sum x + y + z is left with x + y, which is 2. No need to calculate x, y, unless you're just curious, as your name tells it all :). Nice one btw

Anh Vu
- 6 years, 11 months ago

Since z has even power(squared), it will take the values of the form k,-k.

If we sum up all the values, it will always be zero. So we just have to add x and y.

So, answer is x + y, which is 2 as per question.

So, answer is 2.

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you can also have x = 0 and y =2, or x = 2 and y = 0. but as much as it does not satisfy the next condition, it can't be a solution.

Danrey Vallejos
- 6 years, 10 months ago

you can also have x = 0 and y =2, or x = 2 and y = 0. but as much as it satisfies the next condition, it can't be a solution.

Danrey Vallejos
- 6 years, 10 months ago

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since z^2 is always nonnegative we know that xy must be at least 1, meaning it must be positive. this means either x and y are both positive or both negative. the sum of any two negative numbers must be negative and thus cannot equal 2, so x and y have to be both positive. this leave three outcomes for (x,y) based on the first equation. they are (0,2) (1,1) and (2,0). the part established earlier about xy being at least 1 is not met by (0,2) or (2,0), which leaves only (1,1) as our solution for (x,y). some quick substitution on the second equation gives 1*1-z^2=1. z=0. the only solution set is (1,1,0) which, when added together, gives our answer of 2.

that's kind of my "makes sense" solution which is not a mathematically sound proof, which is something I need to work on. my "three outcomes" statement up there is a bit rough, so let me see if I can clean that section up here. rearranging the first equation gives y=2-x. knowing both x and y to be positive gives us the solution set (x,2-x) for all 0<=x<=2. the other part we established earlier (the part about xy at least 1) can be rewritten using the new y as x(2-x)>=1 or 2x-x^2>=1. this inequality is only true at x=1. using the first equation gets us the single solution for (x,y) of (1,1).

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i just thinking about if z = 0, then xy = 1. and to make xy=1, x and y must 1.. and its correct

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From $x + y = 2$ , we can deduce (or differentiate $xy$ ) that the maximum value of $xy$ is $1$ and $z^2$ has to be larger or equal to zero if $z$ is real number.

The maximum of $xy$ minus the minimum of $z^2$ is equal to $1$ , which means the solution to these equations is unique with $x=1, y=1, z=0$ .

Hence $x + y + z = 2$ .

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We have $x + y >= 0$ .

Moreover, $xy - z^2 = 1 \Leftrightarrow z^2 = xy - 1 \Leftrightarrow xy \geq 1 \Leftrightarrow x \geq 1/y$

We know that $y + \dfrac {1}{y} = 2$ with the equality catched when $y = 1 \Leftrightarrow x = y = 1$

Finally, we get $z^2 = xy - 1 = 1 - 1 = 0$

So, the only solution of the system is $(x, y, z) = (1, 1, 0)$ .

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When $y=0$ , we don't have $xy\ge 1\iff x\ge \frac{1}{y}$ . You haven't specified that $y\neq 0$ . As for the later steps, you couldn't have deduced that $y+\frac{1}{y}=2$ just because $x\ge \frac{1}{y}$ . We could have only found out that $2\ge y+\frac{1}{y}$ .

mathh mathh
- 6 years, 11 months ago

to reply ..it can be even x^2+y^2+2z^2=2 ... then the answer is x=0 y=0 and z=1 then the sum is 1 !

Nirmal Parihar
- 6 years, 11 months ago

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$x+y=2\iff y=2-x\implies x(2-x)-z^2=1$ $\iff x^2-2x+(z^2+1)=0$ $\implies D=-4z^2\ge 0\implies z=0.$

Hence $z=0$ . This leaves us with a simple system of equations: \begin{cases}x+y=2\\xy=1\end{cases}\implies x+\frac{1}{x}=2\stackrel{x\neq 0}\iff x^2-2x+1=0\implies x=1\implies y=1.

After checking $(1;1;0)$ , we see it is a solution. Hence $1+1+0=\boxed{2}$ .