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How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$?

  1. $0$
  2. $1$
  3. $2$
  4. infinitely many
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$(x-y)^2 = (x+y)^2 - 4 xy$

$\implies (x-y)^2 = 4 - 4(1+z^2)$

$\implies (x-y)^2 = - 4z^2$

Now, $ \text{LHS} \geq 0$ and $\text{RHS}\leq 0.$ So both of them are equal only when $z = 0$ and $x = y$ .

Solving along with $ x+y = 2$ we get the triplet $(x,y,z) = (1,1,0)$

So $B$ is correct. :)
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We have, x  + y = 2 ----1

xy – z^2 = 1 ------2

Put value of x = (2 – y) from 1 to 2

=> (2 – y)y - z^2 = 1

=> y^2 – 2y + z^2 + 1 = 0

=> now, y = (2 ± sqrt(4 – 4(z^2 + 1))) / 2

=> y = (2 ± sqrt(-4Z^2)) / 2

-4z^2 can’t be negative, so only possible value of z is 0, 

now, y = 1, also x = 1 and z = 0.

So, 1, 1. 0 is the only pair.

 

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Using the idea in theory of equation, we know that if sum of two roots are given and product of two roots are given then we can form a quadratic equation whose roots are required

 

Here, x+y=2 and xy=1+z^2

let us form an equation t^2 – (x+y)xt + yx=0

after solving, you will get x =1+zi or 1-zi

Similarly, for y=1-zi or 1+zi

So,only possible real pair is (1,1)

 

PS: I don’t know latex,so I couldn’t write the equation, if some one any source for latex,please tell

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