Question 1209394:  Let x and y be complex numbers.  If x + y = 2 and xy = 7 - x^2 - y^2, then what is x^2 + y^2? 
 Found 3 solutions by  ikleyn, math_tutor2020, Edwin McCravy: Answer by ikleyn(52903)      (Show Source): 
You can  put this solution on YOUR website! . 
Let x and y be complex numbers.  If x + y = 2 and xy = 7 - x^2 - y^2, then what is x^2 + y^2? 
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. 
x^2 + y^2 = add and subtract 2xy = (x^2 + 2xy + y^2) - 2xy = first parentheses is the square = (x+y)^2 - 2xy = 
          = substitute the given values = 2^2 - 2*(7-x^2-y^2).
Thus you have now
    x^2 + y^2 = 4 - 14 + 2(x^2 + y^2),
    x^2 + y^2 = -10 + 2(x^2 + y^2),
     10 = x^2 + y^2.
At this point, the problem is solved completely.
ANSWER.  x^2 + y^2 = 10.
 
Solved.
 
 
 
 
 Answer by math_tutor2020(3817)      (Show Source): 
You can  put this solution on YOUR website!  
Answer: 10
 
 
 
Work Shown 
xy = 7 - x^2 - y^2 
xy = 7 - (x^2 + y^2) 
x^2 + y^2 = 7 - xy 
2(x^2 + y^2) = 14 - 2xy 
Let's call this equation (3)
 
 
x + y = 2 
(x + y)^2 = 2^2 
x^2 + 2xy + y^2 = 4 
x^2 + y^2 = 4 - 2xy 
Let's call this equation (4)
 
 
Subtract straight down for equations (3) and (4) to arrive at 
x^2 + y^2 = 10
 
 
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An alternative approach would be to rearrange x+y = 2 into y = 2-x 
Then, 
xy = 7-x^2-y^2 
x(2-x) = 7-x^2-(2-x)^2 
Solving that for x yields x = -1 and x = 3 
I'll let the student handle the scratch work.
 
 
If x = -1 then y = 2-x = 2-(-1) = 3 
If x = 3 then y = 2-x = 2-3 = -1 
We get the same pair of values just in a different order. 
Turns out the order doesn't matter.
 
 
Then, 
x^2 + y^2 = (-1)^2 + (3)^2 = 10 
 
 Answer by Edwin McCravy(20064)      (Show Source): 
You can  put this solution on YOUR website! 
As the last tutor showed, 
{x,y} = {3,-1} and x2 + y2 = 9 + 1 = 10
What the student should observe here is that 
All REAL numbers ARE COMPLEX numbers!
3 and -1 are complex numbers because they are 3 + 0i and -1 + 0i respectively.
Nobody said the coefficient of " i " cannot be 0. And if it's 0, the number
is still complex.  It's both REAL and COMPLEX.
Edwin 
  
 
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