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Hard math:

 

If x+y = 2 and xy = 23, then what is x^2 + y^2?

 Jun 23, 2017
 #1
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+1

There are no "real" solutions, just "complex" ones:

x = 1 - i sqrt(22) ≈ 1.00000 - 4.69042 i and y = 1 + i sqrt(22) ≈ 1.00000 + 4.69042 i

x = 1 + i sqrt(22) ≈ 1.00000 + 4.69042 i and y = 1 - i sqrt(22) ≈ 1.00000 - 4.69042 i

So that: ( 1 + i sqrt(22))^2 + (1 - i sqrt(22))^2 

Simplify the following:
(i sqrt(22) + 1)^2 + (-(i sqrt(22)) + 1)^2

(i sqrt(22) + 1)^2 = 1 + i sqrt(22) + i sqrt(22) - 22 = 2 i sqrt(22) - 21:
2 i sqrt(22) - 21 + (-(i sqrt(22)) + 1)^2

(-(i sqrt(22)) + 1)^2 = 1 - i sqrt(22) - i sqrt(22) - 22 = -2 i sqrt(22) - 21:
-21 + 2 i sqrt(22) + -2 i sqrt(22) - 21

-21 + 2 i sqrt(22) - 21 - 2 i sqrt(22) = -42:
Answer: |  -42

 Jun 23, 2017
 #2
avatar+98130 
+2

 

x + y  = 2     square both sides →  x^2 + 2xy + y^2  =  4

 

Since  xy  = 23,, then  2xy  = 46

 

So we have that

 

x^2 + 46 + y^2  = 4       subtract 46 from both sides

 

x^2 + y^2  =  -42

 

 

cool cool cool

 Jun 23, 2017

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