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avatar+921 

Find the unique pair of real numbers (x,y) satisfying
(6x^2 - 18x + 17) + (3y^2 + 6y + 11) = 28

and x + y = 20.


Enter your answer as an ordered pair in the format $(x,y)$, where $x$ and $y$ are replaced by appropriate numbers.

 Feb 19, 2025
 #1
avatar+130462 
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6x^2 -18x + 17  + 3y^2 + 6y  + 11  = 28

 

x + y = 20

 

y  = 20 - x

 

So

 

6x^2 - 18x + 17 + 3(20 -x)^2 + 6(20 -x) + 11  = 28

 

6x^2  -18x + 3(x^2 - 40x + 400)  + 120 -6x + 28  = 28

 

6x^2 - 18x + 3x^2 -120x + 1200 + 120 -6x  =  0

 

9x^2 - 144x + 1320  =  0

 

No real solutions for this since the discriminant   144^2  - 36*1320 = -26784  is negative

 

cool cool cool

 Feb 21, 2025

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