Can you spot the trick?
Let and be four real numbers satisfying the system of equations above.
What is the sum of all possible different values of ?
The answer is 116.
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1 solution
Moderator note:
Very nice question. How did you come up with that?
Let f ( x ) = ( x 2 + a x + c ) ( x 2 + b x + d )
Multiplying we get
f ( x ) = x 4 + ( a + b ) x 3 + ( a b + c + d ) x 2 + ( b c + a d ) x + c d
Substituting the values given we get
f ( x ) = x 4 + 8 x 3 + 2 3 x 2 + 2 8 x + 1 2
⇒ f ( x ) = ( x + 1 ) ( x + 2 ) 2 ( x + 3 )
⇒ f ( x ) = ( x 2 + 4 x + 4 ) ( x 2 + 4 x + 3 ) or ( x 2 + 5 x + 6 ) ( x 2 + 3 x + 2 )
Comparing with original equation we get possible values of ( a , b , c , d ) as
( 5 , 3 , 6 , 2 ) , ( 3 , 5 , 2 , 6 ) , ( 4 , 4 , 4 , 3 ) , ( 4 , 4 , 3 , 4 )
Sum of different possible values = 4 3 + 3 6 + 3 7 = 1 1 6