(Un) Symmetric Quadratic
f ( x ) = a x 2 − c x + b
The coefficients of the above quadratic polynomial satisfy the conditions given below.
{ a , b , c > 0 ( a − c ) 2 + ( b − c ) 2 + 2 a b = c 2
Find the value of f ( 1 ) .
The answer is 0.
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1 solution
( a − c ) 2 + ( b − c ) 2 + 2 a b = c ² a 2 + b 2 + c 2 − 2 a c − 2 b c + 2 a b = 0 ( a − c + b ) 2 = 0
The second condition can be rewritten as :
( a + b + c ) 2 = 4 c ( a + b )
However , by AM - GM inequality :
[ ( a + b ) + c ] 2 ≥ 4 c ( a + b )
Hence , a + b = c
So, f ( 1 ) = a − c + b = 0