(Un) Symmetric Quadratic

Algebra Level 3

f ( x ) = a x 2 c x + b \large f(x)=ax^2-cx+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 \begin{cases} a,b,c >0\\ (a-c)^2+(b-c)^2+2ab=c^2 \end{cases}

Find the value of f ( 1 ) f(1) .


The answer is 0.

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1 solution

The second condition can be rewritten as :

( a + b + c ) 2 = 4 c ( a + b ) (a+b+c)^2=4c(a+b)

However , by AM - GM inequality :

[ ( a + b ) + c ] 2 4 c ( a + b ) [(a+b)+c]^2 \ge 4c(a+b)

Hence , a + b = c a+b=c

So, f ( 1 ) = a c + b = 0 f(1)=a-c+b=\boxed{0}

( 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 (a-c)^2+(b-c)^2+2ab=c² \\ a^2+b^2+c^2-2ac-2bc+2ab=0 \\ (a-c+b)^2=0

Akshat Sharda - 3 years, 2 months ago

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