RMO Practice question 10

Algebra Level 5

Let η ( ω ) = ( ω 2 + 2015 ω − 1 ) 2 + ( 2 ω + 2015 ) 2 \eta(ω)=(ω^{2}+2015ω-1)^{2}+(2ω+2015)^{2} Find the sum of all real solutions to the equation η ( ω ) = min ⁡ ( η ( ω ) ) \eta(ω)=\min(\eta(ω))


The answer is -2015.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Sarthak Behera
Sep 24, 2015

D i f f e r e n t i a t i n g , W e g e t 2 ( x 2 + 2015 x − 1 ) ( 2 x + 2015 ) + 2 ( 2 x + 2015 ) ( 2 ) = 0 ⇒ ( 2 x + 2015 ) ( x 2 + 2015 x + 1 ) = 0 − 2015 2 l i e s b e t w e e n r o o t s o f ( x 2 + 2015 x + 1 ) . S o , − 2015 2 i s a l o c a l m a x i m u m w h i l e r o o t s o f ( x 2 + 2015 x + 1 ) a r e g l o b a l m i n i m u m . ( b y f i r s t d e r i v a t i v e t e s t ) T h u s b y v i e t a ′ s , w e g e t s u m a s − 2015 . Differentiating,\quad We\quad get\quad 2\left( { x }^{ 2 }+2015x-1 \right) \left( 2x+2015 \right) +2\left( 2x+2015 \right) \left( 2 \right) =0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Rightarrow \left( 2x+2015 \right) \left( { x }^{ 2 }+2015x+1 \right) =0\\ \quad -\frac { 2015 }{ 2 } \quad lies\quad between\quad roots\quad of\quad \left( { x }^{ 2 }+2015x+1 \right) .\\ So,\quad -\frac { 2015 }{ 2 } is\quad a\quad local\quad maximum\quad while\quad roots\quad of\quad \left( { x }^{ 2 }+2015x+1 \right) \quad are\quad global\quad minimum.\quad (by\quad first\quad derivative\quad test)\\ Thus\quad by\quad vieta's,\quad we\quad get\quad sum\quad as\quad \boxed { -2015 } .

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