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Let a , b , c , d be real numbers and all zeroes β 1 , β 2 , β 3 , and β 4 of the polynomial P ( x ) = x 4 + a x 3 + b x 2 + c x + d are real. Find the smallest value the product ( β 1 2 + 1 ) ( β 2 2 + 1 ) ( β 3 2 + 1 ) ( β 4 2 + 1 ) can take.
This is a modified AIME problem.
Also try this .
The answer is 1.
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2 solutions
I LITERALLY DID NOT UNDERSTAND ANYTHING.
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Hey Bro, what do ya not understand?
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I understood the ques and even solved it, but it was joel's sol which not much self explanatory to me.
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@Vaibhav Prasad – Come on b'llant hangouts box!
We know, the smallest value of any expression is 0 , So, the least value of the zeros of the polynomial is β 1 = β 2 = β 3 = β 4 = 0 So, the smallest value of the product is 1
A square of a real is at least 0. Hence 1 is the answer. Equality is achieved when a = b = c = d = 0