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If the answers to a quadratic equation were 2+6√3 and 2-6√3, then what was the original polynomial

User Metase
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1 Answer

3 votes

Answer:


p(x) = k[ x^2-4x-104]

Explanation:

Say if alpha and beta are the zeroes of the quadratic polynomial then the quadratic polynomial is given by ,


\implies p(x) = k[x^2-(\alpha+ \beta)x +\alpha\beta ]

  • where k is a constant


\implies p(x) = k[ x^2 - [ (2+6\sqrt3)+(2-6\sqrt3)x] + \{ (2+6\sqrt3)(2-6\sqrt3)\} ] \\\\\implies p(x) = k[ x^2 - 4x + \{ (2)^2-(6\sqrt3)^2\}] \\\\\implies p(x) = k[ x^2 -4x \{ 4-108\} ] \\\\\implies\boxed{\boxed{ p(x) = k[ x^2-4x-104] }}

This is the required answer !

User Natrium
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