Question

find the polynomial whose zeroes are 2+√3 and 2-√3

Answer 1

Ans: The polynomial whose zeroes are 2+√3 and 2-√3 is x^2 - 4x + 1

Answer 2

$\sf{Let,}$

$\sf{ \alpha \ = 2 + {\sqrt{3}} \ \ and \ \ \beta \ = 2 - {\sqrt{3}}}$

$\sf{Now,}$

$\sf{Sum \ of \ zeroes \ -}$

$\sf{= \ \alpha + \beta}$

$\sf{= \ 2 + {\sqrt{3}} + 2 - {\sqrt{3}}}$

$\sf{= \ 4}$

$\sf{Product \ of \ zeroes \ -}$

$\sf{= \ \alpha \beta}$

$\sf{= \ ( \ 2 + {\sqrt{3}} \ ) \ ( \ 2 - {\sqrt{3}} \ )}$


${\large{\boxed{\sf{\red{( \ a + b \ ) \ ( \ a - b \ ) \ = \ a^2 - b^2}}}}}$


$\sf{= \ ( 2 )^2 - ( {\sqrt{3}} )^2}$

$\sf{= \ 4 - 3}$

$\sf{= \ 1}$

$\sf{The \ required \ polynomial \ is \ -}$

$\sf\green{f \ ( \ x \ ) \ = \ x^2 - ( \ Sum \ of \ zeroes \ ) \ x \ +}$
$\sf\green{Product \ of \ zeroes}$

$\sf{f \ ( \ x \ ) \ = \ x^2 - \ ( \ 4 \ ) \ x \ + 1}$

${\boxed{\sf{\blue{f \ ( \ x \ ) \ = \ x^2 - 4x + 1}}}}$