Question

former quadratic polynomial whose zeros are 2 + √3, 2-√3​

Answer 1

Step-by-step explanation:

hope you would understand

Answer 2

$\huge\text{\underline{Answer}}$

Let the roots of the quadratic polynomial be $\bold{\alpha \: \: and \: \beta }$

$\boxed{\sf{Given }}$

$\bold{\alpha = 2 + \sqrt{3} }$

$\bold{\beta =2 - \sqrt{3} }$

$\huge\sf{solution:}$

The standard form of quadratic equation is :-

$\boxed{\sf{{x}^{2} - ( \alpha + \beta ) + \alpha \beta }}$

Putting the value of the alpha and betta.

$\implies$ $\bold{{x}^{2} - (2 + \sqrt{3} + 2 - \sqrt{3} )x +(2 + \sqrt{3} )(2 - \sqrt{3} )}$

$\implies$ $\bold{{x}^{2} - (4)x + {2}^{2} - { (\sqrt{3} )}^{2} }$

$\implies$ $\bold{ {x}^{2} - 4x + 1}$

hence, the required quadratic polynomial is :- $\boxed{\sf{ {x}^{2} - 4x + 1}}$