Question

1 + ✓3, 1- ✓3 give the equation of the quadratic function using zeros​

Answer 1

${x}^{2} - 2x - 2$

Explanation:

$Given, \: \: \alpha = 1 + \sqrt{3} \: \: \: and \: \: \: \beta = 1 - \sqrt{3} \\ Now, \: \: \alpha + \beta = (1 + \sqrt{3} )+( 1 - \sqrt{3} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 + \sqrt{3} + 1 - \sqrt{3} \\ = \boxed{ 2} \\ \alpha \beta = (1 + \sqrt{3} )( 1- \sqrt{3} ) \\ = (1)^{2} - ( \sqrt{3} ) ^{2} \\ = 1 - 3 \\ = \boxed{ - 2} \\ the \: \: general \: \: form \: \: of \: \: the \: \: quadratic \: \: polynomial, \\ \boxed{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta } \\ putting \: \: values , \: \: then \\ \boxed{ {x}^{2} - 2x - 2}$