Question

Find the quadratic polynomial whose zeroes are
$7 + \sqrt{3} and 7 - \sqrt{3}$

Answer 1

$\alpha = 7 + \sqrt{3}$

$\beta = 7 - \sqrt{3}$

$sum \: of \: the \: roots$

$\alpha + \beta = 7 + \sqrt{3} + 7 - \sqrt{3}$
+√3 -√3 gets cancelled

$\alpha + \beta = 7 + 7 = = > 14$

$product \: of \: the \: roots \:$

$\alpha \beta = (7 + \sqrt{3})(7 - \sqrt{3})$

It is of form,
$(a + b)(a - b) \: {a}^{2} - {b}^{2}$
=
${(7)}^{2} - {( \sqrt{3)}}^{2}$
49 - 3 = 46

${x}^{2} \: - x(sum) + product = 0$
${x}^{2} - 14x + 46 = 0$