Question

find the quadratic polynomial with zeroes 3+√2/3-√2 and 3-√2/3+√2​

Answer 1

Answer:

$\boxed{\dfrac{k}{7} (7x^2 - 22x + 7)}$

Step-by-step explanation:

$\text{Given zeroes = \alpha , \beta }\\\\\rule{300}{1}\\\\\alpha = \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}\\\\\\= \dfrac{(3 + \sqrt{2})^2}{(3)^2 - (\sqrt{2})^2}\\\\\\= \dfrac{(3)^2 + (\sqrt{2})^2 + 2(3)(\sqrt{2})}{9 - 2}\\\\\\= \dfrac{9 + 2 + 6\sqrt{2}}{7}\\\\\\= \dfrac{11 + 6\sqrt{2}}{7}\\\\\rule{300}{1}\\\\\\\beta = \dfrac{3 - \sqrt{2}}{3+ \sqrt{2}}\\\\\\= \dfrac{(3-\sqrt{2})^2}{(3)^2 - (\sqrt{2})^2}\\\\\\= \dfrac{9 + 2 - 6\sqrt{2}}{9 - 2}\\\\\\= \dfrac{11 - 6\sqrt{2}}{7}$

$\rule{300}{1}\\\\\text{Now,}\\\\\rule{300}{1}\\\\\alpha + \beta = \dfrac{11 + 6\sqrt{2}}{7} + \dfrac{11 - 6\sqrt{2}}{7} = \dfrac{22}{7}\\\\\\\alpha \beta = \dfrac{(11)^2 - (6\sqrt{2})^2}{7^2} = \dfrac{121 - 72}{49} = \dfrac{49}{49} = 1\\\\\rule{300}{1}\\\\\therefore \: \text{The required polynomial is:}\\\\\\k \left( x^2 - \left( \dfrac{22}{7} \right)x + 1 \right)\\\\\\= \boxed{\dfrac{k}{7} (7x^2 - 22x + 7)}$