Question

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

Answer 1

$given \\ \alpha = \sqrt{2} + 3 \\ \\ \beta = \sqrt{2} - 3 \\ \\ p(x) = k[ { {x}^{2} + ( \alpha + \beta )x - \alpha \times \beta } ]\\ \\ \: \: \: \: = k [{ {x}^{2} + ( \sqrt{2} + 3 - \sqrt{2} + 3)x - ( \sqrt{2} + 3) \times ( \sqrt{2} - 3)} ] \\ \\ \: \: \: \: </p><p> = k [{ {x}^{2} - 6x - ({ \sqrt{2} })^{2} - {3}^{2}} ] \\ \\ k [{ {x}^{2} - 6x - 2 - 3} ] \: \: \: </p><p> = k [{ {x}^{2} - 6x - 5} ]\\ \\ (let \: k = 1) \\ \\ </p><p>p(x) = {x}^{2} - 6x - 5$may be uh like my process....

Answer 2

Answer:

answer=

x²-6x-5

done!!!!!!