Question

find the quadratic polynomial whose zero's are 4± root5
pls answer​

Answer 1

Answer:

${x}^{2} - 8x + 11$

Step-by-step explanation:

Given zeros 4

$let \: \alpha = 4 + \sqrt{5 \: } and \: \beta = 4 - \sqrt{5} \\ \alpha \beta = (4 + \sqrt{5} )(4 - \sqrt{5} ) \\ \alpha \beta = {4}^{2} - { \sqrt{5} }^{2} \\ \alpha \beta = 16 - 5 \\ \alpha \beta = 11 \\ and \: \alpha + \beta = 4 + \sqrt{5 } + 4 - \sqrt{5 } \\ \alpha + \beta = 8 \\ now \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ {x}^{2} - 8x + 11$