Question

The roots are 8 - √5and 8 +√5 ( form the qe in standard form given condition)

Answer 1

Step-by-step explanation:

$x = 8 - \sqrt{5} \: \: \: \: \: \: \: \: \: \: x = 8 + \sqrt{5} \\ x - 8 + \sqrt{5} = 0 \: \: \: \: \: \: \: \: x - 8 - \sqrt{5} = 0 \\ multipling \: both \\( (x - 8) + \sqrt{5} ) \times ((x - 8) - \sqrt{5} ) \\ it \: is \: in \: (a + b)(a - b) = {a}^{2} - {b}^{2} \: form \\ {(x - 8)}^{2} - {( \sqrt{5}) }^{2} \\ {(x - 8)}^{2} - 5 \\ {(x - 8)}^{2} is \: in \: {(a - b)}^{2} form \\ {x}^{2} + 64 - 16x - 5 \\ {x}^{2} - 16x + 59$

the required quadratic equation is x² - 16x + 59

other method

x² - (α+β)x + αβ

$where \: \alpha = 8 - \sqrt{5} \: and \: \beta = 8 + \sqrt{5} \\ \alpha + \beta = 8 - \sqrt{5} + 8 + \sqrt{5} = 16 \\ \alpha \beta = (8 - \sqrt{5} )(8 + \sqrt{5} ) = 64 - 5 = 59 \\ {x}^{2} - 16x + 59$

Hope you get your answer