Question

sin 90 degrees + tan 45 degrees are zeros of quadratic polynomial .find a quadratic polynomial

Answer 1

Step-by-step explanation:

if the roots be=

$\alpha,\beta$

$\red{\alpha = sin90 + tan45 = 1 + \sqrt{2} }$

$\pink{then \:\beta = 1 - \sqrt{2} \:it \:vice\:versa}$

$\orange{by\:formula\:\:x^2 - (\alpha + \beta)x + (\alpha \times \beta)}$

$\green{x^2 - 2x - 1\: your\:answer}$