Question

show that polynomial X^2+ 9 has no real zeros

Answer 1

x^2 + 9

= x^2 + 0 x + 9

a = 1

b = 0

c = 9

Discriminent = b^2 - 4ac

= 0^2 - 4(9)(1)

= 0 - 36

= - 36

Discriminent is negative, so the given polynomial don't have any real zeroes .

Answer 2

the GIVEN polynomial has two distinct points