Question

Show that thus polynomial x²-4x+9 have no zeros​

Answer 1

Answer:

when we find the roots of the equation we consider it as zeros

if roots are real finding roots we get

(4+√20i)/2 , and (4-√20i) /2

which are complex hence the eqn. has no roots

Answer 2

Answer:

We can do this by using the determinant(D).

If D > 0 , there are distinct, real roots

If D = 0 , there are equal real roots

If D < 0 , there are no real roots

D is given by the formula ( b² - 4ac )

In the given expression x² - 4x + 9,

a = 1,   b = -4   , c = 9

( b² - 4ac ) = [(-4)² - 4(1)(9) ]

                 = 16 -36 = -20

Since D < 0, it has no real roots.