Question

show that the polynomial x^2+4x+9 has no real zeroes​

Answer 1

Answer:

It will have roots but imaginary.

It will not have real roots or zeroes.

Step-by-step explanation:

The given equation is x^2-4x+9. Here a=1 , b=-4, C=9.

Now it's discriminant (D)

D=b^2-4ac

putting the values of a,b,c we get,

D=(4)^2 - 4*1*9

=16 - 36

=-20

Here D is less than 0. So, the equation x^2-4x+9 will have no zeroes.

Answer 2

Step-by-step explanation:

Given polynomial is x²+4x+9

here, a= 1 , b = 4, c = 9

Descriminant = b² - 4ac

If D < 0 , then the polynomial has no real zeroes

=> D = (4)² - 4(1)(9) => 16 - 36 = -20 < 0

.•. the polynomial x²+4x+9 has no real zeroes