Question

Example 6. Show that the polynomial 9x2 + 6x + 4 has no real zeroes.
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Answer 1

Step-by-step explanation:

by quadratic formula

a=9 b=6c=4

d=b2 -4ac

= 36-4(9)(4)

=36-144

= -108

so d is less than 0

so polynomial....has no real zeroes

Answer 2

Quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$.

Inside the radical is $b^2-4ac=D$.

No real numbers give negative numbers if it's squared.

$\sqrt{D}$ is not a real number if D is smaller than 0.

But, D makes -108 and is smaller than 0.

Therefore, no real solutions exist.