Question

show the polynomial p(x) =
${x}^{2} - 4x + 9$
have no zeroes ​

Answer 1

Before using quadratic formula, you can show that this polynomial has no zeroes

b2 -4ac

=16-36= - 20<0

So, this polynomial has no real zeroes.

Answer 2

To Show: This Polynomial have no zeroes.

Solution:

If,

$p(x) = 0 \\ {x}^{2} - 4x + 9 = 0$

Using The Quadratic Formula,

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{4 ± \sqrt{16 - 4(1)(9)} }{2(1)} \\ x = \frac{4± \sqrt{16 - 36} }{2} \\ x = \frac{4 ± \sqrt{ - 20} }{2}$

As, the Discriminant is a Negative Number, This Polynomial have no real roots.