Question

Find the zeroes of the quadratic polynomial 16x^2-3

Answer 1

3/4^2 is the p of x

Answer 2

Gonna use 2 methods to do it.


1.

We have p(x) as
$16 {x}^{2} - 3 = 0$
$16 {x}^{2} = 3 \\ {x}^{2} = \frac{3}{16} \\ x = \sqrt{ \frac{3}{16} } = + \frac{ \sqrt{3} }{4} or - \frac{ \sqrt{3} }{4}$

Therefore the roots of this polynomial will be
$+ \frac{ \sqrt{3} }{4} \: and \: - \frac{ \sqrt{3} }{4}$





2.

The given quadratic polynomial could also be written as
${(4x)}^{2} - { (\sqrt{3} )}^{2} = 0 \\ = (4x - \sqrt{3} )(4x + \sqrt{3} ) = 0$




From here,

$4x - \sqrt{3} = 0 \\ 4x = \sqrt{3} \\ x = \frac{ \sqrt{3} }{4}$

and

$4x + \sqrt{3} = 0 \\ 4x = - \sqrt{3} \\ x = - \frac{ \sqrt{3} }{4}$