Question

It's my homework...and the one who solves this correctly will be my brainliest man...or women...
Find the zeroes of polynomial...√3x^2-8x+4√3

Answer 1

↑ Here is your answer ↓
_____________________________________________________________
_____________________________________________________________

$\sqrt{3}x^2 - 8x + 4 \sqrt{3} = 0$

$b^2 - 4ac =$

$(-8)^2 - (4)(\sqrt 3)(4 \sqrt 3)=$

$64 - 48=$

$16$

Now using quadratic formula,

$\frac{-b + \sqrt{b^2 - 4ac} }{2a}$

We get,

$x = 4 \sqrt{3} / 5$

Answer 2

I think this is the most easy way to solve it.