Question

find the zeroes of the polynomial √3x^2 - 8x+4√3

Answer 1

$\huge {\blue {\bold {Bonjour!}}}$

$= > \sqrt{3} {x}^{2} - 8x + 4 \sqrt{3} = 0 \\ \\ = > \sqrt{3} {x}^{2} - 6x - 2x + 4 \sqrt{3} = 0 \\ \\ = > \sqrt{3} x(x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) = 0\\ \\ = > (x - 2 \sqrt{3}) ( \sqrt{3} x - 2) = 0 \\ \\ = > x = 2 \sqrt{3} \: or \: \frac{2}{ \sqrt{3} }$

$\large {\green {\bold {Hope \: this \: helps...:)}}}$