Question

Find the zeroes of the quadratic polynomial
$\sqrt{3}{x}^2-8x+4\sqrt{3}$

Answer 1

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

$\sqrt{3}{x}^2-6x-2x+4\sqrt{3}=0$

$\sqrt{3}x(x-2\sqrt{3})-2(x-2\sqrt{3})=0$

$(\sqrt{3}x-2)(x-2\sqrt{3})=0$

$\sqrt{3}x-2=0;x-2\sqrt{3}=0$

$x=\frac{2}{\sqrt{3}};x=2\sqrt{3}$

$\therefore$Zeroes are $\frac{2}{\sqrt{3}},2\sqrt{3}$