Question

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

Answer 1

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

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

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

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

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