Question

Solve :
$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

Answer 1

$x^2(\sqrt{3}+1)x+\sqrt{3}=0$

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

$x(x-\sqrt{3})-1(x-\sqrt{3})=0$

$(x-\sqrt{3})(x-1)=0$

$\Rightarrow x=1$ or $x=\sqrt{3}$

$\therefore$ roots are $(1,\sqrt{3})$