Question

If $x=\frac{1}{(2-\sqrt{3}),}$ then find the value of $(x^2-4x+1)$.

Answer 1

$x=\frac{1}{(2-\sqrt{3})}\times \frac{(2+\sqrt{3})}{(2+\sqrt{3})}=\frac{(2+\sqrt{3})}{4-3}=(2+\sqrt{3})\Rightarrow (x-2)=\sqrt{3}\therefore (x-2{)}^2=3\Rightarrow x^2-4x+4=3\Rightarrow x^2-4x+1=0[1mark]$