Question

$x=\frac{1}{2+\sqrt{3}}$ Find the values of $3x^4+2x^3+10x^2$

Answer 1

$x=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2-\sqrt{3}}{4-3}3x^4+2x^3+10x^2={(2-\sqrt{3})}^4+2{(2-\sqrt{3})}^3+10{(2-\sqrt{3})}^2={\left({(2-\sqrt{3})}^2\right)}^2+2{(2-\sqrt{3})}^3+10{(2-\sqrt{3})}^2={(4+3-4\sqrt{3})}^2+2(8-3\sqrt{3}-12\sqrt{3}+18)-10(4+3-4\sqrt{3})={(7-4\sqrt{3})}^2+2(26-15\sqrt{3})-10(7-4\sqrt{3})=49+48-56\sqrt{3}+52-30\sqrt{3}-70+40\sqrt{3}=97+52-70+10\sqrt{3}-56\sqrt{3}=79-46\sqrt{3}$