Question

If x = $\frac{1}{2-\sqrt{3}}$ find the value of $x^3-2x^2-7x+5$

Answer 1

$\text{Given:}x=\frac{1}{(2-\sqrt{3})}\text{Rationalising denominator},x=\frac{1}{(2-\sqrt{3})}\times \frac{(2+\sqrt{3})}{(2+\sqrt{3})}\Rightarrow x=\frac{(2+\sqrt{3})}{2^2-(\sqrt{3}{)}^2}=\frac{(2+\sqrt{3})}{4-3}=(2+\sqrt{3})\text{Substituting the value of}x\text{in}{x}^3-2x^2-7x+5,\text{we get,}\Rightarrow x^3-2x^2-7x+5=(2+\sqrt{3}{)}^3-2(2+\sqrt{3}{)}^2-7(2+\sqrt{3})+5=8+12\sqrt{3}+18+3\sqrt{3}-8-8\sqrt{3}-6-14-7\sqrt{3}+5[\because (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3]=15\sqrt{3}-15\sqrt{3}+18-20+5=3\text{Hence,}{x}^3-2x^2-7x+5=3$