Question

If $x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\mathrm{and}y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$, show that $x^2-y^2=-\frac{10\sqrt{3}}{11}$.

Answer 1

Disclaimer: The question is incorrect.

$$\begin{gathered} x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \\ \Rightarrow x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\times \frac{5-\sqrt{3}}{5-\sqrt{3}} \\ \Rightarrow x=\frac{{\left(5-\sqrt{3}\right)}^2}{5^2-{\left(\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{25+3-10\sqrt{3}}{25-3} \\ \Rightarrow x=\frac{28-10\sqrt{3}}{22} \\ \Rightarrow x=\frac{14-5\sqrt{3}}{11} \end{gathered}$$

$$\begin{gathered} y=\frac{5+\sqrt{3}}{5-\sqrt{3}} \\ \Rightarrow y=\frac{5+\sqrt{3}}{5-\sqrt{3}}\times \frac{5+\sqrt{3}}{5+\sqrt{3}} \\ \Rightarrow y=\frac{{\left(5+\sqrt{3}\right)}^2}{5^2-{\left(\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow y=\frac{25+3+10\sqrt{3}}{25-3} \\ \Rightarrow y=\frac{28+10\sqrt{3}}{22} \\ \Rightarrow y=\frac{14+5\sqrt{3}}{11} \end{gathered}$$

$$\begin{gathered} \therefore x^2-y^2 \\ ={\left(\frac{14-5\sqrt{3}}{11}\right)}^2-{\left(\frac{14+5\sqrt{3}}{11}\right)}^2 \\ =\frac{196+75-140\sqrt{3}}{121}-\frac{196+75+140\sqrt{3}}{121} \end{gathered}$$
$$\begin{gathered} =\frac{271-140\sqrt{3}}{121}-\frac{271+140\sqrt{3}}{121} \\ =\frac{271-140\sqrt{3}-271-140\sqrt{3}}{121} \\ =\frac{-280\sqrt{3}}{121} \end{gathered}$$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10\sqrt{3}}{11}$.
$$\begin{gathered} \therefore x-y \\ =\frac{14-5\sqrt{3}}{11}-\frac{14+5\sqrt{3}}{11} \\ =\frac{14-5\sqrt{3}-14-5\sqrt{3}}{11} \\ =\frac{-10\sqrt{3}}{11} \end{gathered}$$