Given that \(a=\frac{1}{2-\sqrt{3}}, b=\frac{1}{2+\sqrt{3}}\). find the value of \(a^{2}+b^{2}\)
A. \(\frac{14}{37}\)
B. \(7\)
C. \(142 \sqrt{3}\)
D. 14
Correct Answer: D
Explanation
Given: \(a=\frac{1}{2-\sqrt{3}}\) and \(b=\frac{1}{2+\sqrt{3}}\)
Required \(a^{2}+b^{2}\)
\(a^{2}=\left(\frac{1}{2-\sqrt{3}}\right)^{2}, b^{2}=\left(\frac{1}{2+\sqrt{3}}\right)^{2}\)
\(a^{2}+b^{2}=\left(\frac{1}{2-\sqrt{3}}\right)^{2}+\left(\frac{1}{2+\sqrt{3}}\right)^{2}\)
we find the L.C.M
\begin{aligned}
\frac{(2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}}{(2-\sqrt{3})^{2}(2+\sqrt{3})^{2}} &=\frac{4+4 \sqrt{3}+3+4-4 \sqrt{3}+3}{[(2-\sqrt{3})(2+\sqrt{3})]^{2}} \\
&=\frac{14}{[4+2 \sqrt{3}-2 \sqrt{3}-3]^{2}} \\
&=\frac{14}{[4-3]^{2}}=\frac{14}{1^{2}}=14
\end{aligned}