Question

Find the values of a and b if
$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b$

Answer 1

We have, $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b$
$\Rightarrow \frac{(7+\sqrt{5}{)}^2-(7-\sqrt{5}{)}^2}{(7-\sqrt{5})(7+\sqrt{5})}=a+\frac{7}{11}\sqrt{5}b\Rightarrow \frac{[7^2+(\sqrt{5}{)}^2+2\times 7\times \sqrt{5}]-(7^2+(\sqrt{5}{)}^2-2\times 7\times \sqrt{5}]}{7^2-(\sqrt{5}{)}^2}=a+\frac{7}{11}\sqrt{5}b$
$\left\{\begin{array}{cc}Usingidentity,& (a+b{)}^2=a^2+2ab+b^2\\ & (a-b{)}^2=a^2-2ab+b^2\\ and& (a-b)(a+b)=a^2-b^2\end{array}\right\}$
$\Rightarrow \frac{49+5+14\sqrt{5}-49-5+14\sqrt{5}}{49-5}=a+\frac{7}{11}\sqrt{5}b\Rightarrow \frac{28\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b\Rightarrow 0+\frac{7}{11}\sqrt{5}=a+\frac{7}{11}\sqrt{5}b$
On comparing both sides, we get
a = 0 and b = 1