Question

find the values of a and b.

(7+ root5/ 7-root5) -(7-root5/7+root5) = a+7/11root5b

Answer 1

$$\begin{gathered} \frac{\left(7+\sqrt{5}\right)}{\left(7-\sqrt{5}\right)}-\frac{\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{{\left(7+\sqrt{5}\right)}^2-{\left(7-\sqrt{5}\right)}^2}{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{\left[\left(7+\sqrt{5}\right)+\left(7-\sqrt{5}\right)\right]\left[\left(7+\sqrt{5}\right)-\left(7-\sqrt{5}\right)\right]}{7^2-{\left(\sqrt{5}\right)}^2}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{14\times 2\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{28}{44}\sqrt{5}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{7}{11}\sqrt{5}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow 0+\frac{7}{11}\sqrt{5}=a+\frac{7}{11}\sqrt{5}b \\ \mathrm{comparing}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{we}\mathrm{get}; \\ a=0\mathrm{and}b=1 \end{gathered}$$