Question

$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7\sqrt{5}}{11}b$
find a & b.

Answer 1

$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7\sqrt{5}}{11}b$

$\Rightarrow \frac{7+\sqrt{5}}{7-\sqrt{5}}\times \frac{7+\sqrt{5}}{7+\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}\times \frac{7-\sqrt{5}}{7-\sqrt{5}}=a+\frac{7\sqrt{5}}{11}b$

$\Rightarrow \frac{49+5+14\sqrt{5}}{49-5}-\frac{49+5-14\sqrt{5}}{49-5}=a+\frac{7\sqrt{5}}{11}b$

$\Rightarrow \frac{49+5+14\sqrt{5}-49-5+14\sqrt{5}}{44}=a+\frac{7\sqrt{5}}{11}b$

$\Rightarrow \frac{28\sqrt{5}}{44}=a+\frac{7\sqrt{5}}{11}b$

$\Rightarrow 0+\frac{7\sqrt{5}}{11}=a+\frac{7\sqrt{5}}{11}b$

Comparing both sides, we get

$a=0,b=1$

$\therefore a=0,b=1$