Question

When denominator is rationalised, then the number $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}$ becomes $a+\frac{7}{11}\sqrt{5}b$. Find $a+b$.

Answer 1

$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b$
$\frac{{(7+\sqrt{5})}^2+{(7-\sqrt{5})}^2}{49-5}=a+\frac{7}{11}\sqrt{5}b$
$\frac{49+5+2\times 7\times \sqrt{5}-49-5+2\times 7\times \sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b$
$\frac{28\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b$
Compare rational number on L.H.S to rational term on R.H.S and Compare Irrational number on L.H.S to Irrational number on R.H.S
Therefore, $a=0$, $b=$ $1$