Question

If $\frac{\sqrt{7}-1}{\sqrt{7}+1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$, find the values of a and b.

Answer 1

$\frac{\sqrt{7}-1}{\sqrt{7}+1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}=\frac{\sqrt{7}-1}{\sqrt{7}+1}\times \frac{\sqrt{7}-1}{\sqrt{7}-1}+\frac{\sqrt{7}+1}{\sqrt{7}-1}\times \frac{\sqrt{7}+1}{\sqrt{7}+1}$
$=\frac{(\sqrt{7}-1{)}^2}{(\sqrt{7}{)}^2-1}+\frac{(\sqrt{7}+1{)}^2}{(\sqrt{7}{)}^2-1}$
$=\frac{7+1-2\sqrt{7}}{7-1}+\frac{7+1+2\sqrt{7}}{7-1}$
$=\frac{8-2\sqrt{7}}{6}+\frac{8+2\sqrt{7}}{6}$
$=\frac{8-2\sqrt{7}+8+2\sqrt{7}}{6}$
$=\frac{16}{6}=\frac{8}{3}+0\sqrt{7}$
$\therefore \frac{8}{3}+0\sqrt{7}=a+b\sqrt{7}\Rightarrow a=\frac{8}{3},b=0$.