Question

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

• A. $a=0$, $b=\frac{-2}{3}$
• B. $a=\frac{-2}{3}$, $b=0$
• C. $a=\frac{2}{3}$, $b=0$
• D. $a=0$, $b=\frac{2}{3}$,
  

Answer 1

Answer: (A)

$a=0$, $b=\frac{-2}{3}$

Consider $\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$ and solve as follows:

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}\Rightarrow \frac{(\sqrt{7}-1)(\sqrt{7}-1)-(\sqrt{7}+1)(\sqrt{7}+1)}{(\sqrt{7}+1)(\sqrt{7}-1)}=a+b\sqrt{7}\Rightarrow \frac{[{\left(\sqrt{7}\right)}^2+{\left(1\right)}^2-2\sqrt{7}]-[{\left(\sqrt{7}\right)}^2+{\left(1\right)}^2+2\sqrt{7}]}{{\left(\sqrt{7}\right)}^2-{\left(1\right)}^2}=a+b\sqrt{7}\Rightarrow \frac{[7+1-2\sqrt{7}]-[7+1+2\sqrt{7}]}{7-1}=a+b\sqrt{7}\Rightarrow \frac{[8-2\sqrt{7}]-[8+2\sqrt{7}]}{7-1}=a+b\sqrt{7}\Rightarrow \frac{-4\sqrt{7}}{6}=a+b\sqrt{7}\Rightarrow 0+\frac{-2}{3}\sqrt{7}=a+b\sqrt{7}\Rightarrow a=0,b=-\frac{2}{3}$

Hence, $a=0,b=-\frac{2}{3}$.