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. $a=0,b=\frac{3}{2}$
• B. $a=\frac{-3}{2},b=1$
• C. $a=0,b=\frac{-2}{3}$
• D. $a=4,b=2$

Answer 1

Answer: (C)

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

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$

$\frac{(\sqrt{7}-1{)}^2-(\sqrt{7}+1{)}^2}{(\sqrt{7}+1)(\sqrt{7}-1)}=a+b\sqrt{7}$

$\frac{7+1-2\sqrt{7}-7-1-2\sqrt{7}}{(\sqrt{7}{)}^2-1^2}=a+b\sqrt{7}$

$\frac{7+1-2\sqrt{7}-7-1-2\sqrt{7}}{(\sqrt{7}{)}^2-1^2}=a+b\sqrt{7}$

$0+\frac{-2\sqrt{7}}{3}=a+b\sqrt{7}$

So, from the above equation,

we can find, $a=0$ and $\frac{-2\sqrt{7}}{3}=b\sqrt{7}$

$b=\frac{-2}{3}$

Answer (C) $a=0,b=\frac{-2}{3}$