Question

Determine $a$ and $b$, if $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=a+7\sqrt{5}b$.

• A. $a=0,b=\frac{2}{\sqrt{7}}$
• B. $a=1,b=\frac{1}{\sqrt{7}}$
• C. $a=\frac{2}{\sqrt{7}},b=0$
• D. $a=\frac{1}{\sqrt{7}},b=1$
  

Answer 1

The correct option is A $a=0,b=\frac{2}{\sqrt{7}}$

firstly we solve the L.H.S of the expression,

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$

take the LCM,

$=>\frac{(\sqrt{7}+\sqrt{5}{)}^2-(\sqrt{7}-\sqrt{5}{)}^2}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}$

$=>\frac{\left(\right(\sqrt{7}{)}^2+(\sqrt{5}{)}^2+2(\sqrt{5}\left)\right(\sqrt{7}\left)\right)-\left(\right(\sqrt{7}{)}^2+(\sqrt{5}{)}^2-2(\sqrt{7}\left)\right(\sqrt{5}\left)\right)}{(\sqrt{7}{)}^2-({\sqrt{5}}^2)}$

$=>\frac{7+5+2\sqrt{5}\sqrt{7}-7-5+2\sqrt{5}\sqrt{7}}{7-5}$. (cancel $7and5$)

$=>\frac{4\sqrt{7}\sqrt{5}}{2}$

$=>2\sqrt{7}\sqrt{5}$ ---------$\left(1\right)$

If we taken R.H.S

$a+7\sqrt{5}b$

from the above option only option A will satisfy the condition to make L.H.S=R.H.S

put $a=0,b=\frac{2}{\sqrt{7}}$ in R.H.S

R.H.S $=2\sqrt{7}\sqrt{5}$ ------------$\left(2\right)$

from equation 1 and 2 ,

L.H.S $=$ R.H.S

answer will be A