Question

please explain
if (7+√5)/(7-√5) - (7-√5)/(7+√5) = a+(7/11)√5b find value of a and b

Answer 1

heya user !!

refer to given attachment please

Answer 2

Answer:

$a=\frac{27}{22}\:\:and\:\:b=\frac{1}{2}$

Step-by-step explanation:

Given: $\frac{7+\sqrt{5}}{7-\sqrt{5}}=a+\frac{5}{11}\sqrt{5}b$

To find: value of a and b

First we rationalize the denominator of LHS and them compare the answer with RHS to find value of a & b

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

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

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

      $=\frac{7^2+(\sqrt{5})^2+14\sqrt{5}}{7^2-(\sqrt{5})^2}$

      $=\frac{49+5+14\sqrt{5}}{49-5}$

      $=\frac{54+14\sqrt{5}}{44}$

      $=\frac{54}{44}+\frac{14\sqrt{5}}{44}$

      $=\frac{27}{22}+\frac{7\sqrt{5}}{22}$

$RHS=a+\frac{7}{11}\sqrt{5}b$

$\implies a=\frac{27}{22}$

$\implies \frac{7}{11}\sqrt{5}b=\frac{7\sqrt{5}}{22}$

$b=\frac{1}{2}$

Therefore, $a=\frac{27}{22}\:\:and\:\:b=\frac{1}{2}$