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If $\frac{7+\sqrt{5}}{7-\sqrt{5}}=a+b \sqrt{5}$, find a and b.
Given :
The given expression is $\frac{7+\sqrt{5}}{7-\sqrt{5}}=a+b \sqrt{5}$
To do :
We have to find the values of a and b.
Solution :
$\frac{7+\sqrt{5}}{7-\sqrt{5}}$
To factorize the denominator, multiply and divide by $7+\sqrt{5}$
LHS
$\frac{7+\sqrt{5}}{7-\sqrt{5}} = \frac{(7+\sqrt{5}) \times (7+ \sqrt{5})}{(7-\sqrt{5}) \times (7 + \sqrt{5})}$
$=\frac{(7+\sqrt{5})^2}{7^2-\sqrt{5}^2} $
$ = \frac{7^2 + \sqrt{5}^2 + 2 \times 7\sqrt{5}}{49-5} $
$ = \frac{49 + 5 + 14\sqrt{5}}{44}$
$ = \frac{54 + 14\sqrt{5}}{44}$
$ = \frac{54}{44} + \frac{14 \sqrt{5}}{44}$
$ = \frac{27}{22} + \frac{7\sqrt{5}}{22}$
RHS
$ a + b\sqrt{5}$
On comparing,
$ a + b\sqrt{5} = \frac{27}{22} + \frac{7\sqrt{5}}{22}$
$a = \frac{27}{22}$
$b = \frac{7}{22}$.
Therefore, the values of a and b are $\frac{27}{22}, \frac{7}{22}$.