If $$\frac{3\sqrt5-5}{3\sqrt5+5}=a+b\sqrt5$$​, then find the value of $$b$$​.

Correct option is D

Given:
If $$\frac{3\sqrt{5} - 5}{3\sqrt{5} + 5}$$​ = a + b$$\sqrt 5$$​, then find the value of b.
Formula Used:
Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.
Conjugate of (3√5 + 5) is (3√5 − 5).
Use the identity: (a − b)(a + b) = a² − b²
Solution:
​$$\frac{3\sqrt{5} - 5}{3\sqrt{5} + 5}$$​
Multiply numerator and denominator by (3√5 − 5):
​$$\frac{3\sqrt{5} - 5}{3\sqrt{5} + 5}\\= \frac{(3\sqrt{5} - 5)^2}{(3\sqrt{5} + 5)(3\sqrt{5} - 5)}\\= \frac{9 \cdot 5 - 2 \cdot 3\sqrt{5} \cdot 5 + 25}{9 \cdot 5 - 25}\\= \frac{45 - 30\sqrt{5} + 25}{45 - 25}\\= \frac{70 - 30\sqrt{5}}{20}\\= \frac{70}{20} - \frac{30\sqrt{5}}{20}\\= \frac{7}{2} - \frac{3\sqrt{5}}{2}\\$$​​
Comparing with a + b√5, we get b = -3/2