In each of the following determine rational numbers $a$ and $b$:$ \frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=a+b \sqrt{5} $

Given:

\( \frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=a+b \sqrt{5} \)

To do: 

We have to determine rational numbers $a$ and $b$.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

LHS $=\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=\frac{(4+3 \sqrt{5})(4+3 \sqrt{5})}{(4-3 \sqrt{5})(4+3 \sqrt{5})}$

$=\frac{(4+3 \sqrt{5})^{2}}{(4)^{2}-(3 \sqrt{5})^{2}}$

$=\frac{16+45+24 \sqrt{5}}{16-45}$

$=\frac{61+24 \sqrt{5}}{-29}$

$=\frac{-61}{29}+\frac{-24}{29}\sqrt5$

Therefore,

$a+b \sqrt{5}=\frac{-61}{29}+\frac{-24}{29}\sqrt{5}$

Comparing both sides, we get,

$a=\frac{-61}{29}$ and $b=\frac{-24}{29}$

Hence, $a=\frac{-61}{29}$ and $b=\frac{-24}{29}$.