Question

If $\frac{4+3\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}$, then $(a,b)=$.........

• A. $(\frac{61}{29},-\frac{-24}{29})$
• B. $(\frac{-61}{29},-\frac{24}{29})$
• C. $(\frac{61}{29},-\frac{24}{29})$
• D. $(\frac{-61}{29},-\frac{-24}{29})$

Answer 1

Answer: (B)

$(\frac{-61}{29},-\frac{24}{29})$

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

Rationalizing the L.H.S

$\frac{4+3\sqrt{5}}{4-3\sqrt{5}}\times \frac{4+3\sqrt{5}}{4+3\sqrt{5}}$

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

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

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

$=-\frac{61}{29}+\left(\frac{-24\sqrt{5}}{29}\right)=a+b\sqrt{5}$

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

Hence, $(a,b)=(\frac{-61}{29},\frac{-24}{29})$