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

The correct option is D $(\frac{-61}{29},\frac{-24}{29})$

(D) Given, $a+b\sqrt{5}=\frac{4+3\sqrt{5}}{4-3\sqrt{5}}$

$=\frac{4+3\sqrt{5}}{4-3\sqrt{5}}\times \frac{4+3\sqrt{5}}{4+3\sqrt{5}}$ {Rationalizing}
$=\frac{(4+3\sqrt{5}{)}^2}{16-45}$

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

Hence,

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