Question

4+3root 5/4-3root5= a+b root 5 find the value of A and B​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ a \ is \ -\frac{61}{29} \ and \ b \ is \ -\frac{24}{29}}$

$\sf\orange{Given:}$

$\sf{\implies{\frac{4+3\sqrt5}{4-3\sqrt5}=a+b\sqrt5}}$

$\sf\pink{To \ find:}$

$\sf{The \ values \ of \ a \ and \ b.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{\implies{\frac{4+3\sqrt5}{4-3\sqrt5}=a+b\sqrt5}}$

___________________________________

$\sf{\implies{\frac{4+3\sqrt5}{4-3\sqrt5}}}$

$\sf{On \ rationalising \ the \ denominator}$

$\sf{\implies{\frac{4+3\sqrt5(4+3\sqrt5)}{4-3\sqrt5(4+3\sqrt5)}}}$

$\sf{\implies{\frac{(4+3\sqrt5)^{2}}{4^{2}-(3\sqrt5)^{2}}}}$

$\sf{\implies{\frac{16+24\sqrt5+45}{16-45}}}$

$\sf{\implies{\frac{61+24\sqrt5}{-29}}}$

$\sf{\implies{-\frac{61}{29}-\frac{24\sqrt5}{29}}}$

___________________________________

$\sf{\therefore{a+b\sqrt5=-\frac{61}{29}-\frac{24\sqrt5}{29}}}$

$\sf{On \ comparing}$

$\boxed{\sf{a=-\frac{61}{29}}}$

$\sf{b\sqrt5=-\frac{24\sqrt5}{29}}$

$\sf{\therefore{b=-\frac{24\sqrt5}{29\times\sqrt5}}}$

$\boxed{\sf{\therefore{b=-\frac{24}{29}}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ -\frac{61}{29} \ and \ b \ is \ -\frac{24}{29}}}}$