Question

If
$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = x + \sqrt{16} y$
find the value of x and y​

Answer 1

$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = x + \sqrt{16} y$

Consider Left Hand Side

$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} }$

The simplest Rationalising factor of 3√2 - 2√3 is 3√2 + 2√3. So, multiply the numerator and denominator of the given fraction by Rationalising factor.

$= \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} }$

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

Since (x + y)(x - y) = x² - y²

$= \frac{ \sqrt{2}(3 \sqrt{2} + 2 \sqrt{3}) + \sqrt{3}(3 \sqrt{2} + 2 \sqrt{3}) }{9(2) - 4(3)}$

$= \frac{3(2) + 2 \sqrt{3 \times 2} + 3 \sqrt{3 \times 2} + 2(3)}{18 - 12}$

$= \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{6}$

$= \frac{12 + 5 \sqrt{6} }{6}$

$= \frac{12}{6} + \frac{5 \sqrt{6} }{6}$

$= 2 + \frac{5 \sqrt{6} }{ \sqrt{6} \times \sqrt{6} }$

$= 2 + \frac{5}{ \sqrt{6} }$

$now \: consider \\ \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = x + \sqrt{16} y$

$i.e \:2 + \frac{5}{ \sqrt{6} } = x + \sqrt{16}y$

Equating corresponding rational and irrational factors, we have

x = 2

$\sqrt{16} y = \frac{5}{ \sqrt{6} }$

$y = \frac{ \frac{5}{ \sqrt{6} } }{ \sqrt{16} }$

$y = \frac{5}{ \sqrt{6} \times \sqrt{16} }$

$y = \frac{5}{ \sqrt{6} \times 4}$

$y = \frac{5}{ 4\sqrt{6} }$

$\boxed{\tt{x = 2}}$

$\boxed{\tt{ y = \frac{5}{ 4\sqrt{6} } }}$