Question

√2+√3/3√2+2√3=a-b√6
Find the value of a and b

Answer 1

Answer :-

a = 1

b = - 1/6

Explanation :-

Given :-

$\tt \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a - b \sqrt{6}$

To find :-

Values of a and b

Solution :-

$\tt \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a - b \sqrt{6}$

Consider LHS

$\tt \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3}}$

Rationalise the denominator.

The rationalising factor of 3√2 + 2√3 is 3√2 - 2√3. So multiply both numerator and denominator with rationalising factor.

$\tt = \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3}} \times \dfrac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} }$

$\tt = \dfrac{ \sqrt{2} + \sqrt{3}(3 \sqrt{2} - 2 \sqrt{3})}{(3 \sqrt{2} + 2 \sqrt{3})(3 \sqrt{2} + 2 \sqrt{3}) }$

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

[Since (x + y)(x - y) = x² - y² ]

$\tt = \dfrac{3(4) - 2 \sqrt{6} + 3 \sqrt{6} - 2(3)}{9(2) - 4(3)}$

$\tt = \dfrac{12 - 2 \sqrt{6} + 3 \sqrt{6} - 6}{18 - 12}$

$\tt = \dfrac{6 - 2 \sqrt{6} + 3 \sqrt{6}}{6}$

$\tt = \dfrac{6 + \sqrt{6}}{6}$

$\tt = \dfrac{6}{6} + \dfrac{ \sqrt{6} }{6}$

$\tt = 1+ \dfrac{ \sqrt{6} }{6}$

Now consider

$\bf \dfrac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a - b \sqrt{6}$

$\tt \implies 1 + \dfrac{ \sqrt{6} }{6} = a - b \sqrt{6}$

Comparing on both sides

$\tt a = 1$

$\tt - b \sqrt{6} = \dfrac{ \sqrt{6} }{6}$

Cancelling √6 on both sides

$\tt - b = \dfrac{1}{6}$

$\tt b = - \dfrac{1}{6}$