Question

if a and b are rational numbers, find a and b √2+√3/3√2+2√3=a+b√6​

Answer 1

Given -

$a \: and \: b \: are \: rational \: numbers \: and \: \\ \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a + b \sqrt{6}$

To find -

The values of a and b

Solution -

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

By rationalising the denominator,

$\frac{( \sqrt{2} + \sqrt{3})(3 \sqrt{2} - 2 \sqrt{3} ) }{(3 \sqrt{2} + 2 \sqrt{3})(3 \sqrt{2} - 2 \sqrt{3} ) } = a + b \sqrt{6}$

$\frac{ \sqrt{2} (3 \sqrt{2} - 2 \sqrt{3}) + \sqrt{3} (3 \sqrt{2} - 2 \sqrt{3} ) }{ {(3 \sqrt{2} ) }^{2} - {(2 \sqrt{3}) }^{2} } = a + b \sqrt{6}$

$\frac{6 - 2 \sqrt{6} + 3 \sqrt{6} - 6}{18 - 12} = a + b \sqrt{6}$

$\frac{ \sqrt{6} }{6} = a + b \sqrt{6}$

$\sqrt{6} \times \frac{1}{6} = a + b \sqrt{6}$

We know that,

Rational number is always equal to other rational numbers and irrational numbers is always equal to other irrational numbers

$a = \frac{1}{6} and \: \sqrt{6} = b \sqrt{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b =1$

Therefore the value of a is 1/6 and b is 1