Question

Find the rational number A and B such that

Answer 1

Hey friend,
Here is the answer you were looking for:

$(i) \: \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} = a + b \sqrt{2}$



On rationalizing the denominator we get,

$= \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 } \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1 }$

Using the identity :

${(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} - 2( \sqrt{2})(1) }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } \\ \\ = \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} \\ \\ 3 - 2 \sqrt{2} = a + b \sqrt{2} \\ \\ a = 3 \: : \: b = - 2$
$(ii) \frac{2 - \sqrt{5} }{2 + \sqrt{5} } = a \sqrt{5} + b$

On rationalizing the denominator we get,

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

Using same identities as above :

$= \frac{ {(2)}^{2} + {( \sqrt{5} )}^{2} - 2(2)( \sqrt{5} )}{ {(2)}^{2} - {( \sqrt{5} )}^{2} } \\ \\ = \frac{4 + 5 - 4 \sqrt{5} }{4 - 5} \\ \\ = \frac{9 - 4 \sqrt{5} }{ - 1} \\ \\ = - 9 + 4 \sqrt{5} \\ \\ 4 \sqrt{5} - 9 = a \sqrt{2} + b \\ \\ a = 4 \: : \: b = - 9$
$(iii) \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

On rationalizing the denominator we get,

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

Using the denominator we get,

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} + 2( \sqrt{3} )( \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \\ a = 5 \: : \: b = 2$
$(iv) \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

On rationalizing the denominator we get,

$= \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

Using the identity :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{5(7 - 4 \sqrt{3}) + 2 \sqrt{3} (7 - 4 \sqrt{3}) }{ {(7)}^{2} - {(4 \sqrt{3} })^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48} \\ \\ 11 - 6 \sqrt{3} = a + b \sqrt{6} \\ a = 11 \: : \: b = - 6$


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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