Question

√(6)−√(5)÷√(6)+√(5)=a-b=√(30) Find the values of a and b.


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Answer 1

Solution:-

$\to \rm \: \dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} + \sqrt{5} } = a \: - b \sqrt{30}$

$\to\dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} + \sqrt{5} } \times \dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5} }$

$\to \rm \: \dfrac{( \sqrt{6} - \sqrt{5} ){}^{2} }{( \sqrt{6}) {}^{2} - ( \sqrt{5} ) {}^{2} }$

$\to \rm \: \dfrac{( \sqrt{6}) {}^{2} + (\sqrt{5} ) {}^{2} - 2 \times \sqrt{6} \times \sqrt{5} }{6 - 5}$

$\to \rm \: {6 + 5 - 2 \sqrt{30} }$

$\to \rm11 - 2 \sqrt{30}$

So a = 11 and b = - 2

we use identity

( a - b )² = a²+ b² - 2ab

(a - b )(a + b ) = a² - b²

Answer 2

Solution:-

$\to \bf \frac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} + \sqrt{5} } = a - b \sqrt{30}$

$\to \bf \frac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} + \sqrt{5} } \times \frac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5} }$

$\to \bf \frac{ {( \sqrt{6} - \sqrt{5} )}^{2} }{ { \sqrt{6} }^{2} - { \sqrt{5} }^{2} }$

$\to \bf \frac{ { \sqrt{6} }^{2} - 2( \sqrt{6} \times \sqrt{5}) + { \sqrt{5} }^{2} }{6 - 5}$

$\to\bf \frac{ {6} - 2( \sqrt{30} ) + { \ {5} } }{1}$

$\bf\to6 - 2 \sqrt{30} + 5$

$\bf\to11 - 2 \sqrt{30}$

Therefore:-

• The value of (a) = 11

• The value of (b) = -2

Identities used:-

• $\sf {\boxed{ \sf{{ {(x - y)}^{2} = {x}^{2} - 2xy - {y}^{2} }}}}$

• $\sf {\boxed{ \sf{{ {x}^{2} - {y}^{2} = (x + y) \times (x - y) }}}}$