Question

solve this equation ​

Answer 1

Given:

• $\sf a = \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} + \sqrt{6} }$
• $\sf b = \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} - \sqrt{6} }$

To find:

The value of a² + b² + ab

Solution:

Let's firstly simplify the values of a and b by rationalising the denominator.

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$\tt \implies a = \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} + \sqrt{6} }$

$\tt \implies a = \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} + \sqrt{6} } \times \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} - \sqrt{6} }$

$\tt \implies a = \dfrac{( \sqrt{7} - \sqrt{6})^{2} }{ (\sqrt{7} + \sqrt{6})(\sqrt{7} - \sqrt{6} ) }$

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Now use the following identities:

• $\boxed{\sf (A+B)(A-B) = A^2 - B^2 }$
• $\boxed{\sf (A-B)^2 = A^2 + B^2 - 2AB}$

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$\tt \implies a = \dfrac{( 7 + 6 - 2 \sqrt{42} )}{7 - 6 }$

$\underline{ \tt \implies a = 13- 2 \sqrt{42}}$

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Similarly, rationalising b.

$\tt \implies b = \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} - \sqrt{6} }$

$\tt \implies b = \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} - \sqrt{6} } \times \dfrac{ \sqrt{7} + \sqrt{6} }{ \sqrt{7} + \sqrt{6} }$

$\tt \implies b = \dfrac{ (\sqrt{7} + \sqrt{6})^2}{ (\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6}) }$

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Now use the following identities:

• $\boxed{\sf (A+B)(A-B) = A^2 - B^2 }$
• $\boxed{\sf (A+B)^2 = A^2 + B^2 + 2AB}$

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$\tt \implies b = \dfrac{ 7 + 6 + 2 \sqrt{42}}{7 - 6}$

$\underline {\tt \implies b =13 + 2 \sqrt{42}}$

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So the required value of given expression is given by:

$\tt \implies {a}^{2} + {b}^{2} + ab$

$\tt \implies {(13 - 2 \sqrt{42}) }^{2} + {(13 + 2 \sqrt{42}) }^{2} + (13 - 2 \sqrt{42}) (13 + 2 \sqrt{42})$

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Now use the following identities:

• $\boxed{\sf (A-B)^2 = A^2 + B^2 - 2AB}$
• $\boxed{\sf (A+B)(A-B) = A^2 - B^2 }$
• $\boxed{\sf (A+B)^2 = A^2 + B^2 + 2AB}$

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$\tt \implies(169 + 84 - 52 \sqrt{42} ) + (169 + 84 + 52 \sqrt{42} ) + 169- 84$

$\tt \implies169 + 84 + 169 + 84 + 85$

$\tt \implies 591$

So the required answer is 591.