Question

Solve for a & b


$\sf \dfrac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - \sqrt{3} } = a - b \sqrt{21}$
​

Answer 1

By rationalizing the denominator:-

$\sf :\implies \purple{ \dfrac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - \sqrt{3} } \times \dfrac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} + \sqrt{3} } = a - b \sqrt{21} }$

$\sf \:\:\: :\implies \dfrac{ (\sqrt{7} + \sqrt{3})( \sqrt{7} + \sqrt{3}) }{ (\sqrt{7} - \sqrt{3}) ( \sqrt{7} + \sqrt{3}) } = a - b \sqrt{21}$

$\sf \:\:\:\:\: :\implies \dfrac{ (\sqrt{7} + \sqrt{3}) ^{2} }{ (\sqrt{7})^{2} - (\sqrt{3})^{2} } = a - b \sqrt{21}$

$\sf \:\:\:\:\:\: :\implies \dfrac{ (\sqrt{7}) ^{2} + (\sqrt{3})^{2} + 2( \sqrt{7})( \sqrt{3} ) }{ 7 - 3} = a - b \sqrt{21}$

$\sf \:\:\:\:\:\: :\implies \dfrac{ 7 + 3 + 2 \sqrt{21} }{7 - 3} = a - b \sqrt{21}$

$\sf \:\:\:\:\:\:\: :\implies \dfrac{ 10 + 2 \sqrt{21} }{4} = a - b \sqrt{21}$

$\sf \: \:\:\:\:\:\: : \implies \dfrac{ 2(5+ \sqrt{21}) }{ 4} = a - b \sqrt{21}$

$\sf \:\:\:\:\: \:\:\:\:\:\: :\implies \dfrac{ 5+ \sqrt{21} }{2} = a - b \sqrt{21}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\: :\implies \purple {\dfrac{ 5}{2}+ \dfrac{\sqrt{21} }{2} = a - b \sqrt{21}}$

Comparing LHS and RHS:-

$\sf : \implies \red {a = \dfrac{5}{2} \: \: \: and \: \: \: b = - \dfrac{1}{2} }$

Answer 2

Answer:

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