Question

I will mark brainliest plss answer ;if a and b are rational numbers and √7+√3/2√7-3√3=a-b√21 then find the values of a and b

Answer 1

Formula used :

(a + b)(a - b) = a² - b²

(√a)² = a

Step by step explanation :

L.H.S.

$\frac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - 3 \sqrt{3} }$

Rationalizing the denominator we get:

$= \frac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - 3 \sqrt{3} } \times \frac{ \sqrt{7} + 3 \sqrt{3} }{ \sqrt{7} + 3 \sqrt{3} } \\ \\ = \frac{ \sqrt{7}( \sqrt{7} + 3 \sqrt{3} ) + \sqrt{3}( \sqrt{7} + 3 \sqrt{3} )}{( \sqrt{7} - 3 \sqrt{3} )( \sqrt{7} + 3 \sqrt{3} ) } \\ \\ = \frac{7 + 3 \sqrt{21} + \sqrt{21} + 3 \sqrt{ {3}^{2} } }{ {( \sqrt{7}) }^{2} - {(3 \sqrt{3} )}^{2} } \\ \\ = \frac{7 + 3 \times 3 + 4 \sqrt{21} }{7 - 27} \\ \\ = \frac{7 + 9 + 4 \sqrt{21} }{ - 20} \\ \\ = \frac{ - 16 - 4 \sqrt{21} }{20} \\ \\ = \frac{ - 4 - \sqrt{21} }{5}$

Equating L.H.S and R.H.S we get :

$- \frac{4}{5} - \frac{ \sqrt{21} }{5} = a - b \sqrt{21} \\ \\ ( - \frac{4}{5} ) - \frac{1}{5} . \sqrt{21} = a - b \sqrt{21} \\ \\ \bf a = - \frac{4}{5} \: \: and \: \: b = \frac{1}{5}$

Answer 2

Formula used :

(a + b)(a - b) = a² - b²

(√a)² = a

Step by step explanation :

L.H.S.

$\frac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - 3 \sqrt{3} }$

Rationalizing the denominator we get:

$= \frac{ \sqrt{7} + \sqrt{3} }{ \sqrt{7} - 3 \sqrt{3} } \times \frac{ \sqrt{7} + 3 \sqrt{3} }{ \sqrt{7} + 3 \sqrt{3} } \\ \\ = \frac{ \sqrt{7}( \sqrt{7} + 3 \sqrt{3} ) + \sqrt{3}( \sqrt{7} + 3 \sqrt{3} )}{( \sqrt{7} - 3 \sqrt{3} )( \sqrt{7} + 3 \sqrt{3} ) } \\ \\ = \frac{7 + 3 \sqrt{21} + \sqrt{21} + 3 \sqrt{ {3}^{2} } }{ {( \sqrt{7}) }^{2} - {(3 \sqrt{3} )}^{2} } \\ \\ = \frac{7 + 3 \times 3 + 4 \sqrt{21} }{7 - 27} \\ \\ = \frac{7 + 9 + 4 \sqrt{21} }{ - 20} \\ \\ = \frac{ - 16 - 4 \sqrt{21} }{20} \\ \\ = \frac{ - 4 - \sqrt{21} }{5}$

Equating L.H.S and R.H.S we get :

$- \frac{4}{5} - \frac{ \sqrt{21} }{5} = a - b \sqrt{21} \\ \\ ( - \frac{4}{5} ) - \frac{1}{5} . \sqrt{21} = a - b \sqrt{21} \\ \\ \bf a = - \frac{4}{5} \: \: and \: \: b = \frac{1}{5}$