Question

Find the value of a and bif (7 + sqrt(48))/(7 - sqrt(48)) = a + b * sqrt(3) .​

Answer 1

$\bf\purple{Answer :- }$

• a= 97
• b=56

$\bf\purple{Given :- }$

$\dfrac{7 + \sqrt{48} }{7 - \sqrt{48} } = a + b \sqrt{3}$

$\bf\purple{To\: find :-}$

Value of a, b

$\bf\purple{Solution:-}$

》 Firstly lets write the sqrt 48 in simplest form

$\sqrt{48} = \sqrt{2 {}^{4} \times 3 }$

$\sqrt{48} = \sqrt{(2 {}^{2} ) {}^{2} } \times \sqrt{3}$

$\sqrt{48} = 2 {}^{2} \sqrt{3}$

$\red{ \sqrt{48} = 4 \sqrt{3} }$

$\cfrac{7 + 4 \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3}$

》 Now , lets rationalize the denominator .In order to rationalize the denominator multiply and divide with its Rationalizing factor i.e conjuagate to the denominator .

》 Rationalizing factor of

$7 - 4 \sqrt{3} \: is \: 7 + 4 \sqrt{3}$

So, multiply and divide with this .

$\cfrac{7 + 4 \sqrt{3} }{7 - 4 \sqrt{3} } \times \cfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

$= \cfrac{(7 + 4 \sqrt{3} )(7 + 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3} )} = a + b \sqrt{3}$

》 Simplifying the numerator by Algebraic Identitiy (a+b)² = a² + 2ab + b² .

》 Simplifying the denominator by Algebraic Identitiy (a+b)(a-b) = a²-b² .

$\cfrac{(7) {}^{2} + (4 \sqrt{3} ) {}^{2} + 2(7)(4 \sqrt{3} )}{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} }$

$\cfrac{49 + 48 +56 \sqrt{3} }{49 - 48} = a + b \sqrt{3}$

$\dfrac{97 +56 \sqrt{3} }{1} = a + b \sqrt{3}$

$\red{97 + 56 \sqrt{3} = a + b \sqrt{3}}$

》 By comparing L.H.S and R.H.S

$\red{a = 97} \\ \red{b = 56}$

$\bf\purple{Know\: more:-}$

》 Algebraic Identities :-

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

$\bf\purple{Answer :- }$

a= 97

b=56

$\bf\purple{Given :- }$

$\dfrac{7 + \sqrt{48} }{7 - \sqrt{48} } = a + b \sqrt{3}$

$\bf\purple{To\: find :-}$

Value of a, b

$\bf\purple{Solution:-}$

》 Firstly lets write the sqrt 48 in simplest form

$\sqrt{48} = \sqrt{2 {}^{4} \times 3 }$

$\sqrt{48} = \sqrt{(2 {}^{2} ) {}^{2} } \times \sqrt{3}$

$\sqrt{48} = 2 {}^{2} \sqrt{3}$

$\red{ \sqrt{48} = 4 \sqrt{3} }$

$\cfrac{7 + 4 \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3}$

》 Now , lets rationalize the denominator .In order to rationalize the denominator multiply and divide with its Rationalizing factor i.e conjuagate to the denominator .

》 Rationalizing factor of

$7 - 4 \sqrt{3} \: is \: 7 + 4 \sqrt{3}$

So, multiply and divide with this .

$\cfrac{7 + 4 \sqrt{3} }{7 - 4 \sqrt{3} } \times \cfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

$= \cfrac{(7 + 4 \sqrt{3} )(7 + 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3} )} = a + b \sqrt{3}$

》 Simplifying the numerator by Algebraic Identitiy (a+b)² = a² + 2ab + b² .

》 Simplifying the denominator by Algebraic Identitiy (a+b)(a-b) = a²-b² .

$\cfrac{(7) {}^{2} + (4 \sqrt{3} ) {}^{2} + 2(7)(4 \sqrt{3} )}{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} }$

$\cfrac{49 + 48 +56 \sqrt{3} }{49 - 48} = a + b \sqrt{3}$

$\dfrac{97 +56 \sqrt{3} }{1} = a + b \sqrt{3}$

$\red{97 + 56 \sqrt{3} = a + b \sqrt{3}}$

》 By comparing L.H.S and R.H.S

$\red{a = 97} \\ \red{b = 56}$

$\bf\purple{Know\: more:-}$

》 Algebraic Identities :-

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc