Question

$\frac{5-4\sqrt{3} }{7-4\sqrt{3} } = a+b\sqrt{3}$
What are the values of a and b?

Answer 1

Topic :-

Rationalisation

Given :-

$\dfrac{5-4\sqrt{3} }{7-4\sqrt{3} } = a+b\sqrt{3}$

To find :-

Value of a and b

Solution :-

In order to find the value of a and b, we will rationalise the denominator of the given fraction by multiplying both numerator and denominator with the conjugate of the denominator.

Conjugate is basically the change of sign of the term whose coefficient is y.

In the given expression, denominator is 7 - 4 √3, it's conjugate is 7 + 4 √3.

So multiply both numerator and denominator with 7 + 4 √3 inorder to rationalise it.

$\implies\dfrac{5-4\sqrt{3} }{7-4\sqrt{3} } = a+b\sqrt{3}$

${\implies\dfrac{5-4\sqrt{3} }{7-4\sqrt{3} } \times \dfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } = a+b\sqrt{3}}$

${\implies\dfrac{(5-4\sqrt{3}) (7 + 4\sqrt{3})}{(7-4\sqrt{3})( 7 + 4\sqrt{3})} = a+b\sqrt{3}}$

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Now apply algebraic identity :-

• ( A + B ) ( A - B ) = A² - B²

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In the denominator :

• A = 7
• B = 4√3

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${\implies\dfrac{5(7 + 4\sqrt{3}) - 4 \sqrt{3}(7 + 4 \sqrt{3}) }{(7)^{2} -(4\sqrt{3})^{2}} = a+b\sqrt{3}}$

${\implies\dfrac{35 + 20\sqrt{3} - 28 \sqrt{3} - 48 }{49 - 48} = a+b\sqrt{3}}$

${\implies\dfrac{-13-8\sqrt3 }{1} = a+b\sqrt{3}}$

${\implies -13-8\sqrt3 = a+b\sqrt{3}}$

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By comparing LHS and RHS, we get :

• a = $-13$
• b = $-8$

Hence these are the required values of a and b.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{5-4\sqrt{3} }{7-4\sqrt{3} } = a+b\sqrt{3}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{5-4\sqrt{3} }{7-4\sqrt{3} } \red{ \times \dfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } }= a+b\sqrt{3}$

We know,

$\red{ \boxed{ \sf{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{5(7 + 4 \sqrt{3}) - 4 \sqrt{3} (7 + 4 \sqrt{3} )}{ {7}^{2} - {(4 \sqrt{3} )}^{2}} = a + b \sqrt{3}$

$\rm :\longmapsto\:\dfrac{35 + 20 \sqrt{3} - 28 \sqrt{3} - 48 }{ 49 - 48} = a + b \sqrt{3}$

$\rm :\longmapsto\:\dfrac{ - 13 - 8 \sqrt{3}}{1} = a + b \sqrt{3}$

$\rm :\longmapsto\: - 13 - 8 \sqrt{3} = a + b \sqrt{3}$

So, on comparing, we get

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \:a = - 13}}} \: \: \: \: \: \: \: \green{and} \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \:b = - 8 \: }}}$

Additional Information :-

$\red{ \boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\red{ \boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\red{ \boxed{ \sf{ \: {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\red{ \boxed{ \sf{ \: {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\red{ \boxed{ \sf{ \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$

$\red{ \boxed{ \sf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}}$

$\red{ \boxed{ \sf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}}$