Question

3/ under root 3 + 1 + 5/ under 3 -1 = a+b under root 3​

Answer 1

$\tt \: If \: \dfrac{3}{ \sqrt{3} + 1} + \dfrac{5}{ \sqrt{3} - 1 } = a + b \sqrt{3}$

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

Concept Used :-

Rationalization is the process of eliminating a radical or imaginary number from the denominator or numerator of an algebraic fraction. That is, remove the radicals in a fraction so that the denominator or numerator only contains a rational number.

• The denominator of the above fraction has a binomial radical i.e., is the sum of two terms, one of which is an irrational number.

• Multiply the numerator and denominator of the fraction with the conjugate of the radical.

Let's solve the problem!!

Consider,

$\rm :\longmapsto\:\dfrac{3}{ \sqrt{3} + 1 }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{3}{ \sqrt{3} + 1 } \times \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1}$

$\rm :\longmapsto\:\dfrac{3 \sqrt{3} - 3 }{ {( \sqrt{3}) }^{2} - {(1)}^{2} }$

$\rm :\longmapsto\:\dfrac{3 \sqrt{3} - 3 }{3 - 1}$

$\rm :\longmapsto\:\dfrac{3 \sqrt{3} - 3 }{2}$

$\rm :\implies\: \boxed{ \bf \: \dfrac{3}{ \sqrt{3} + 1} = \dfrac{3 \sqrt{3} - 3 }{2} }$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{5}{ \sqrt{3} - 1 }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{5}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

$\rm :\longmapsto\:\dfrac{5 \sqrt{3} + 5}{ {( \sqrt{3} )}^{2} - {(1)}^{2} }$

$\rm :\longmapsto\:\dfrac{5 \sqrt{3} + 5}{3 - 1}$

$\rm :\longmapsto\:\dfrac{5 \sqrt{3} + 5 }{2}$

Now,

$\large \underline{\tt\:{ According \: to \: statement }}$

$\rm :\longmapsto\: \tt \:\: \dfrac{3}{ \sqrt{3} + 1} + \dfrac{5}{ \sqrt{3} - 1 } = a + b \sqrt{3}$

On substituting the values evaluated above, we get

$\rm :\longmapsto\:\dfrac{3 \sqrt{3} - 3}{2} + \dfrac{5 \sqrt{3} + 5}{2} = a + b \sqrt{3}$

$\rm :\longmapsto\:\dfrac{3 \sqrt{3} - 3 + 5 \sqrt{3} + 5}{2} = a + b \sqrt{ 3}$

$\rm :\longmapsto\:\dfrac{8\sqrt{3} + 2}{2} = a + b \sqrt{ 3}$

$\rm :\implies\:4 \sqrt{3} + 1 = a + b \sqrt{3}$

So,

On comparing we get,

$\rm :\implies\: \boxed{ \bf \: a \: = \: 1 \: \: and \: \: b \: = \: 4}$