Question

if
$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ find \: the \: value \: of \: \: a \: and \: b$

Answer 1

Given,



$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$



To find ,


the value a and b



Main solution :

Let's simplify the L.H.S by rationalising its denominator


$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$


We will use identity (a+b)(a-b)=a^2-b^2 in the denominator.


$\frac{5(7 - 4 \sqrt{3}) + 2 \sqrt{3}(7 - 4 \sqrt{3} ) }{ {7}^{2} - {4 \sqrt{3} }^{2} }$


$\frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48}$


$\frac{11 - 6 \sqrt{3} }{1}$


$11 - 6 \sqrt{3}$


$11 + ( - 6 \sqrt{3} )$


Now we will compare it with RHS,Upon comparing we get,


$\textbf{a = 11 , and}$


$\bold{b = -6}$

Answer 2

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

Given :

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

To find the value of a and b in the given expression.

On simplifying it's L.H.S by rationalising it's denominator.

We get ;

$\frac{ 5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \frac{ 5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{ 7 - 4\sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ \frac{5(7 - 4 \sqrt{3} ) + 2 \sqrt{3} (7 - 4 \sqrt{3} )}{(7) {}^{2} - (4 \sqrt{3}) {}^{2} } \\ \\ \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} -24 }{49 - 48} \\ \\ \frac{11 - 6 \sqrt{3} }{1}$

$11 + (-6) \sqrt{3}$

On comparing the result of L.H.S with R.H.S , we get;

Therefore,

$\therefore{ \bold{ \: a = 11 \: \: and \: \: b = - 6}}$

Thanks for the question!