Question

find the 'a' and 'b' in 5+2√3/7+4√3 = a-b√3

Answer 1

5+2√3/7+4√3 = 5+2√3 ×7-4√3 /7+4√3 ×7-4√3
                      =35-20√3+14√3-24/ (7)²-(4√3)²
                      =11-6√3/49-48 =11-6√3

              so a=11 and
 -b√3= -6√3
 b=6
                  

Answer 2

Step-by-step explanation:

$\bf{ \underline{Given-}}$

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

$\bf \underline{To \: find \: out-}$

$\rm{Value \: of \: a \: and \: b \: in \: given \: expression.}$

$\bf \underline{Solution-}$

Given expression

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

The denominator is 7 + 4√3.

We know that

Rationalising factor of a + b√c = a - b√c.

So, the rationalising factor of 7 +4√3 = 7-4√3.

On rationalising the denominator them

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

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

Now, applying algebraic identity in denominator because it is in the form of;

(a+b)(a-b) = a² - b²

Where, we have to put in our expression: a = 7 and b = 4√3 , we get

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

$\longrightarrow \: \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{49 - 48}$

Subtract 49 from 48 in denominator to get 1.

$\longrightarrow \: \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{1}$

$\longrightarrow \: (5 + 2 \sqrt{3} )(7 - 4 \sqrt{3} )$

Now, multiply both term left side to right side.

$\longrightarrow \: 35 + 14 \sqrt{3} - 20 \sqrt{3} - 8 \sqrt{3 \times 3 }$

$\longrightarrow \: 35 + 14 \sqrt{3} - 20 \sqrt{3} - 24$

$\longrightarrow \: (35 - 24) - 6 \sqrt{3}$

$\longrightarrow \: 11 - 6 \sqrt{3}$

$\bf\therefore \: 11 - 6 \sqrt{3} = a - b \sqrt{3}$

On, comparing with R.H.S , we have

a = 11 and b = -6

$\bf \underline{Hence, value \: of \: a = 11 \: and \: b = - 6.}$

Used Formulae:

(a+b)(a-b) = a² - b

Rationalising factor of a + b√c = a - b√c.