Question

rationalize the denominator of 5+2√3/7+4√3=a+b√3

Answer 1

Shikharrai114p1ekhp Expert

Hi ,

LHS = (5 + 2√3 ) / ( 7 + 4√3 )

rationalize the denominator

= (5 + 2√3 ) ( 7 - 4√3 ) / [ ( 7 + 4√3 ) ( 7 - 4√3 ) 

= [ 5 ×7 - 5 × 4√3 + 2√3 × 7 - 2√3 × 4√3 ] / [ (7 )² - (4√3 )² ]

here we used ( x + y ) (x - y ) = x² - y²  identity

= [35 -20√3 + 14√3 -24 ] / [ 49 - 48 ]

= (11 - 6√3 )

therefore ,

LHS = RHS

11 - 6√3 = A - B√3

comparing both sides

A = 11,

B = 6

i hope this is useful to you.

****

mark as brainliest!!

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.