Question

if a and b r rational no.s and 7-4 √3/7+ 4 √3 = a+c√ 3, find the values of a and b

Answer 1

Given :-

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

Required to find :-

• Values of " a " and " b " ?

Solution :-

Given information :-

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

we need to find the values of " a " and " b " .

Consider the LHS part ;

$\rm \dfrac{7 - 4 \sqrt{3} }{7 + 4 \sqrt{3} }$

we need to rationalize the denominator .

Rationalising factor of 7 + 4√3 = 7 - 4√3

Multiply the both numerator and denominator with the Rationalising factor

$\tt \dfrac{7 - 4 \sqrt{3} }{7 + 4 \sqrt{3} } \times \dfrac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

Here we need to use some algebraic identities .

They are ;

• 1. ( x - y ) ( x - y ) = ( x - y )²

• 2. ( x + y ) ( x - y ) = x² - y²

• 3. ( x - y )² = x² + y² - 2xy

Using 1st and 2nd identities we get ;

$\dfrac{(7 - 4 \sqrt{3} {)}^{2} }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2} }$

Using the 3rd identity expand the numerator ;

$\dfrac{(7 {)}^{2} + (4 \sqrt{3} {)}^{2} - 2(7)(4 \sqrt{3}) }{49 - 16 \times 3}$

$\dfrac{49 + 16 \times 3 - \: 56\sqrt{3} }{49 - 48}$

$\dfrac{49 + 48 - 56 \sqrt{3} }{1}$

$\dfrac{97 - 56 \sqrt{3} }{1}$

$\implies 97 - 56 \sqrt{3}$

Equal the LHS side and RHS side

$\rm 97 - 56 \sqrt{3} = a + b \sqrt{3}$

$\implies \rm 97 + ( - 56 \sqrt{3} ) = a + b \sqrt{3}$

From the above we can conclude that ;

The LHS is in the form of RHS

Hence,

➾ Value of a is 97

➾ Value of b is - 56

Answer 2

$\frac{7 - 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ = \frac{7 - 4 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ ( {7 - 4 \sqrt{3} )}^{2} \div {7}^{2} - {4 \sqrt{3} }^{2} \\ \frac{49 + 48 - 56 \sqrt{3} }{49 - 48} \\ 97 - 56 \sqrt{3}$

so, a=97 ,b=-56