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

1.	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» 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