Question

$(3 + \sqrt{7}) \div( 3 - \sqrt{7}) = a + b \sqrt{7}$
find the value of A and B​

Answer 1

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Answer 2

Given : -

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

To find : -

Value of a, b

Solution :-

We shall do rationalizing the denominator  for rationalizing the denominator we need to multiply and divide with its rationalizing factor

Rationalizing factor for

$3-\sqrt{7} \: is \: 3+\sqrt{7}$ So multiply and divide with this

$\dfrac{3+\sqrt{7} }{3-\sqrt{7} }\; \times \dfrac{3+\sqrt{7} }{3+\sqrt{7} }$

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

Numerator is in form of (a+b)² and denominator is in form of (a+b)(a-b)

$(a+b)^2 = a^2 + b^2 +2ab \\(a+b)(a-b) = a^2 - b^2$

Applying these two formulae

$\dfrac{9+7+2\times 3\sqrt{7} }{(3)^2-(\sqrt{7})^2 }$

$\dfrac{16+6\sqrt{7} }{9-7}$

$\dfrac{16+6\sqrt{7} }{2}$

$\dfrac{16}{2} +\dfrac{6\sqrt{7} }{2} = a+b\sqrt{7}$

$8+3\sqrt{7} = a+b\sqrt{7}$

by comparision a= 8 b=3

so the value of a, b are 8,3

Know more some algebraic identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc