Question

3-√5/3+2√5=a√5-19/11 b​

Answer 1

$\large\underline{\bold{Given \:Question -}}$

$\rm :\longmapsto\:\dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } =a \sqrt{5} - \dfrac{19}{11}b$

find the values of a and b.

Concept Used :-

Method of Rationalization of denominator :-

• Rationalization is the process of eliminating a radicals from the denominator. That is, to remove the radicals in a fraction so that the denominator must be a rational number.

• We multiply and divide by the term with opposite sign in between binomial of the denominator.

$\large\underline{\bf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } =a \sqrt{5} - \dfrac{19}{11}b$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \dfrac{3 - 2 \sqrt{5} }{3 - 2 \sqrt{5} } =a \sqrt{5} - \dfrac{19}{11}b$

$\rm :\longmapsto\:\dfrac{9 - 6\sqrt{5} - 3 \sqrt{5} + 10}{ {3}^{2} - {(2 \sqrt{5})}^{2} } =a \sqrt{5} - \dfrac{19}{11}b$

$\: \: \: \: \: \: \: \: \boxed{ \red{ \bf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\rm :\longmapsto\:\dfrac{19 - 9\sqrt{5}}{9 - 20} =a \sqrt{5} - \dfrac{19}{11}b$

$\rm :\longmapsto\:\dfrac{19 - 9\sqrt{5}}{ -11} =a \sqrt{5} - \dfrac{19}{11}b$

$\rm :\longmapsto\: - \dfrac{19}{11} + \dfrac{9}{11} \sqrt{5} =a \sqrt{5} - \dfrac{19}{11}b$

$\rm :\longmapsto\:\dfrac{9}{11} \sqrt{5}- \dfrac{19}{11} \times 1 =a \sqrt{5} - \dfrac{19}{11}b$

So,

On comparing, we get

$\bf\implies \:a = \dfrac{9}{11} \: \: \: and \: \: \: b = 1$

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)