Question

Rationalizing the denominator  3√5+4√3/ 3√5−4√3​

Answer 1

In this question , we have to rationalise the denominator of

$\\ \sf \: \dfrac{3 \sqrt{5 } + 4 \sqrt{3} }{3 \sqrt{5} - 4 \sqrt{3} }$

SOLUTION :-

We have ,

$\\ \sf \: \dfrac{3 \sqrt{5} + 4 \sqrt{3} }{3 \sqrt{5} - 4 \sqrt{3} }$

Rationalising factor is [3√5 + 4√3].

Multiplying numerator and denominator by (3√5 + 4√3) ,

$\\ \implies \sf \: \dfrac{3 \sqrt{5} + 4 \sqrt{3} }{3 \sqrt{5} - 4 \sqrt{3} } \times \dfrac{3 \sqrt{5} + 4 \sqrt{3} }{3 \sqrt{5} + 4 \sqrt{3} } \\ \\ \\ \implies \sf \: \dfrac{(3 \sqrt{5} + 4 \sqrt{3})^{2} }{(3 \sqrt{5} - 4 \sqrt{3} )(3 \sqrt{5} + 4 \sqrt{3}) }$

In numerator ,

★ (a+b)² = a² + b² + 2ab

• a = 3√5
• b = 4√3

In denominator ,

★ (a-b)(a+b) = a² - b²

• a = 3√5
• b = 4√3

$\\ \implies \sf \: \dfrac{( {3 \sqrt{5}) }^{2} + ( {4 \sqrt{3}) }^{2} + 2(3 \sqrt{5} )(4 \sqrt{3}) }{( {3 \sqrt{5}) }^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ \\ \implies \sf \: \dfrac{45 + 48 + 24 \sqrt{15} }{45 - 48} \\ \\ \\ \implies \sf \: \dfrac{93 + 24 \sqrt{15} }{ - 3} \\ \\ \\ \implies \sf \: \cancel \dfrac{ - 93}{3} + \dfrac{ \cancel{24 }\sqrt{15} }{ \cancel3} \\ \\ \\ \sf \implies \: \underline{- 31 + 8 \sqrt{15} }$

$\\ \therefore \underbrace{\boxed{\mathfrak{ \dfrac{3 \sqrt{5} + 4 \sqrt{3} }{3 \sqrt{5} - 4 \sqrt{3} } = - 31 + 8\sqrt{15} }}}$

MORE IDENTITIES :-

★ (a-b)² = a² + b² - 2ab

★ (a+x)(a+y) = a² + (x+y)a + xy

★ (a+b)³ = a³ + 3a²b + 3ab² + b³

★ (a-b)³ = a³ - 3a²b + 3ab² - b³