Question

rationalise the denominatos 2+3 oot 5 / 2-3root5

Answer 1

Given to Rationalize the denominator:-

$\dfrac{2 + 3 \sqrt{5} }{2 - 3 \sqrt{5} }$

Solution:-

For rationalizing the denominator we have to multiply and divide with its Rationalizing factor

Rationalizing factor for

$2 - 3 \sqrt{5} \: \: is \: 2 + 3 \sqrt{5}$

Just conjugate for the denominator

$\dfrac{2 + 3 \sqrt{5} }{2 - 3 \sqrt{5} } \times \dfrac{2 + 3 \sqrt{5} }{2 + 3 \sqrt{5} }$

$\dfrac{(2 + 3 \sqrt{5}) {}^{2} }{(2 - 3 \sqrt{5})(2 + 3 \sqrt{5}) }$

Numerator is in form of (a+b)² = a² + 2ab + b²

Denominator is in form of (a+b)(a-b) = a² - b²

$\dfrac{(2) {}^{2} + (3 \sqrt{5} ) {}^{2} + 2 \times 3 \sqrt{5} }{(2) {}^{2} - (3 \sqrt{5}) {}^{2} }$

$\dfrac{4 + 45 +6 \sqrt{5} }{4 - 45}$

$\dfrac{49 + 6 \sqrt{5} }{ - 41}$

$\dfrac{ - 49 - 6 \sqrt{5} }{41}$

Hence denomiantor rationalized !

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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