Question

Rationalize the denominator 2√3-√5/2√2+3√3

Answer 1

Rationalising denominator

2√3—√5/2√2+3√3

=(2√3—√5)/(2√2+3√3) x (2√2— 3√3)/(2√2—3√3)

=(2√3—√5)*(2√2—3√3)/(2√2)^2—(3√3)^2

(by using (a—b)(a+b)=a^2—b^2)

=4√6—18—2√10+3√15/8—27

=4√6—18—2√10+3√15/19

Answer 2

Step-by-step explanation:

Given that:

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

What to do:-

To rationalised the denominator.

Solution:-

We have,

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

The denominator is 2√2+3√3. Multiplying the numerator and denominator by 2√2-3√3,

We get,

$⟹ \frac{2 \sqrt{3} - \sqrt{5} }{2 \sqrt{2} + 3 \sqrt{3} } \times \frac{2 \sqrt{2} - 3 \sqrt{3} }{2 \sqrt{2} - 3 \sqrt{3} }$

$⟹ \frac{(2 \sqrt{3} - \sqrt{5} )(2 \sqrt{2} - 3 \sqrt{3} )}{(2 \sqrt{2 } + 3 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3} ) }$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

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

$⟹ \frac{4 \sqrt{6} - 6 \sqrt{ {3}^{2} } - 2 \sqrt{10} + 3 \sqrt{15} }{8 - 27}$

$⟹ \frac{4 \sqrt{6} - 18 - 2 \sqrt{10} + 3 \sqrt{15} }{ - 19}$

$⟹ \frac{ - (18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} )}{ - 19}$

$⟹ \frac{ \cancel{- }(18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} )}{ \cancel{- }19}$

$⟹ \frac{ 18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} }{ 19} \: \: \tt \red{Ans}.$

Hence the denominator is rationalised.

Know more 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

I hope it's help with...☺