Question

Express 3/√3-√2+√5 with rational denominator

Answer 1

Solution :-

We have ,

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

Multiplying both numerator and denominator with same values , we get ;-

$= > \frac{3}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \frac{( \sqrt{3} - \sqrt{2}) + \sqrt{5} }{( \sqrt{3} - \sqrt{2) + \sqrt{5} } }$

$= > \frac{3( \sqrt{3} - \sqrt{2} - \sqrt{5} ) }{( \sqrt{3} - \sqrt{2}) ^{2} - ( \sqrt{5})^{2} }$

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

$= > \frac{3( \sqrt{3 } - \sqrt{2} -\sqrt{5)} }{(3 + 2 - 2 \sqrt{6}) - 5 }$

$= > \frac{3( \sqrt{3} - \sqrt{2} - \sqrt{5}) }{ - 2 \sqrt{6} } \times \frac{ \sqrt{6} }{ \sqrt{6} }$

$= > \frac{3( \sqrt{18} - \sqrt{12} - \sqrt{30}) }{ - 2 \times 6}$

$= > \frac{3 \sqrt{2} - 2 \sqrt{3} - \sqrt{30} }{ - 4}$

$= > \frac{2 \sqrt{3} + \sqrt{30} - 3 \sqrt{2} }{4}$

$= > \frac{3(3 \sqrt{2} - 2 \sqrt{3} - \sqrt{30}) }{ - 12}$

Hence , Answer is :-

$= > \frac{3(3 \sqrt{2} - 2 \sqrt{3} - \sqrt{30}) }{ - 12}$

Answer 2

Concept

In elementary algebra, root rationalisation is a process by which radicals in the denominator of an algebraic fraction are eliminated.

Given

3/(√3-√2+√5)

Find

Express 3/(√3-√2+√5) with rational denominator

Solution

We are given  3/(√3-√2+√5)

Rationalising

Multiplying Numerator and denominator with ((√3-√2) - √5)

We get

= 3 * ((√3-√2) - √5)   /   ((√3-√2)+√5) * ((√3-√2) - √5)

= 3((√3-√2) - √5)   /   (√3-√2)² - (√5)²

= 3((√3-√2) - √5)   / ( 3 + 2 - 2√6) - 5

= 3((√3-√2) - √5)   / (-2√6)

Again Rationalising to remove √6

Multiplying Numerator and Denominator with √6

= 3((√3-√2) - √5)  * √6   /   (-2√6) * √6

= 3(√18 - √12 - √30)  / (-12)  

= (√18 - √12 - √30) / (-4)

= (-) (√30 +  √12 - √18) / (-) 4

= (√30 +  √12 - √18) / 4

= 1/4 ( √30  + 2√3 - 3√2)

After rationalising  3/(√3-√2+√5)  =  1/4 ( √30  + 2√3 - 3√2)

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