Math, asked by karsimar, 5 months ago

2/√5+√3+2 , rationalize the denominator.​

Answers

Answered by hemanthkumar76
4

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Given

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

To find

Rationalization of the given number

Solution

Conjugate of (5+3)+2 = (5+3)-2

For rationalize the denominator we want to multiple it's conjugate to the numerator and denominator

 \frac{2}{( \sqrt{5} +  \sqrt{3})   +  2  }  \times  \frac{( \sqrt{5} +  \sqrt{3})    -   2  }{( \sqrt{5} +  \sqrt{3})    -   2  }  =  \frac{2[( \sqrt{5} +  \sqrt{3})    -   2]}{[( \sqrt{5} +  \sqrt{3})   +  2][\sqrt{5} +  \sqrt{3})    -   2]}

  =  \frac{2[( \sqrt{5} +  \sqrt{3})    -   2]}{[ {( \sqrt{5} +  \sqrt{3})}^{2}     -    {2}^{2} ]}  =  \frac{{2[( \sqrt{5} +  \sqrt{3})    -   2]}}{ {( \sqrt{5}} )^2 + 2 \times  \sqrt{5} \times  \sqrt{3}  +  { (\sqrt{3}) }^{2}  }

  =  \frac{2[( \sqrt{5} +  \sqrt{3})    -   2]}{5 + 2 \sqrt{8}  + 3  - 4} =\frac{2[( \sqrt{5} +  \sqrt{3})    -   2]}{4 + 4 \sqrt{2} }

 = \frac{2[( \sqrt{5} +  \sqrt{3})    -   2]}{2(2 + 2 \sqrt{2}) } = \frac{( \sqrt{5} +  \sqrt{3})    -   2}{2 + 2 \sqrt{2} }

The sum is not yet completed.

Now again we want to rationalize the denominator.

Conjugate of 2+22 is 2-22

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

 =\frac{2√5+2√3-2√10-2√6+4√2}{2²-(2√2)²}

= \frac{2√5+2√3-4-2√10-2√6+4√2}{4-8}\\ = \frac{2(√5+√3-2-√10-√6+2√2)}{-4}

= \frac{√5+√3-2-√10-√6+2√2}{-2}

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