Question

Rationalize
1) 1/(4 + √10)
2) 1/(sqrt(6) - sqrt(5))

Answer 1

Step-by-step explanation:

$\large{ \sf \: Solution \: 1:}$

★ The given fraction consists of irrational denominator. To make the calculations more simplified we will have to rationalize the denominator of the given fraction. To do so, we'll have to multiply both numerator and denominator by conjugate of the given denominator, i.e ,

$\sf\frac{4- \sqrt{10} }{4- \sqrt{10}}$

So,

$\sf \: \frac{1}{4 + \sqrt{10} }\times \frac{4- \sqrt{10} }{4- \sqrt{10}} \\ \\ \sf\Rightarrow \frac{4- \sqrt{10} }{4^2 - \sqrt{10^ 2}} \: \: \: \: \: \: \: \: \: \: \: \:$

By using the identity

$\sf(a + b)(a - b) = (a) ^ 2 - (b) ^ 2$

$\sf\Rightarrow \frac{4- \sqrt{10}}{ 16-10} \\ \\ \sf\Rightarrow \frac{4- \sqrt{10} }{6}$

So the required rationalized fraction is:

$\bf \red{ \implies\frac{4- \sqrt{10} }{6}}$

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$\large \sf \: Solution \: 2:$

★ Since the given fraction has irrational denominato in it. So, to make it more simplified we will have to rationalize the denominator of the given fraction. To do so, we'll have to multiply both numerator an denominator of the fraction by

$\sf\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6} + \sqrt{5}}$

So,

$\sf\frac{1}{\sqrt{6} - \sqrt{5} }\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6} + \sqrt{5}} \\ \\ \sf\Rightarrow\frac{\sqrt 6 + \sqrt 5 }{\sqrt 6^ 2 - \sqrt 5^ 2} \: \: \: \: \: \: \: \: \: \: \: \: \:$

By using the identity

$\sf(a + b)(a - b) = (a) ^ 2 - (b) ^ 2$

$\sf\Rightarrow \frac{ \sqrt 6 + \sqrt 5 }{1} \\ \\ \sf\Rightarrow \sqrt{6} + \sqrt{5}$

So, the required rationalized fraction is :

$\sf \red{ \implies\sqrt{6} + \sqrt{5}}$

Answer 2

1)

$\frac{1}{4+\sqrt{10} } = \frac{1}{4+\sqrt{10} } \times \frac{4-\sqrt{10}}{4-\sqrt{10}} \\\\= \frac{4-\sqrt{10}}{{(4)}^{2} -{(\sqrt{10})}^{2} } = \frac{4-\sqrt{10}}{16-10} = \frac{4-\sqrt{10}}{6}$

2)

$\frac{1}{\sqrt{6} -\sqrt{5} } = \frac{1}{\sqrt{6} -\sqrt{5} } \times \frac{\sqrt{6} +\sqrt{5}}{\sqrt{6} +\sqrt{5}} \\\\= \frac{\sqrt{6} +\sqrt{5}}{{(\sqrt{6} )}^{2} -{(\sqrt{5} )}^{2}} = \frac{\sqrt{6} +\sqrt{5}}{6-5} = \frac{\sqrt{6} +\sqrt{5}}{1} \\\\= \sqrt{6} +\sqrt{5}$

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