Question

rationalize root3+root2/5+root2​

Answer 1

Solution:-

$\rm : \implies \: \dfrac{ \sqrt{3} + \sqrt{2} }{5 + \sqrt{2} }$

For rationalization multiply and divide by 5 - √2

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

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

Using this identities

=> ( a - b )(a + b ) = a² - b²

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

$\rm \to \: \dfrac{5 \sqrt{3} - \sqrt{6} + 5 \sqrt{2} - 2 }{25 - 2}$

$\rm \to \: \dfrac{5 \sqrt{3} + 5 \sqrt{2} - \sqrt{6} - 2 }{23}$

Answer

$\rm \to \: \dfrac{5 \sqrt{3} + 5 \sqrt{2} - \sqrt{6} - 2 }{23}$

More about Rationalization

We know that irrational numbers are those which can’t be expressed in the ‘p/q’ form where ‘p’ and ‘q’ are integers.

But these rational numbers can be used in rational fractions as either numerator or denominator.

When these numbers are present in numerators of fractions, calculations can be done. But when these exist in denominators of fractions, they make calculations more difficult and complicated.

To avoid such complications in the numeric calculations, we use method of rationalization.

Hence, rationalization can be defined as the process by which we eliminate radicals present in the denominators of fractions

Answer 2

Step-by-step explanation:

Required answer:-

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

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

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

${:}\dashrightarrow$${\dfrac {5 {\sqrt {3}}-{\sqrt {6}}+5 {\sqrt {2}}-2}{25-2}}$

${:}\dashrightarrow$${\dfrac {5 {\sqrt {3}}+5 {\sqrt {2}}-{\sqrt{6}}-2}{23}}$

Identity used:-

${\boxed{(a-b)(a+b)={a}^{2}-{b}^{2}}}$