Question

Rationalise the denominator of
6 upon √5 + √2​

Answer 1

Rationalise the denominator :-

$\longrightarrow \sf{ \dfrac{6}{ \sqrt{5} + \sqrt{2} } }$

$\Large {\underline { \sf \orange{Clarification :}}}$

Here, we have to rationalise the denominator of the given fraction. In order to rationalise the denominator, we multiply the denominator's rationalising factor with the numerator and the denominator of the fraction so that it's denominator become rational.

Here, let √5 as 'a' and √2 as 'b'. We know that, (√a + √b) × (√a - √b) = a - b, therefore rationalising factor of (√a + √b) is (√a - √b), so rationalising factor of (√5 + √2) is (√5 - √2). We'll multiply (√5 - √2) with both numerator and denominator.

$\Large {\underline { \sf \orange{Explication \: of \: steps :}}}$

$\longrightarrow \sf{ \dfrac{6}{ \sqrt{5} + \sqrt{2} } }$

$\longrightarrow \sf{ \dfrac{6}{ \sqrt{5} + \sqrt{2} } \times \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} }$

$\longrightarrow \sf{ \dfrac{6(\sqrt{5} - \sqrt{2})}{ (\sqrt{5} + \sqrt{2} )(\sqrt{5} - \sqrt{2})} }$

• (√a + √b)(√a - √b) = a - b

$\longrightarrow \sf{ \dfrac{6(\sqrt{5} - \sqrt{2})}{5 -2 } }$

$\longrightarrow \sf{ \dfrac{ \not6(\sqrt{5} - \sqrt{2})}{ \not3 } }$

$\longrightarrow$ ❝ $\boxed{ \sf \orange { \sf{ 2(\sqrt{5} - \sqrt{2}) } }}$ ❞

So, 2(√5 - √2) is the answer.

$\Large {\underline { \sf \orange{A \: Little \: Further. . . . !}}}$

More identities :

• (√a)² = a

• √a√b = √ab

• √a/√b = √a/b

• (√a + √b)(√a - √b) = a - b

• (a + √b)(a - √b) = a² - b

• (√a ± √b)² = a ± 2√ab + b

• (√a + √b)(√c + √d) = √ac + √ad + √bc + √bd