Question

Rationalisation of denominator
$\frac{16}{ \sqrt{41} - 5}$
​

Answer 1

Required Answer:-

The above fraction given so far is:

$\Large{\frac{16}{ \sqrt{41} - 5}}$

Here,

• The denominator is √41 - 5 which is to be converted into a rational number. (This is what known as rationalization).
• Hence, we need to make changes. We can multiply it with √41 + 5, because it will become an identity i.e. (a + b)(a - b) = a² - b².
• Also we have to square √41 to get a rational number. Okay, So let's do it.

Rationalizing,

$= \Large{\frac{16}{ \sqrt{41} - 5 }}$

Multiplying √41 + 5 in both numerator and denominator:

$= \Large{\frac{16( \sqrt{41} + 5) }{( \sqrt{41} - 5)( \sqrt{41} + 5) } }$

By using the identity:

$= \Large{\frac{16( \sqrt{41} + 5)}{( \sqrt{41}) {}^{2} - (5) {}^{2} }}$

$= \Large{\frac{16( \sqrt{41} + 5)}{41 - 25} }$

$= \Large{\frac{16( \sqrt{41} + 5)}{16} }$

$= \Large{\sqrt{41} + 5}$

Hence:-

• Rationalizing the denominator of the above fraction, w get √41 + 5 as the final result.

Answer 2

$\\ \sf{:}\longrightarrow \dfrac{16}{\sqrt{41}-5}$

$\\ \sf{:}\longrightarrow \dfrac{16(\sqrt{41}+5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$

$\\ \sf{:}\longrightarrow \dfrac{16(\sqrt{41}+5)}{(\sqrt{41})^2-(5)^2}$

$\\ \sf{:}\longrightarrow \dfrac{16(\sqrt{41}+5)}{41-25}$

$\\ \sf{:}\longrightarrow \dfrac{\bcancel{16}(\sqrt{41}+5)}{\bcancel{16}}$

$\\ \boxed{\bf{:}\longrightarrow \sqrt{41}+5}$