Question

When denominator is rationalised, then the number $\frac{16}{\sqrt{41}-5}$ becomes

• A. $\sqrt{41}+5$
• B. $\sqrt{5}+41$
• C. $\sqrt{41}-5$
• D. $\sqrt{5}-41$

Answer 1

The correct option is A $\sqrt{41}+5$

Given, $\frac{16}{\sqrt{41}-5}$

Rationalizing the denominator, we get

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

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

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

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

$=(\sqrt{41}+5)$