Question

rationalise the denominator :-
16/√41-5

Answer 1

Given
16/(root41 - 5)
=16(root41+5)/[(root 41 -5)(root41+5)]
=[16(root41+5)]/[(root41)^2 -(5)^2]
=[16(root41+5)]/(41-25)
=16(root41+5]/16
=root41+5

Answer 2

Answer:

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

Step-by-step explanation:

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

WE have to rationalize the denominator.

Consider the given fraction  $\frac{16}{\sqrt{41}-5}$

Multiply and divide by $(\sqrt{41}+5)$, we get,

$\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$

Thus,

Denominator becomes $(a+b)(a-b)=a^2-b^2$

We have,

$(\sqrt{41}+5)(\sqrt{41}-5)=(\sqrt{41})^2-5^2=41-25=16$

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

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