Question

rationalise the denominator.....please answer this question with full process

Answer 1

Mark this answer as brainliest answer

Answer 2

Hey friend, Harish here.

Here is your answer:

To Rationalize:

[tex] \frac{b^{2}}{ \sqrt{a^{2}+b^{2}} + a } [/tex]

Solution:

To rationalize the denominator we must multiply and divide the number by the conjugate of the denominator.

$The\ Conjugate\ is \ (\sqrt{a^{2}+ b^{2}} - a)$

Then,

$\frac{b^{2}}{ \sqrt{a^{2}+b^{2}} + a } \times \frac{\sqrt{a^{2}+b^{2}} - a}{\sqrt{a^{2}+b^{2}} - a}$

⇒ $\frac{b^{2}(\sqrt{a^{2}+b^{2}} - a)}{(\sqrt{a^{2}+b^{2}} + a)(\sqrt{a^{2}+b^{2}} - a)}$

We know that, x² - y² = (x+y)(x-y)

Then,

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

So,

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

Here b² will get canceled :

So,

⇒  $\frac{b^{2}(\sqrt{a^{2}+b^{2}} - a)}{b^{2}} = (\sqrt{a^{2}+b^{2}} - a)$

Hence the denominator is rationalized.
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Hope my answer is helpful to you.