Question

simplify by rationalizing the denominator: 4 over the square root of 10 minus square room of 6 ​

Answer 1

Answer:

$\large\boxed{\dfrac{4}{\sqrt{10}-\sqrt6}=\sqrt{10}+\sqrt6}$

Step-by-step explanation:

$\dfrac{4}{\sqrt{10}-\sqrt6}\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\ \text{and}\ (\sqrt{a})^2=a\\\\=\dfrac{4}{\sqrt{10}-\sqrt6}\cdot\dfrac{\sqrt{10}+\sqrt6}{\sqrt{10}+\sqrt6}=\dfrac{4(\sqrt{10}+\sqrt6)}{(\sqrt{10})^2-(\sqrt6)^2}=\dfrac{4(\sqrt{10}+\sqrt6)}{10-6}\\\\=\dfrac{4(\sqrt{10}+\sqrt6)}{4}=\sqrt{10}+\sqrt6$

Answer 2

The rationalization of the given expression will be equal to

$\dfrac{4}{\sqrt{10}-\sqrt{6}}=\sqrt{10}+\sqrt{6}$

What is rationalization?

Rationalization is a procedure used in elementary mathematics to remove the irrational number from the denominator.

The rationalization of the given fraction will be :-

$=\dfrac{4}{\sqrt{10}-\sqrt{6}}\times \dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{10}+\sqrt{6}}$

$\dfrac{4\sqrt{10}+\sqrt{6}}{\sqrt{10}^2-\sqrt{6}^2}=\dfrac{4\sqrt{10}+\sqrt{6}}{10-6}$

$\dfrac{4\sqrt{10}+\sqrt{6}}{\sqrt{10}^2-\sqrt{6}^2} =\dfrac{4\sqrt{10}+\sqrt{6}}{4}=\sqrt{10}+\sqrt{6}$

Hence rationalization of the given expression will be equal to

$\dfrac{4}{\sqrt{10}-\sqrt{6}}=\sqrt{10}+\sqrt{6}$

To know more about rationalization follow

https://brainly.com/question/14261303

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