Question

What is the rationalizing factor for \[\sqrt{xy}\]?

Answer 1

Hint: We are given to find the rationalizing factor of \[\sqrt{xy}\]. Rationalizing factor is nothing but when two surds are multiplied with each other, gives us the product in the form of a rational number. So we have to find a term such that, when multiplied by \[\sqrt{xy}\] gives us a rational number.

Complete step-by-step solution:
Now let us have a brief regarding the rationalizing factor. The factor of multiplication by which rationalization is done is known as rationalizing factor. If a product of two surds is a rational number, then each of them is a rationalizing factor to others. This is the basic principle involved in the surds.
The general rule of a surd under rationalization is \[\sqrt{a}\times \sqrt{a}=a\].
Since our given \[\sqrt{xy}\] is in the form of the above mentioned rule, we will be proceeding with the same method.
On rationalizing \[\sqrt{xy}\] by considering \[\sqrt{xy}\] as rationalizing factor, we get-
\[\sqrt{xy}\times \sqrt{xy}=xy\]
\[\therefore \] The rationalizing factor of \[\sqrt{xy}\] is \[\sqrt{xy}\] itself since it is a surd.

Note: In order to rationalize the complex surds where the surds are in order two, conjugates are used to rationalize the surds. This comes from the formula \[{{a}^{2}}-{{b}^{2}} = \left( a+b \right)\left( a-b \right)\]. So in complex surds order 2 surds get squared off and denominators are converted to rational numbers.
For example, consider the complex surd \[\dfrac{1}{\sqrt[2]{2}-1}\].
The denominator \[\sqrt[2]{2}-1\] is to be converted into a rational number. Let us apply the formula \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
We get,
\[\begin{align}
  & \dfrac{1}{\sqrt[2]{2}-1} \\
 & =\dfrac{\left( \sqrt[2]{2}+1 \right)}{\left( \sqrt[2]{2}-1 \right)\left( \sqrt[2]{2}+1 \right)} \\
 & =\dfrac{\sqrt[2]{2}+1}{2-1} \\
 & =\sqrt[2]{2}+1 \\
\end{align}\]
Hence the rationalizing factor is obtained.