Question

Rationalize the following numbers
(a) \[\dfrac{1}{\sqrt{2}}\]
(b) \[\dfrac{1}{\sqrt{5}-\sqrt{6}}\]
(c) \[\dfrac{1}{\sqrt{7}-\sqrt{2}}\]

Answer 1

Hint: To rationalize numbers with denominators having only one term, we have to multiply the numerator and denominator with the denominator and simplify. To rationalize numbers with denominators having more than one term, we have to multiply the numerator and denominator with the conjugate of the denominator and simplify.

Complete step by step answer:
We have to rationalize each of the given numbers. Let us consider one by one.
(a) To rationalize $\dfrac{1}{\sqrt{2}}$ , we have to multiply the numerator and denominator with the denominator.
$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{1\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$
We know that $\sqrt{a}\times \sqrt{a}={{\left( \sqrt{a} \right)}^{2}}=a$ . Therefore, we can write the above result as
$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$
(b) Let us consider \[\dfrac{1}{\sqrt{5}-\sqrt{6}}\] . We have to multiply the numerator and denominator with the conjugate of the denominator.
\[\Rightarrow \dfrac{1}{\sqrt{5}-\sqrt{6}}=\dfrac{\sqrt{5}+\sqrt{6}}{\left( \sqrt{5}-\sqrt{6} \right)\left( \sqrt{5}+\sqrt{6} \right)}\]
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Thus, we can simplify the above result as
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{5}+\sqrt{6}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}} \\
 & =\dfrac{\sqrt{5}+\sqrt{6}}{5-6} \\
 & =\dfrac{\sqrt{5}+\sqrt{6}}{-1} \\
 & =-\sqrt{5}-\sqrt{6} \\
\end{align}\]
Hence, the rationalized form of \[\dfrac{1}{\sqrt{5}-\sqrt{6}}\] is \[-\sqrt{5}-\sqrt{6}\] .
(c) Now, let us rationalize \[\dfrac{1}{\sqrt{7}-\sqrt{2}}\] . We have to multiply the numerator and denominator with the conjugate of the denominator.
\[\Rightarrow \dfrac{1}{\sqrt{7}-\sqrt{2}}=\dfrac{\sqrt{7}+\sqrt{2}}{\left( \sqrt{7}-\sqrt{2} \right)\left( \sqrt{7}+\sqrt{2} \right)}\]
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Thus, we can simplify the above result as
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{7}+\sqrt{2}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & =\dfrac{\sqrt{7}+\sqrt{2}}{7-2} \\
 & =\dfrac{\sqrt{7}+\sqrt{2}}{5} \\
\end{align}\]
Hence, the rationalized form of \[\dfrac{1}{\sqrt{7}-\sqrt{2}}\] is \[\dfrac{\sqrt{7}+\sqrt{2}}{5}\] .

Note: Rationalization method of moving the radical (i.e., square root or cube root) from the bottom (denominator) of the fraction to the top (numerator). Conjugate of a denominator is obtained by changing the sign that separates the numbers in the denominator. For example, we can write the conjugate of $\left( a+b \right)$ and $\left( a-b \right)$ as $\left( a-b \right)$ and $\left( a+b \right)$ respectively. We can also write the conjugate of denominators having 3 terms also. Let us consider $a+b+c$ . The conjugate of this number can be found by grouping the numbers as follows.
$\Rightarrow a+b+c=\left( a+b \right)+c$
Conjugate can be written as $\left( a+b \right)-c$ .