Question

Rationalize the denominators of the following:
1. $\dfrac{1}{\sqrt{7}}$
2. \[\dfrac{1}{\sqrt{7}-\sqrt{6}}\]
3. \[\dfrac{1}{\sqrt{5}+\sqrt{2}}\]
4. \[\dfrac{1}{\sqrt{7}-2}\]

Answer 1

Hint: First of all we will have to know the significance of rationalizing the denominator which simply means to simplify the calculation. Now, doing rationalization of fraction having irrational part in denominator is as shown below:
\[\dfrac{a~+\sqrt{b}}{a-\sqrt{b}}\times \dfrac{a+\sqrt{b}}{a+\sqrt{b}}=\dfrac{{{a}^{2}}+b+2a\sqrt{b}}{{{a}^{2}}-b}\]
Here the conjugate of denominator, i.e algebraic sign changed between rational and irrational part is multiplied by its numerator as well as denominator.

Complete step-by-step answer:
Now, we will move further to rationalize the give question one by one as shown below:

1. $\dfrac{1}{\sqrt{7}}$
The conjugate of the denominator is \[\sqrt{7}\] itself as there is no rational part added or subtracted. On multiplying the numerator and denominator by \[\sqrt{7}\], we get as follows:
 \[\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{\sqrt{7}}{{{\left( \sqrt{7} \right)}^{2}}}=\dfrac{\sqrt{7}}{7}\]

Now, we have option 2.
2. \[\dfrac{1}{\sqrt{7}-\sqrt{6}}\]
The conjugate of the denominator is \[\left( \sqrt{7}+\sqrt{6} \right)\]. Now we will multiply the conjugate to numerator as well as denominator, we get as follows:
\[\dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\]
Since we know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
$\Rightarrow \dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}=\dfrac{\sqrt{7}+\sqrt{6}}{7-6}=\dfrac{\sqrt{7}+\sqrt{6}}{1}=\sqrt{7}+\sqrt{6}$

Similarly we have option 3.
3. $\dfrac{1}{\sqrt{5}+\sqrt{2}}$
Here, the conjugate of the denominator will be $\left( \sqrt{5}-\sqrt{2} \right)$. Now we will multiply the conjugate with numerator as well as denominator of the number and we get as follows:
$\dfrac{1}{\sqrt{5}+\sqrt{2}}=\dfrac{1}{\sqrt{5}+\sqrt{2}}\times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
Since we know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
\[\Rightarrow \dfrac{1}{\sqrt{5}+\sqrt{2}}\times \dfrac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}=\dfrac{\sqrt{5}-\sqrt{2}}{5-2}=\dfrac{\sqrt{5}-\sqrt{2}}{3}\]

Again we have option 4.
4. $\dfrac{1}{\sqrt{7}-2}$
Here, the conjugate of the denominator will be $\left( \sqrt{7}+2 \right)$. Now we will multiply the numerator and denominator by this conjugate to the number and we get as follows:
$\dfrac{1}{\sqrt{7}-2}=\dfrac{1}{\sqrt{7}-2}\times \dfrac{\sqrt{7}+2}{\sqrt{7}+2}$
Since we know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
$\Rightarrow \dfrac{1}{\sqrt{7}-2}=\dfrac{\sqrt{7}+2}{{{\left( \sqrt{7} \right)}^{2}}+{{\left( 2 \right)}^{2}}}=\dfrac{\sqrt{7}+2}{7-4}=\dfrac{\sqrt{7}+2}{3}$
Therefore, we have rationalised each of the given numbers.

Note: Be careful while finding the conjugate of denominator of a given fraction also take care of the calculation part because there is a chance that due to sign or any other reason you may get the incorrect answer of the given question. Remember the way of rationalizing a fraction having irrational denominators which have already shown in above.