In this chapter we will learn to simplify fraction whose denominator contain numbers with roots like square root or cube root.
By rationalization we mean to simplify the number by removing square root from the denominator.
Here we will rationalize two types of numbers;
(a) Denominator is monomial
(b) Denominator is binomial ( contain 2 elements)
Let’s understand each of the methods step by step;
Monomial denominator with square root
In this case, simply multiply denominator number on both numerator & denominator.
Let’s understand this with example;
Example 01
Rationalize \mathtt{\frac{5}{\sqrt{13}}}
Solution
Here denominator is monomial.
Multiply \mathtt{\sqrt{13}} on both numerator and denominator.
\mathtt{\Longrightarrow \ \frac{5}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{5\sqrt{13}}{13}}
Hence, \mathtt{\frac{5\sqrt{13}}{13}} is the simplified form of given expression.
Example 02
Rationalize the expression \mathtt{\frac{1}{\sqrt{7}}}
Solution
Multiply numerator & denominator by \mathtt{\sqrt{7}}
\mathtt{\Longrightarrow \ \frac{1}{\sqrt{7}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1\times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{7}}{7}}
Hence, \mathtt{\frac{\sqrt{7}}{7}} is the rationalized form of given expression.
Binomial denominator with square root
In this case we have to take denominator’s conjugate and then multiply it with numerator & denominator.
In binomials, we can get conjugate by changing the middle sign of the expression.
(a) \mathtt{x+\sqrt{y}} conjugate is \mathtt{x-\sqrt{y}} ;
(b) \mathtt{x-\sqrt{y}} conjugate is \mathtt{x+\sqrt{y}}
Just change the middle sign.
I hope you understood the concept. Let us solve examples for better understanding.
Example 01
Rationalize the expression \mathtt{\ \frac{5}{6+\sqrt{3}}}
Solution
Conjugate of \mathtt{6+\sqrt{3}} is \mathtt{6-\sqrt{3}} .
Multiplying numerator & denominator by conjugate.
\mathtt{\Longrightarrow \ \frac{5}{6+\sqrt{3}} \times \frac{\mathtt{6-\sqrt{3}}}{\mathtt{6-\sqrt{3}}}}
Referring the formula;
\mathtt{( a+b)( a-b) =a^{2} -b^{2}}
Now simplifying the expression.
\mathtt{\Longrightarrow \ \frac{30-5\sqrt{3}}{6^{2} -\left(\sqrt{3}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{30-5\sqrt{3}}{36-3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{30-5\sqrt{3}}{33}}
The above expression is the solution.
Example 02
Rationalize the below expression.
\mathtt{\frac{\sqrt{3}}{-\sqrt{7} -\sqrt{2}}}
Solution
Conjugate of \mathtt{-\sqrt{7} -\sqrt{2}} is \mathtt{-\sqrt{7} +\sqrt{2}}
Multiplying numerator & denominator by conjugate.
\mathtt{\Longrightarrow \frac{\sqrt{3}}{-\sqrt{7} -\sqrt{2}} \times \frac{-\sqrt{7} +\sqrt{2}}{-\sqrt{7} +\sqrt{2}}}
For denominator multiplication, we will use the formula;
\mathtt{( a+b)( a-b) =a^{2} -b^{2}}
Simplifying the expression we get;
\mathtt{\Longrightarrow \ \frac{-\sqrt{3}\sqrt{7} +\sqrt{3}\sqrt{2}}{\left( -\sqrt{7}\right)^{2} -\left(\sqrt{2}\right)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-\sqrt{21} +\sqrt{6}}{7-2}}\\\ \\ \mathtt{\Longrightarrow \ \frac{\sqrt{6} -\sqrt{21}}{5}}
Hence, the above expression is the solution.
Next chapter : Problems on rationalizing the denominator