5We know that real numbers are a combination of rational numbers and irrational numbers. In a number system, irrational numbers are the numbers that can’t be expressed in the form of a fraction and these are non-terminating numbers. Rationalization is a process of eliminating irrational numbers like square roots in the denominator of a function. Check the detailed process of rationalizing the denominator from the following sections.
Have a look at the practice problems on rationalizing the denominator and get the method of it. The denominator of the fraction may have a single number or a function. So, know the process of rationalizing the denominator process and solve the questions easily.
Also, Check:
Solved Problems on Rationalizing the Denominator
Problem 1:
Rationalize the fraction \(\frac { 1 }{ √7 } \).
Solution:
Given fraction is \(\frac { 1 }{ √7 } \)
Since the given fraction has an irrational denominator, so we need to rationalize this and make it more simple. So, to rationalize this, we will multiply the numerator and denominator of the given fraction by root 7, i.e., √7. So,
\(\frac { 1 }{ √7 } \) = \(\frac { 1 }{ √7 } \) x \(\frac { √7 }{ √7 } \)
= \(\frac { √7 }{ 7 } \)
So, the required rationalized form is \(\frac { √7 }{ 7 } \).
Problem 2:
Rationalize the following denominators.
(i) \(\frac { √75 }{ √18 } \)
(ii) \(\frac { 2√3 }{ √28 } \)
Solution:
(i) Given fraction is \(\frac { √75 }{ √18 } \) = \(\frac { 5√3 }{ 3√2 } \)
Since the given fraction has an irrational denominator, so we need to rationalize this and make it more simple. So, to rationalize this, we will multiply the numerator and denominator of the given fraction by root 2, i.e., √2. So,
\(\frac { 5√3 }{ 3√2 } \) = \(\frac { 5√3 }{ 3√2 } \) x \(\frac { √2 }{ √2 } \)
= \(\frac { 5√6 }{ 2 } \)
So, the required rationalized form is \(\frac { 5√6 }{ 6 } \).
(ii) Given fraction is \(\frac { 2√3 }{ √28 } \)
Since the given fraction has an irrational denominator, so we need to rationalize this and make it more simple. So, to rationalize this, we will multiply the numerator and denominator of the given fraction by root 28, i.e., √28. So,
\(\frac { 2√3 }{ √28 } \) = \(\frac { 2√3 }{ √28 } \) x \(\frac { √28 }{ √28 } \)
= \(\frac { 2√21 }{ 28 } \) = \(\frac { √21 }{ 14 } \)
So, the required rationalized form is \(\frac { √21 }{ 14 } \).
Problem 3:
Rationalize the denominator and simplify \(\frac { 2 }{ √11 + √2 } \).
Solution:
Given fraction is \(\frac { 2 }{ √11 + √2 } \)
Since, the given problem has an irrational term in the denominator with addition format. So we need to rationalize using the method of multiplication by the conjugate. So,
\(\frac { 2 }{ √11 + √2 } \) = \(\frac { 2 }{ √11 + √2 } \)] x \(\frac { √11 – √2 }{ √11 – √2 } \)
[(a + b)(a – b) = a² – b²]
= \(\frac { 2(√11 – √2) }{ √11² – √2² } \)
= \(\frac { 2√11 – √2 }{ 11 – 2 } \) = \(\frac { 2√11 – √2 }{ 9 } \)
So, the rationalized number is \(\frac { 2√11 – √2 }{ 9 } \).
Problem 4:
Rationalize \(\frac { 1 }{ √3 + 2 } \)
Solution:
Given fraction is \(\frac { 1 }{ √3 + 2 } \)
Since the given fraction contains an irrational denominator, so we need to convert it into a rational denominator so that calculations may become easier and simplified. To do so we will multiply both numerator and denominator by the conjugate of the denominator. So,
\(\frac { 1 }{ √3 + 2 } \) = \(\frac { 1 }{ √3 + 2 } \) x \(\frac { √3 – 2 }{ √3 – 2 } \)
= \(\frac { √3 – 2 }{ 3 – 2 } \) = √3 – 2
= √3 – 2
So, the rationalized fraction is √3 – 2.
Problem 5:
Rationalize \(\frac { 2√6 – √5 }{ 3√5 – 5√6 } \)
Solution:
Given fraction is \(\frac { 2√6 – √5 }{ 3√5 – 5√6 } \)
To rationalize the denominator, multiply both numerator and denominator by (3√5 + 5√6)
\(\frac { 2√6 – √5 }{ 3√5 – 5√6 } \) = \(\frac { 2√6 – √5 }{ 3√5 – 5√6 } \) x \(\frac { 3√5 + 5√6 }{ 3√5 + 5√6 } \)
[(a + b)(a – b) = a² – b²]
= \(\frac { (2√6 – √5)(3√5 + 5√6) }{ (3√5)² – (5√6)² } \)
= \(\frac { 6√30 + 10(6) – 3(5) – 5√30 }{ 45 – 150 } \)
= \(\frac { √30 + 60 – 15 }{ -105 } \)
= \(\frac { -√30 – 45 }{ 105 } \)
So, the rationalized fraction is \(\frac { -√30 – 45 }{ 105 } \).
Problem 6:
Rationalize \(\frac { √13 – √17 }{ √13 + √7 } \)
Solution:
Given fraction is \(\frac { √13 – √17 }{ √13 + √7 } \)
To rationalize the denominator, multiply both numerator and denominator by \(\frac { √13 – √17 }{ √13 – √17 } \)
\(\frac { √13 – √17 }{ √13 + √7 } \) = \(\frac { √13 – √17 }{ √13 + √7 } \) x \(\frac { √13 – √17 }{ √13 – √7 } \)
= \(\frac { 13 + 17 – 2√91 }{ 13 – 7 } \)
= \(\frac { 20 – 2√91 }{ 6 } \)
= \(\frac { 10 – √91 }{ 3 } \)
So, the rationalized fraction is \(\frac { 10 – √91 }{ 3 } \).
Problem 7:
Find the value of a and b if
\(\frac { √7 – 2 }{ √7 + 2 } \) = a√7 + b
Solution:
Given that,
\(\frac { √7 – 2 }{ √7 + 2 } \) = a√7 + b
Multiply numerator, denominator by √7 – 2
\(\frac { √7 – 2 }{ √7 + 2 } \) = \(\frac { √7 – 2 }{ √7 + 2 } \) x \(\frac { √7 – 2 }{ √7 – 2 } \)
= \(\frac { (√7 – 2)² }{ 7 – 4 } \)
= \(\frac { 7 + 4 – 4√7 }{ 3 } \)
= \(\frac { 11 – 4√7 }{ 3 } \)
= \(\frac { -4√7 }{ 3 } \) + \(\frac { 11 }{ 3 } \)
Therefore, the values of a = \(\frac { -4 }{ 3 } \), b = \(\frac { 11 }{ 3 } \).