Rationalization of Surds

We will discuss about the rationalization of surds. When the denominator of an expression is a surd which can be reduced to an expression with rational denominator, this process is known as rationalizing the denominator of the surd.

In other words, the process of reducing a given surd to a rational form after multiplying it by a suitable surd is known as rationalization.

When the product of two surds is a rational number, then each of the two surds is called rationalizing factor of the other.

In other words, if the product of two surds is rational, then each is called a rationalizing factor of the other and each is said to be rationalized by the other.

 

Examples of rationalization of surds:

1. For example, the rationalizing factor of √5 is √5 and rationalizing factor of ∛2 is ∛2^2 or ∛4. Since, √5 × √5 = 5 and ∛2 × ∛2^2 = ∛(2 × 2^2) = ∛2^3 = 2

2. (a√z) × (b√z) = (a × b) × (√z × √z) = ab(√z)^2 = abz, which is rational. Therefore, each of the surds a√z and b√z is a rationalizing factor of the other.

3. √5 × 2√5 = 2 × (√5)^2 = 3 × 5 = 15, which is rational. Therefore, each of the surds √5 and 2√5 is a rationalizing factor of the other.

4. (√a + √b) × (√a - √b) = (√a)^2 - (√b)^2 = a - b, which is rational. Therefore, each of the surds (√a + √b)  and (√a - √b) is a rationalizing factor of the other.

5. (x√a + y√b) × (x√a - y√b) = (x√a)^2 - (y√b)^2 = ax - by, which is rational. Therefore, each of the surds (x√a + y√b) and (x√a - y√b) is a rationalizing factor of the other.

6. (4√7 + √3) × (4√7 - √3) = (4√7)^2 - (√3)^2 = 112 - 3 = 109, which is rational. Therefore, each of the surd factors (4√7 + √3) and (4√7 - √3) is a rationalizing factor of the other.

7. Also rationalizing factor of ∛(ab^2c^2) is ∛(a^2bc) because ∛(ab^2c^2) × ∛(a^2bc) = abc.

 

8. Find the rationalizing factor of (√x - ∛y).

Solution:

Let, √x = x^1/2 = a and ∛y = y^1/3 = b.

Now, the order of the surds √x and ∛y are 2 and 3 respectively and the L.C.M. of 2 and 3 is 6.

Therefore,

a^6 = (x^1/2)^6 = x^3 and b^6 = (y^1/3)^6 = y^2.

Therefore, a^6 and b^6 both are rational and as such (a^6 - b^6) is also rational.

Now, a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)

Therefore the rationalizing factor of (a - b) = (√x - ∛y) is (a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5) = x^5/2 + x^2y^1/3 + x^3/2y^2/3 + xy + x^1/2y^4/3 + y^5/3






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