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
`11 and 12 Grade Math
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