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.

If a surd or surd with rational numbers present in the denominator of an equation, to simplify it or to omit the surds from the denominator, rationalization of surds is used. Surds are irrational numbers but if multiply a surd with a suitable factor, result of multiplication will be rational number. This is the basic principle involved in rationalization of surds. The factor of multiplication by which rationalization is done, is called as rationalizing factor. If the product of two surds is a rational number, then each surd is a rationalizing factor to other. Like if \(\sqrt{2}\) is multiplied with  \(\sqrt{2}\), it will 2, which is rational number, so \(\sqrt{2}\) is rationalizing factor of \(\sqrt{2}\).

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.

For example, \(\frac{\sqrt{5}}{\sqrt{2}}\) is a surd where \(\sqrt{5}\) is numerator and \(\sqrt{2}\) is denominator. Now for rationalization of surds, if we multiply both numerator and denominator by \(\sqrt{2}\), then denominator will be a rational number.

\(\frac{\sqrt{5}}{\sqrt{2}}\) = \(\frac{\sqrt{5}\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\) 

= \(\frac{\sqrt{10}}{2}\).

So after rationalization of surd \(\frac{\sqrt{5}}{\sqrt{2}}\), it is becoming \(\frac{\sqrt{10}}{2}\) where \(\sqrt{2}\) is used as rationalization factor. 

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.

For complex surds where the surds are in order 2, conjugates are used to rationalize the surds. This comes the from the formula \(a^{2} - b^{2}=(a + b)(a - b)\). So in complex surds order 2 surds get squared off and denominators are converted to a rational number.

Like for example, if rationalization of the complex surd \(\frac{1}{\sqrt[2]{2} - 1}\) to be done, denominator \(\sqrt[2]{2} - 1\) is to converted to a rational number. If a = \(\sqrt[2]{2}\) and b = 1, then denominator is (a - b), if we multiply (a + b), it will \(a^{2} - b^{2}\) and \(\sqrt[2]{2}\) will be squared off.

\(\frac{1}{\sqrt[2]{2} - 1}\)

= \(\frac{(\sqrt[2]{2} + 1)}{(\sqrt[2]{2} - 1)(\sqrt[2]{2} + 1)}\)

= \(\frac{(\sqrt[2]{2} + 1)}{2 - 1}\)

= \(\sqrt[2]{2}\) + 1.

For complex surds in the denominator in other forms or in order more than 2, can be rationalized by using suitable multiplication factors.


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. Rationalize the following surds.

\(\frac{\sqrt[2]{3}}{\sqrt[2]{2}}\), \(\frac{6}{\sqrt{3}}\), \(\frac{\sqrt[3]{5}}{\sqrt[3]{3}}\)

Solution:

\(\frac{\sqrt[2]{3}}{\sqrt[2]{2}}\)

= \(\frac{\sqrt{3}\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\)

= \(\frac{\sqrt{6}}{2}\).

\(\frac{6}{\sqrt[2]{3}}\)

= \(\frac{6\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\)

= \(\frac{6\sqrt{3}}{3}\)

= \(2\sqrt{3}\).

\(\frac{\sqrt[3]{5}}{\sqrt[3]{3}}\)

Here the denominator is \(\sqrt[3]{3}\) or \(3^{\frac{1}{3}}\), for this surd of order 3, rationalization factor will be \(3^{\frac{2}{3}}\).

= \(\frac{\sqrt[3]{5}\times 3^{\frac{2}{3}}}{3^{\frac{1}{3}}\times 3^{\frac{2}{3}}}\)

= \(\frac{\sqrt[3]{5}\times \sqrt[3]{3^{2}}}{3}\)

= \(\frac{\sqrt[3]{45}}{3}\).


9. 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


10. Rationalize the surd \(\frac{\sqrt[2]{2}+1}{\sqrt[2]{2}-1}\).

Solution:

= \(\frac{\sqrt[2]{2} + 1}{\sqrt[2]{2} - 1}\)

As the denominator is \(\sqrt[2]{2} - 1\), for rationalization of the surd, we need to multiply both numerator and denominator by the rationalizing factor \(\sqrt[2]{2} + 1\).

= \(\frac{(\sqrt[2]{2} + 1)(\sqrt[2]{2} + 1)}{(\sqrt[2]{2} - 1)(\sqrt[2]{2} + 1)}\)

= \(\frac{(\sqrt[2]{2} + 1)^{2}}{2 - 1}\)….. as we know \((a + b)(a - b)=a^{2} - b^{2}\)

= \((\sqrt[2]{2})^{2} + 2\times \sqrt[2]{2}\times 1 + 1^{2}\)

= \(3 + 2\sqrt[2]{2}\).


11. Rationalize the surd \(\frac{\sqrt{x}}{\sqrt{x} - \sqrt{y}}\).

Solution:

\(\frac{\sqrt{x}}{\sqrt{x} - \sqrt{y}}\)

As the denominator is \((\sqrt{x} - \sqrt{y})\), if \(\sqrt{x}\) = a and \(\sqrt{y}\) = b, denominator (a - b) is multiplied with (a + b) or \((\sqrt{x} + \sqrt{y})\), it will be rationalized.

= \(\frac{\sqrt{x}\times (\sqrt{x} + \sqrt{y})}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}\)

= \(\frac{x + \sqrt{xy}}{x - y}\)


12. Rationalize the surd \(\frac{2\sqrt{5}}{5\sqrt{3} - 3\sqrt{5}}\).

Solution:

\(\frac{2\sqrt{5}}{5\sqrt{3} - 3\sqrt{5}}\)

As the denominator is \((5\sqrt{3} - 3\sqrt{5})\), if \(5\sqrt{3}\) = a and \(3\sqrt{5}\) = b, denominator (a - b) is multiplied with (a + b) or \((5\sqrt{3} + 3\sqrt{5})\), it will be rationalized.

= \(\frac{2\sqrt{5}\times (5\sqrt{3} + 3\sqrt{5})}{(5\sqrt{3} - 3\sqrt{5})(5\sqrt{3} + 3\sqrt{5})}\)

= \(\frac{10\sqrt{15} + 6\times 5}{(5\sqrt{3})^{2}-(3\sqrt{5})^{2}}\)

= \(\frac{10\sqrt{15} + 30}{25\times 3 - 9\times 5}\)

= \(\frac{10\sqrt{15} + 30}{75 - 45}\)

= \(\frac{10(\sqrt{15} + 3)}{30}\)

= \(3 + \sqrt{15}\)

● Surds






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