The sum and difference of two simple quadratic surds are said to be conjugate surds to each other.
Conjugate surds are also known as complementary surds.
Thus, the sum and the difference of two simple quadratic surds 4√7and √2 are 4√7 + √2 and 4√7  √2 respectively. Therefore, two surds (4√7 + √2) and (4√7  √2) are conjugate to each other.
Similarly, two surds (2√5 + √3) and (2√5  √3) are conjugate to each other.
In general, two binomial quadratic surds (x√a + y√b) and (x√a  y√b) are conjugate to each other.
In complex or binominal surds, if sum of two quadratic surds or a quadratic surd and a rational number is multiplied with difference of those two quadratic surds or quadratic surd and rational number, then rational number under root of surd is get squared off and it becomes a rational number as product of sum and difference of two numbers is difference of the square of the two numbers.
\(a^{2}  b^{2} = (a + b)(a  b)\).
The sum and difference of two quadratic surds is called as conjugate to each other. For example \(\sqrt{x}\) = a and \(\sqrt{y}\) = b, a and b are two quadratic surds, if (a + b) or \((\sqrt{x} + \sqrt{y})\) is multiplied with (a  b) or \((\sqrt{x}  \sqrt{y})\), the result will \((\sqrt{x})^{2}\)  \((\sqrt{y})^{2}\) or (x  y) which is rational number. Here \((\sqrt{x} + \sqrt{y})\) and \((\sqrt{x}  \sqrt{y})\) are conjugate surds to each other and the process is called as rationalization of surds as the result becomes a rational number. This process is used for fraction expression of complex surds, where the denominator needs to converted to a rational number eliminating the roots of surds, conjugate surds multiplied to both numerator and denominator and denominator becomes rational.
Like for example, if simplification of the complex surd \(\frac{6}{\sqrt{3}  1}\) is to be done, denominator \(\sqrt{3}  1\) is to be converted to a rational number. If a = \(\sqrt{3}\) and b = 1, then denominator is (ab), if we multiply (a + b) or \(\sqrt{3} + 1\), it will \(a^{2}  b^{2}\) and \(\sqrt{3}\) will be squared off.
\(\frac{6}{\sqrt{3}  1}\)
= \(\frac{6(\sqrt{3} + 1)}{(\sqrt{3}  1)(\sqrt{3} + 1)}\)
= \(\frac{6(\sqrt{3} + 1)}{3  1}\)
= \(\frac{6(\sqrt{3} + 1)}{2}\)
= 2(\sqrt{3} + 1).
In the above example \(\sqrt{3} + 1\) is used as rationalizing factor which is a conjugate to \(\sqrt{3}  1\).
Note:
1. Since 3 + √5 = √9 + √5 and surd conjugate to √9 + √5 is √9  √5, hence it is evident that surds 3 + √5 and 3  √5 are conjugate to each other.
In general, surds (a + x√b) and (a  x√b) are complementary to each other.
2. The product of two binomial quadratic surds is always rational.
For example,
(√m + √n)(√m  √n) = (√m)^2  (√n)^2 = m  n, which is rational.
Here are some examples of conjugates in the following table.
\((\sqrt{2} + \sqrt{3})\) \((\sqrt{5} + \sqrt{3})\) \(\sqrt{2} + 1\) \((4\sqrt{2} + 2\sqrt{3})\) \((\sqrt{x} + y)\) \((a\sqrt{x} + b\sqrt{y})\) 
\((\sqrt{2}  \sqrt{3})\) \((\sqrt{5}  \sqrt{3})\) \(\sqrt{2}  1\) \((4\sqrt{2}  2\sqrt{3})\) \((\sqrt{x}  y)\) \((a\sqrt{x}  b\sqrt{y})\) 
Problems on conjugate surds:
1. Find the conjugates of the following surds.
\((\sqrt{5} + \sqrt{7})\), \((4\sqrt{11}  3\sqrt{7})\), \(3\sqrt{17} + 19\), \((a\sqrt{b}  b\sqrt{a})\).
Solution:
Given Surds \((\sqrt{5} + \sqrt{7})\) \((4\sqrt{11}  3\sqrt{7})\) \(3\sqrt{17} + 19\) \((a\sqrt{b}  b\sqrt{a})\) 
Conjugate \((\sqrt{5}  \sqrt{7})\) \((4\sqrt{11} + 3\sqrt{7})\) \(3\sqrt{17}  19\) \((a\sqrt{b} + b\sqrt{a})\) 
2. Simplify the surd \(\frac{\sqrt[2]{5}  1}{\sqrt[2]{5} + 1}\) by using conjugate surd.
Solution:
= \(\frac{\sqrt[2]{5}  1}{\sqrt[2]{5} + 1}\)
As the denominator is \(\sqrt[2]{5} + 1\), for rationalization of the surd, we need to multiply both numerator and denominator by the conjugate surd \(\sqrt[2]{5}  1\).
= \(\frac{(\sqrt[2]{5}  1)(\sqrt[2]{5}  1)}{(\sqrt[2]{5} + 1)(\sqrt[2]{5}  1)}\)
= \(\frac{(\sqrt[2]{5}  1)^{2}}{5  1}\)….. as we know \((a + b)(a  b) = a^{2}  b^{2}\)
= \(\frac{((\sqrt[2]{5})^{2}  2\times \sqrt{5} + 1^{2})}{4}\)
= \(\frac{5  2\sqrt{5} + 1}{4}\)
= \(\frac{6  2\sqrt{5}}{4}\)
= \(\frac{2(3  \sqrt{5})}{4}\)
= \(\frac{3  \sqrt{5}}{2}\)
3. Rationalize the surd \(\frac{\sqrt{2}}{\sqrt{x}\sqrt{2}}\).
Solution:
\(\frac{\sqrt{2}}{\sqrt{x}  \sqrt{2}}\)
As the denominator is \((\sqrt{x}  \sqrt{2})\), the conjugate surd is \((\sqrt{x} + \sqrt{2})\), we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= \(\frac{(\sqrt{2})(\sqrt{x} + \sqrt{2})}{(\sqrt{x}  \sqrt{2})(\sqrt{x} + \sqrt{2})}\)
= \(\frac{\sqrt{2x} + 2}{x  2}\).
4. Rationalize the surd \(\frac{\sqrt{5}}{2\sqrt{7}3\sqrt{5}}\).
Solution:
\(\frac{\sqrt{5}}{2\sqrt{7}3\sqrt{5}}\)
As the denominator is \((2\sqrt{7}  3\sqrt{5})\), the conjugate surd is \((2\sqrt{7} + 3\sqrt{5})\), we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= \(\frac{\sqrt{5}\times (2\sqrt{7} + 3\sqrt{5})}{(2\sqrt{7}  3\sqrt{5})(2\sqrt{7} + 3\sqrt{5})}\)
= \(\frac{2\sqrt{35} + 3\times 5}{(2\sqrt{7})^{2}  (3\sqrt{5})^{2}}\)
= \(\frac{2\sqrt{35} + 15}{4\times 7  9\times 5}\)
= \(\frac{2\sqrt{35} + 15}{28  45}\)
= \(\frac{(2\sqrt{35} + 15)}{17}\)
● Surds
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