In this section, we will discuss about conjugate surds. For the basics of surds, please visit the page an introduction to surds.

Table of Contents

**Definition of Conjugate Surds**

A surd is said to be a conjugate surd to another surd if they are the sum and difference of two simple quadratic surds. In other words, the sum and the difference of two simple quadratic surds are conjugate to each other.

For example, we consider two simple quadratic surds $\sqrt{2}$ and $7\sqrt{3}.$ According to the above definition, the two binomial surds $\sqrt{2}+7\sqrt{3}$ and $\sqrt{2}-7\sqrt{3}$ are conjugate (or complementary) to each other. In a similar way, we have the following examples of conjugate surds:

(i) $5\sqrt{2}+2\sqrt{7}$ and** **$5\sqrt{2}-2\sqrt{7}$

(ii) $1+\sqrt{3}$ and $1-\sqrt{3}$

**Properties of Conjugate Surds**

**• **The general form of two conjugate surds are $a\sqrt{x}+b\sqrt{y}$ and $a\sqrt{x}-b\sqrt{y}.$

**• **The product of two conjugate surds is always a rational number.

Proof. The product of two general conjugate surds is given by

$(a\sqrt{x}+b\sqrt{y})$$(a\sqrt{x}-b\sqrt{y})$

$=(a\sqrt{x})^2-(b\sqrt{y})^2$ $[\because (m+n)(m-n)=m^2-n^2]$

$=a^2x-b^2y,$

which is a rational number.

**Importance of Conjugate Surds**

To rationalize the denominator of a fraction containing surds, we need to take the help of conjugate surds. In this case, we need to multiply the denominator with its conjugate surd. For example,

Question: Rationalize the denominator of $\dfrac{1}{1+\sqrt{2}}$

Answer:

See that the denominator 1+√2 is not a rational number. To rationalize the denominator we have to multiply with the conjugate of 1+√2 which is 1-√2. By doing so we get that

$\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}$

$=\dfrac{1-\sqrt{2}}{1^2-\sqrt{2}^2}$ $[\because (a-b)(a+b)=a^2-b^2]$

$=\dfrac{1-\sqrt{2}}{1-2}$

$=-(1-\sqrt{2})=-1+\sqrt{2}$

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