Division of Surds

In division of surds we need to divide a given surd by another surd the quotient is first expressed as a fraction. Then by rationalizing the denominator the required quotient is obtained with a rational denominator. For this the numerator and the denominator are multiplied by appropriate rationalizing factor. In rationalization of surds the multiplying surd-factor is called the rationalizing factor of the given surd.

Division of surds in general can be obtained by following the law of indices.

\(\sqrt[a]{x}\div \sqrt[b]{x}\)= \(\frac{\sqrt[a]{x}}{\sqrt[b]{x}}\)= \(x^{(\frac{1}{a}-\frac{1}{b})}\).

From the above equation we can understand that if surds of rational number x are in different orders, then the indices are expressed in fraction and division is obtained by the subtraction of indices of the surds. Here surds of rational number x are in order a and b, so the indices of the surds are \(\frac{1}{a}\) and \(\frac{1}{b}\) and after division the result index of x is \({(\frac{1}{a}-\frac{1}{b})}\).

If the surds are in same order, then division of surds can be done by following rule.

\(\sqrt[a]{x}\div \sqrt[a]{y}\)= \(\frac{\sqrt[a]{x}}{\sqrt[a]{y}}\)= \(\sqrt[a]{\frac{x}{y}}\).

From the above equation we can understand that if two or more rational numbers like x and y are in a same order a, then division of those surds can be obtained by division of the radicands or rational numbers of the surds.

In division if the surds are not in same order, we can convert them in same order to obtain the result of a division problem. But first we should try to express the surds in simplest forms and compare with other surds that they are similar surds or equiradical or dissimilar. Whatever the surds are, we can multiply the rational coefficients.

Sometimes for division of surds, we need to rationalize the denominator to get a simpler form and obtain a result. For this both numerator and denominator need to be multiplied by appropriate rationalizing factor.

Like for example \(\frac{\sqrt[2]{x}}{\sqrt[2]{y}}\)

= \(\frac{\sqrt[2]{x}\times \sqrt[2]{y}}{\sqrt[2]{y}\times \sqrt[2]{y}}\) 

=\(\frac{\sqrt[2]{xy}}{y}\)

In the above example \(\sqrt[2]{y}\) is the denominator and rationalizing factor for \(\sqrt[2]{y}\) is \(\sqrt[2]{y}\). So \(\sqrt[2]{y}\) is multiplied to both the nominator and denominator to rationalize the surd.


Now we will solve some problems to understand more on division of surds:

1. Find the division of \(\sqrt[2]{12}\) by \(\sqrt[2]{3}\).

Solution:

\(\sqrt[2]{12}\) ÷ \(\sqrt[2]{3}\)

= \(\sqrt[2]{\frac{12}{3}}\)

= \(\sqrt[2]{\frac{4\times 3}{3}}\)

= \(\sqrt[2]{4}\)

= \(\sqrt[2]{2^{2}}\)

= 2.   


2. Divide: √x by √y

Solution:

√x by √y

= √x ÷ √y

= √x/√y

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


3. Find the division of \(\sqrt[2]{5}\) by \(\sqrt[2]{3}\).

Solution:

\(\sqrt[2]{5}\) ÷ \(\sqrt[2]{3}\)

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

= \(\frac{\sqrt[2]{5}\times \sqrt[2]{3}}{\sqrt[2]{3}\times \sqrt[2]{3}}\) ….multiplying \(\sqrt[2]{3}\) as rationalizing factor

= \(\frac{\sqrt[2]{15}}{3}\).


4. Divide the first surd by the second surd: √32, √8

Solution:

√32 divided by √8

= √32 ÷ √8

= \(\sqrt{\frac{32}{8}}\)

= √4

= 2.


5. Find the division of \(\sqrt[2]{3}\) by \(\sqrt[2]{2}-1\).

Solution:

\(\sqrt[2]{3}\) ÷ \(\sqrt[2]{2} - 1\)

= \(\frac{\sqrt[2]{3}}{\sqrt[2]{2} - 1}\)

As the denominator is \(\sqrt[2]{2} - 1\), for the division, we need to multiply it with a rationalizing factor \(\sqrt[2]{2} + 1\).

= \(\frac{\sqrt[2]{3}(\sqrt[2]{2} + 1)}{(\sqrt[2]{2} - 1)(\sqrt[2]{2} + 1)}\)

= \(\frac{\sqrt[2]{3}\times \sqrt[2]{2} + \sqrt[2]{3}}{2 - 1}\)….. as we know \((a + b)(a - b) = a^{2} - b^{2}\)

=  \(\sqrt[2]{6}\) + \(\sqrt[2]{3}\).


6. Find the quotient dividing the surd √96 by the surd √16.

Solution:

Required quotient

= √96 ÷ √16

= \(\sqrt{\frac{96}{16}}\)

= √6.


7. Find the division of \((x-1)\) by \(\sqrt[2]{x}-1\).

Solution:

\((x - 1)\) ÷ \(\sqrt[2]{x} - 1\)

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

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

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

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


8. Divide: √5 by √7

Solution:

√5 divided by √7

= √5 ÷ √7

= \(\sqrt{\frac{5}{7}}\)

= \(\frac{\sqrt{5}\times \sqrt{7}}{\sqrt{7}\times \sqrt{7}}\), [Rationalization of denominator of surds]

= √35/7.

 Surds






11 and 12 Grade Math

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