Multiplication of Surds

In multiplication of surds we will learn how to find the product of two or more surds.

Follow the following steps to find the multiplication of two or more surds.

Step I: Express each surd in its simplest mixed form.

Step II: Observe whether the given surds are of the same order or not.

Step III: If they are of the same order then the required product is obtained by multiplying the product of the rational co-efficient by the product of surd-factors.

If they are of different orders then the product is obtained by the above method after reducing them to surds of the same order.

If different order surds have the same base then their product can easily be obtained using the laws of indices.

Multiplication of surds can be obtained by simply following the law of indices.

\(\sqrt[a]{x}\times \sqrt[b]{x} = x^{\frac{1}{a}}\times x^{\frac{1}{b}} = x^{(\frac{1}{a} + \frac{1}{b})}\)

From the above equation we can understand that if surds of rational number x are in different orders, then the product of those surds can be obtained by the sum 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 multiplication the result index of x is \({(\frac{1}{a} + \frac{1}{b})}\).

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

\(\sqrt[a]{x}\times \sqrt[a]{y} = \sqrt[a]{xy}\)

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 product of those surds can be obtained by product of the radicands or rational numbers.

If the surds are not in same order, we can express them in same order to obtain the result of a multiplication 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. Products of surds can rational or irrational, depending upon the situations.

Like \(\sqrt[2]{3}\)×\(\sqrt[2]{3}\) = 3, so the product of two similar surds is rational number.

But \(\sqrt[2]{3}\)×\(\sqrt[3]{3}\) = \(3^{(\frac{1}{2} + \frac{1}{3})}\) = \(3^{\frac{5}{6}}\)


Now we will solve some problems on multiplication of surds to understand more on this.

Examples of multiplication of surds:

1. Find the product of \(5\sqrt[2]{5}\) and \(\sqrt[2]{45}\).

Solution:

\(5\sqrt[2]{5}\) × \(\sqrt[2]{45}\) = \(5\sqrt[2]{5\times 5\times 3\times 3}\) = 5 × 5 × 3 = 75.

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2. Find the product of 7∜4 and 5∜3

Solution:

 The product of 7∜4 and 5∜3

= (7∜4) × (5∜3)

= (7 × 5) × (∜4 × ∜3)

= 35 × \(\sqrt[4]{4\cdot 3}\)

= 35 × ∜12

= 35∜12


3. Find the product of \(3\sqrt[2]{2}\) and \(4\sqrt[6]{3}\).

Solution:

\(3\sqrt[2]{2}\) and \(4\sqrt[6]{3}\) are in order 2 and 6. As the LCM of 2 and 6 is 6, we can convert \(3\sqrt[2]{2}\) into a surd of order 6.

\(3\sqrt[2]{2}\) × \(4\sqrt[6]{3}\) = \(3\times 2^{\frac{1}{2}}\) × \(4\sqrt[6]{3}\)

= \(3\times 2^{\frac{3}{6}}\) × \(4\sqrt[6]{3}\)

= \(3\times 8^{\frac{1}{6}}\) × \(4\sqrt[6]{3}\)

= \(3\sqrt[6]{8}\) × \(4\sqrt[6]{3}\)

= 3 × 4 × \(\sqrt[6]{8}\) × \(\sqrt[6]{3}\)

= 12 × \(\sqrt[6]{8\times 3}\)

= \(12\sqrt[6]{24}\).


4. Find the product of 2√12, 7√20 and √32

Solution:

The product of 2√12, 7√20 and √32

= (2√12) × (7√20) × (√32)

= (2\(\sqrt{2\cdot 2\cdot 3}\)) × (7\(\sqrt{2\cdot 2\cdot 5}\)) × (\(\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2}\))

= (4√3) × (14√5) × (4√2)

= (4 × 14 × 4) × (√3 × √5 × √2)

= 224 × \(\sqrt{3\cdot 5\cdot 2}\)

= 224 × √30

= 224√30


5. Find the product of \(3\sqrt[2]{12}\), \(\sqrt[2]{98}\) and \(5\sqrt[2]{27}\).

Solution:

\(3\sqrt[2]{12}\) × \(\sqrt[2]{98}\) × \(5\sqrt[2]{27}\)

= \(3\sqrt[2]{2\times 2\times 3}\) × \(\sqrt[2]{7\times 7\times 2}\) × \(5\sqrt[2]{3\times 3\times 3}\)

= \(12\sqrt[2]{3}\) × \(7\sqrt[2]{2}\) × \(15\sqrt[2]{3}\)

= 12 × 7 × 15 × \(\sqrt[2]{3\times 2\times 3}\)

= 1260 × 3 × \(\sqrt[2]{2}\)

= \(3780\sqrt[2]{2}\).


6. Simplify: 2√2 × 7∛5 × 3∜3.

Solution:

3∜3 × 2√2 × 7∛5

The orders of the given surds are 4, 2, 3 respectively and L.C.M. of 4, 2 and 3 is 12.

∜3 = 3\(^{1/4}\) = 3\(^{3/12}\) = \(\sqrt[12]{3^{3}}\) = \(\sqrt[12]{27}\)

√2 = 2\(^{1/2}\) = 2\(^{6/12}\) = \(\sqrt[12]{2^{6}}\) = \(\sqrt[12]{64}\)

∛5 = 5\(^{1/3}\) = 5\(^{4/12}\) = \(\sqrt[12]{5^{4}}\) = \(\sqrt[12]{625}\)

Therefore, the given expression 3∜3 × 2√2 × 7∛5

= (3 × 2 × 7) × (∜3 × √2 × ∛5)

= 42 × (\(\sqrt[12]{27}\) × \(\sqrt[12]{64}\) × \(\sqrt[12]{625}\))

= 42 × (\(\sqrt[12]{27 × 64 × 625}\))

= 42 × (\(\sqrt[12]{1080000}\))

= 42\(\sqrt[12]{1080000}\)

 

7. Find the product of \(3\sqrt[2]{2}\), \(5\sqrt[3]{4}\) and \(2\sqrt[4]{8}\).

Solution:

\(3\sqrt[2]{2}\) × \(5\sqrt[3]{4}\) × \(2\sqrt[4]{8}\)

= 3 × \(2^{\frac{1}{2}}\) × 5 × \(\sqrt[3]{2^{2}}\) × 2 × \(\sqrt[4]{2^{3}}\)

= 3 × 5 × 2 × \(2^{\frac{1}{2}}\) × \(2^{\frac{2}{3}}\) × \(2^{\frac{3}{4}}\)

= 30 × \(2^{(\frac{1}{2} + \frac{2}{3} + \frac{3}{4})}\)

= 30 × \(2^{\frac{23}{12}}\)

= \(30\sqrt[12]{2^{23}}\)

= \(30\sqrt[12]{2^{(12 + 11)}}\)

= 30 × \(2\sqrt[12]{2^{11}}\)

= \(60\sqrt[12]{2048}\).


8. Simplify: 4√3 × 2∛9 × 5∜27

Solution:

4√3 × 2∛9 × 5∜27

= (4 × 2 × 5) × (3\(^{1/2}\) × 9\(^{1/3}\) × 27\(^{1/4}\))

= 40 × (3\(^{1/2}\) × 3\(^{2/3}\) × 3\(^{3/4}\))

= 40 × 3\(^{1/2 + 2/3 + 3/4}\)

= 40 × 3\(^{23/12}\)

= 40 × \(\sqrt[12]{3^{23}}\)

= 40 × \(\sqrt[12]{3^{12}\cdot 3^{11}}\)

= 40 × 3\(\sqrt[12]{3^{11}}\)

= 120\(\sqrt[12]{177147}\)


9. Find the product of \(\sqrt[2]{x}\), \(\sqrt[4]{x}\) and \(\sqrt[2]{y}\).

Solution:

\(\sqrt[2]{x}\) × \(\sqrt[4]{x}\) × \(\sqrt[2]{y}\)

As the surds are in order 2, 4 and 2, their LCM is 4, we need to convert \(\sqrt[2]{x}\) and \(\sqrt[2]{y}\) into order 4.

= \(x^{\frac{2}{4}}\) × \(x^{\frac{1}{4}}\) × \(y^{\frac{2}{4}}\)

= \(x^{(\frac{2}{4}+\frac{1}{4})}\) × \(\sqrt[4]{y^{2}}\)

= \(x^{\frac{3}{4}}\) × \(\sqrt[4]{y^{2}}\)

= \(\sqrt[4]{x^{3}}\) × \(\sqrt[4]{y^{2}}\)

= \(\sqrt[4]{x^{3}y^{2}}\).

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11 and 12 Grade Math

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