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.

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.

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

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

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

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

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






11 and 12 Grade Math

From Multiplication of Surds to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Worksheet on Money | Conversion of Money from Rupees to Paisa

    Dec 03, 24 01:29 AM

    Worksheet on Money
    Practice the questions given in the worksheet on money. This sheet provides different types of questions where students need to express the amount of money in short form and long form

    Read More

  2. 2nd Grade Money Worksheet | Conversion of Money | Word Problems

    Dec 03, 24 01:19 AM

    Match the following Money
    In 2nd grade money worksheet we will solve the problems on writing amount in words and figures, conversion of money and word problems on money. 1. Write T for true and F for false. (i) Rs. is written…

    Read More

  3. Subtraction of Money | Subtraction with Conversion, without Conversion

    Dec 02, 24 01:47 PM

    Subtraction of Money
    In subtraction of money we will learn how to subtract the amounts of money involving rupees and paise to find the difference. We carryout subtraction with money the same way as in decimal numbers. Whi…

    Read More

  4. Word Problems on Addition of Money |Money Word Problems|Money Addition

    Dec 02, 24 01:26 PM

    Word Problems on Addition of Money
    Let us consider some of the word problems on addition of money. We have solved the problems in both the methods i.e., with conversion into paise and without conversion into paise. Worked-out examples

    Read More

  5. Addition of Money | Add The Amounts of Money Involving Rupees & Paisa

    Nov 29, 24 01:26 AM

    3rd Grade Addition of Money
    In addition of money we will learn how to add the amounts of money involving rupees and paisa together. We carryout with money the same way as in decimal numbers. While adding we need to follow that t…

    Read More