Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.6

Question 1. Simplify the following using addition and subtraction properties of surds:
(i) \( 5\sqrt{3} + 18\sqrt{3} – 2\sqrt{3} \)
(ii) \( 4\sqrt[3]{5} + 2\sqrt[3]{5} – 3\sqrt[3]{5} \)
(iii) \( 3\sqrt{75} + 5\sqrt{48} – \sqrt{243} \)
(iv) \( 5\sqrt[3]{40} + 2\sqrt[3]{625} – 3\sqrt[3]{320} \)
Answer:
(i) To simplify \( 5\sqrt{3} + 18\sqrt{3} – 2\sqrt{3} \), we can combine the coefficients because they all have the same surd, \( \sqrt{3} \).
\( (5 + 18 – 2)\sqrt{3} \)
\( (23 – 2)\sqrt{3} \)
\( 21\sqrt{3} \)
(ii) To simplify \( 4\sqrt[3]{5} + 2\sqrt[3]{5} – 3\sqrt[3]{5} \), we combine the coefficients as they share the common surd \( \sqrt[3]{5} \).
\( (4 + 2 – 3)\sqrt[3]{5} \)
\( (6 – 3)\sqrt[3]{5} \)
\( 3\sqrt[3]{5} \)
(iii) To simplify \( 3\sqrt{75} + 5\sqrt{48} – \sqrt{243} \), first simplify each surd by finding perfect square factors.
We find the prime factors for each number:

This means \( \sqrt{75} = \sqrt{5^2 \times 3} = 5\sqrt{3} \), \( \sqrt{48} = \sqrt{2^4 \times 3} = 2^2\sqrt{3} = 4\sqrt{3} \), and \( \sqrt{243} = \sqrt{3^5} = \sqrt{3^4 \times 3} = 3^2\sqrt{3} = 9\sqrt{3} \).
Now, substitute these back into the expression:
\( 3(5\sqrt{3}) + 5(4\sqrt{3}) – 9\sqrt{3} \)
\( 15\sqrt{3} + 20\sqrt{3} – 9\sqrt{3} \)
\( (15 + 20 – 9)\sqrt{3} \)
\( (35 – 9)\sqrt{3} \)
\( 26\sqrt{3} \)
(iv) To simplify \( 5\sqrt[3]{40} + 2\sqrt[3]{625} – 3\sqrt[3]{320} \), we simplify each cube root by finding perfect cube factors.

This means \( \sqrt[3]{40} = \sqrt[3]{2^3 \times 5} = 2\sqrt[3]{5} \), \( \sqrt[3]{625} = \sqrt[3]{5^3 \times 5} = 5\sqrt[3]{5} \), and \( \sqrt[3]{320} = \sqrt[3]{2^6 \times 5} = 2^2\sqrt[3]{5} = 4\sqrt[3]{5} \).
Now, substitute these back into the expression:
\( 5(2\sqrt[3]{5}) + 2(5\sqrt[3]{5}) – 3(4\sqrt[3]{5}) \)
\( 10\sqrt[3]{5} + 10\sqrt[3]{5} – 12\sqrt[3]{5} \)
\( (10 + 10 – 12)\sqrt[3]{5} \)
\( (20 – 12)\sqrt[3]{5} \)
\( 8\sqrt[3]{5} \)
In simple words: To simplify expressions with surds, first simplify each individual surd by taking out any perfect square or cube factors. Then, if the remaining surds are the same, you can add or subtract their numbers in front.

🎯 Exam Tip: Always look for perfect square or cube factors inside the surd first to simplify it completely before combining terms. This is a common mistake that can lead to incorrect answers.

Question 2. Simplify the following using multiplication and division properties of surds:
(i) \( \sqrt{3} \times \sqrt{5} \times \sqrt{2} \)
(ii) \( \sqrt{35} \div \sqrt{7} \)
(iii) \( \sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125} \)
(iv) \( (7\sqrt{a} – 5\sqrt{b}) (7\sqrt{a} + 5\sqrt{b}) \)
(v) \( (\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}) \div \sqrt{\frac{16}{81}} \)
Answer:
(i) When multiplying surds with the same root index, we can multiply the numbers inside the surd.
\( \sqrt{3} \times \sqrt{5} \times \sqrt{2} = \sqrt{3 \times 5 \times 2} = \sqrt{30} \)
(ii) When dividing surds with the same root index, we can divide the numbers inside the surd.
\( \sqrt{35} \div \sqrt{7} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{\frac{35}{7}} = \sqrt{5} \)
(iii) To multiply these cube roots, we can combine the numbers inside the cube root.
\( \sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125} = \sqrt[3]{27 \times 8 \times 125} \)
We know that \( 27 = 3^3 \), \( 8 = 2^3 \), and \( 125 = 5^3 \).
So, \( \sqrt[3]{3^3 \times 2^3 \times 5^3} = 3 \times 2 \times 5 = 30 \)
(iv) This expression is in the form \( (A – B)(A + B) \), which simplifies to \( A^2 – B^2 \).
Here, \( A = 7\sqrt{a} \) and \( B = 5\sqrt{b} \).
So, \( (7\sqrt{a})^2 – (5\sqrt{b})^2 = (7^2 \times (\sqrt{a})^2) – (5^2 \times (\sqrt{b})^2) = 49a – 25b \)
(v) For \( (\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}) \div \sqrt{\frac{16}{81}} \), first simplify each fraction under the square root.

So, we have: \( \sqrt{\frac{225}{729}} = \sqrt{\frac{15^2}{27^2}} = \frac{15}{27} = \frac{5}{9} \)
\( \sqrt{\frac{25}{144}} = \sqrt{\frac{5^2}{12^2}} = \frac{5}{12} \)
\( \sqrt{\frac{16}{81}} = \sqrt{\frac{4^2}{9^2}} = \frac{4}{9} \)
Substitute these values back into the expression:
\( (\frac{5}{9} – \frac{5}{12}) \div \frac{4}{9} \)
First, subtract the fractions in the parenthesis. The LCM of 9 and 12 is 36.
\( (\frac{5 \times 4}{9 \times 4} – \frac{5 \times 3}{12 \times 3}) \div \frac{4}{9} \)
\( (\frac{20}{36} – \frac{15}{36}) \div \frac{4}{9} \)
\( \frac{5}{36} \div \frac{4}{9} \)
To divide fractions, multiply by the reciprocal of the second fraction:
\( \frac{5}{36} \times \frac{9}{4} \)
\( \frac{5 \times 9}{36 \times 4} = \frac{5 \times 1}{4 \times 4} = \frac{5}{16} \)
In simple words: When multiplying or dividing surds with the same type of root (like square root or cube root), you can multiply or divide the numbers inside the root. If there are fractions, simplify them first, then perform the operations. Remember the difference of squares formula for expressions like \( (A-B)(A+B) \).

🎯 Exam Tip: Always look for perfect square or cube factors inside the surd first to simplify it completely before combining terms. This is a common mistake that can lead to incorrect answers.

Question 3. If \( \sqrt{2} = 1.414 \), \( \sqrt{3} = 1.732 \), \( \sqrt{5} = 2.236 \), \( \sqrt{10} = 3.162 \), then find the values of the following correct to 3 places of decimals.
(i) \( \sqrt{40} – \sqrt{20} \)
(ii) \( \sqrt{300} + \sqrt{90} – \sqrt{8} \)
Answer:
(i) First, simplify each surd by finding perfect square factors:
\( \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} \)
\( \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)
Now, substitute the given values:
\( 2\sqrt{10} – 2\sqrt{5} = 2 \times 3.162 – 2 \times 2.236 \)
\( = 6.324 – 4.472 = 1.852 \)
(ii) First, simplify each surd by finding perfect square factors:
\( \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3} \)
\( \sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10} \)
\( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
Now, substitute the given values:
\( 10\sqrt{3} + 3\sqrt{10} – 2\sqrt{2} = 10 \times 1.732 + 3 \times 3.162 – 2 \times 1.414 \)
\( = 17.32 + 9.486 – 2.828 \)
\( = 26.806 – 2.828 \)
\( = 23.978 \)
In simple words: To solve these, break down the numbers inside the square roots into parts where one part is a perfect square. Then use the given decimal values for the basic square roots and do the math. Always round your final answer to the required number of decimal places.

🎯 Exam Tip: Remember to simplify the surds first before substituting the decimal values. This makes calculations easier and reduces the chance of errors, ensuring you get the correct answer to the specified decimal places.

Question 4. Arrange surds in descending order
(i) \( \sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3} \)
(ii) \( \sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}} \)
Answer:
(i) To arrange \( \sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3} \) in descending order, we need to make their root indices the same. The least common multiple (LCM) of 3, 9, and 6 is 18.
We convert each surd to have an index of 18:

For \( \sqrt[3]{5} \): Multiply index 3 by 6, so we raise 5 to the power of 6.
\( \sqrt[3]{5} = \sqrt[3 \times 6]{5^6} = \sqrt[18]{15625} \)
For \( \sqrt[9]{4} \): Multiply index 9 by 2, so we raise 4 to the power of 2.
\( \sqrt[9]{4} = \sqrt[9 \times 2]{4^2} = \sqrt[18]{16} \)
For \( \sqrt[6]{3} \): Multiply index 6 by 3, so we raise 3 to the power of 3.
\( \sqrt[6]{3} = \sqrt[6 \times 3]{3^3} = \sqrt[18]{27} \)
Now, comparing the numbers inside the surds with the same index (18):
\( 15625 > 27 > 16 \)
Therefore, in descending order:
\( \sqrt[18]{15625} > \sqrt[18]{27} > \sqrt[18]{16} \)
This means: \( \sqrt[3]{5} > \sqrt[6]{3} > \sqrt[9]{4} \)
(ii) To arrange \( \sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}} \) in descending order, first simplify the nested surds. When dealing with nested roots, we multiply their indices.
\( \sqrt[2]{\sqrt[3]{5}} = \sqrt[2 \times 3]{5} = \sqrt[6]{5} \)
\( \sqrt[3]{\sqrt[4]{7}} = \sqrt[3 \times 4]{7} = \sqrt[12]{7} \)
\( \sqrt{\sqrt{3}} = \sqrt[2 \times 2]{3} = \sqrt[4]{3} \)
Now we have \( \sqrt[6]{5}, \sqrt[12]{7}, \sqrt[4]{3} \). The LCM of the indices 6, 12, and 4 is 12.
We convert each surd to have an index of 12:

For \( \sqrt[6]{5} \): Multiply index 6 by 2, so we raise 5 to the power of 2.
\( \sqrt[6]{5} = \sqrt[6 \times 2]{5^2} = \sqrt[12]{25} \)
For \( \sqrt[12]{7} \): The index is already 12, so it stays as \( \sqrt[12]{7} \).
For \( \sqrt[4]{3} \): Multiply index 4 by 3, so we raise 3 to the power of 3.
\( \sqrt[4]{3} = \sqrt[4 \times 3]{3^3} = \sqrt[12]{27} \)
Now, comparing the numbers inside the surds with the same index (12):
\( 27 > 25 > 7 \)
Therefore, in descending order:
\( \sqrt[12]{27} > \sqrt[12]{25} > \sqrt[12]{7} \)
This means: \( \sqrt{\sqrt{3}} > \sqrt[2]{\sqrt[3]{5}} > \sqrt[3]{\sqrt[4]{7}} \)
In simple words: To compare surds, change them so they all have the same root number (index). Find the smallest common number for all the roots, then adjust each surd and the number inside it. Once all roots are the same, just compare the numbers inside to put them in order.

🎯 Exam Tip: When arranging surds, always ensure all surds have the same order (root index) before comparing the radicands (the numbers inside the root). For nested surds, multiply the indices to simplify them into a single surd first.

Question 5. Can you get a pure surd when you find:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Answer:
(i) Yes, we can get a pure surd when adding two surds. A pure surd is a surd that has no rational coefficient other than 1 or -1.
Example (a): \( 3\sqrt{2} + 5\sqrt{2} = (3 + 5)\sqrt{2} = 8\sqrt{2} \)
Example (b): \( 3\sqrt{6} + 2\sqrt{6} = (3 + 2)\sqrt{6} = 5\sqrt{6} \)
These results \( 8\sqrt{2} \) and \( 5\sqrt{6} \) are pure surds.
(ii) Yes, we can get a pure surd when finding the difference of two surds.
Example (a): \( \sqrt{75} - \sqrt{48} \)
First, simplify the surds: \( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \) and \( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \).
So, \( 5\sqrt{3} - 4\sqrt{3} = (5 - 4)\sqrt{3} = \sqrt{3} \)
Example (b): \( \sqrt{98} – \sqrt{72} \)
First, simplify the surds: \( \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2} \) and \( \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \).
So, \( 7\sqrt{2} - 6\sqrt{2} = (7 - 6)\sqrt{2} = \sqrt{2} \)
These results \( \sqrt{3} \) and \( \sqrt{2} \) are pure surds.
(iii) Yes, we can get a pure surd when multiplying two surds.
Example (a): \( \sqrt{8} \times \sqrt{6} = \sqrt{8 \times 6} = \sqrt{48} \). This can be simplified to \( \sqrt{16 \times 3} = 4\sqrt{3} \), which is a pure surd.
Example (b): \( \sqrt{11} \times \sqrt{3} = \sqrt{11 \times 3} = \sqrt{33} \). This is a pure surd.
(iv) Yes, we can get a pure surd when finding the quotient of two surds.
Example (a): \( \sqrt{55} \div \sqrt{5} = \frac{\sqrt{55}}{\sqrt{5}} = \sqrt{\frac{55}{5}} = \sqrt{11} \). This is a pure surd.
Example (b): \( \sqrt{65} \div \sqrt{13} = \frac{\sqrt{65}}{\sqrt{13}} = \sqrt{\frac{65}{13}} = \sqrt{5} \). This is a pure surd.
In simple words: Yes, for all four operations (add, subtract, multiply, divide), you can sometimes end up with a pure surd. A pure surd is just a number under a root sign that cannot be simplified further to remove the root entirely, like \( \sqrt{2} \) or \( \sqrt{33} \).

🎯 Exam Tip: A pure surd has only the number inside the radical, while a mixed surd has a rational number outside (e.g., \( 2\sqrt{3} \)). Make sure to simplify all surds in your examples to clearly show if the result is pure or not.

Question 6. Can you get a rational number when you compute:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Answer:
(i) Yes, the sum of two surds can sometimes give a rational number. A rational number is any number that can be written as a simple fraction (a whole number or a terminating/repeating decimal).
Example (a): \( (2 + \sqrt{3}) + (2 – \sqrt{3}) = 2 + \sqrt{3} + 2 – \sqrt{3} = 4 \)
Here, the \( \sqrt{3} \) terms cancel out, leaving a rational number, 4.
Example (b): \( (\sqrt{5} + 4) + (7 – \sqrt{5}) = \sqrt{5} + 4 + 7 – \sqrt{5} = 11 \)
(ii) Yes, the difference of two surds can sometimes give a rational number.
Example (a): \( (5 + \sqrt{7}) – (-5 + \sqrt{7}) = 5 + \sqrt{7} + 5 – \sqrt{7} = 10 \)
Here, the \( \sqrt{7} \) terms cancel out, and the negative sign with -5 changes to a positive, resulting in a rational number, 10.
Example (b): \( (\sqrt{11} + 5) – (-3 + \sqrt{11}) = \sqrt{11} + 5 + 3 – \sqrt{11} = 8 \)
(iii) Yes, the product of two surds can sometimes give a rational number.
Example (a): \( \sqrt{125} \times \sqrt{45} \)
Simplify the surds first: \( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \) and \( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \).
So, \( 5\sqrt{5} \times 3\sqrt{5} = (5 \times 3) \times (\sqrt{5} \times \sqrt{5}) = 15 \times 5 = 75 \)
The result 75 is a rational number.
Example (b): \( \sqrt{150} \times \sqrt{6} \)
Simplify the surd: \( \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \).
So, \( 5\sqrt{6} \times \sqrt{6} = 5 \times (\sqrt{6} \times \sqrt{6}) = 5 \times 6 = 30 \)
The result 30 is a rational number.
(iv) Yes, the quotient of two surds can sometimes give a rational number.
Example (a): \( \sqrt{32} \div \sqrt{8} = \frac{\sqrt{32}}{\sqrt{8}} = \sqrt{\frac{32}{8}} = \sqrt{4} = 2 \)
The result 2 is a rational number.
Example (b): \( \sqrt{50} \div \sqrt{2} = \frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \)
The result 5 is a rational number.
In simple words: Yes, it is possible to get a rational number (a regular number without a root sign) from adding, subtracting, multiplying, or dividing surds. This usually happens when the root parts cancel each other out or simplify to a whole number.

🎯 Exam Tip: To show that an operation on surds results in a rational number, provide examples where the irrational parts (the surds) either cancel out (in addition/subtraction) or simplify to an integer (in multiplication/division). This demonstrates a clear understanding of rational numbers.