Maharashtra Board Class 9 Maths Chapter 2 Set 2.3 Algebra Standard Part 1 Real Numbers Solutions

Question 1. State the order of the surds given below.
(i) \( \sqrt[3]{7} \)
(ii) \( 5\sqrt{12} \)
(iii) \( \sqrt[4]{10} \)
(iv) \( \sqrt{39} \)
(v) \( \sqrt[3]{18} \)
Answer:
(i) The order of the surd \( \sqrt[3]{7} \) is 3.
(ii) The order of the surd \( 5\sqrt{12} \) is 2.
(iii) The order of the surd \( \sqrt[4]{10} \) is 4.
(iv) The order of the surd \( \sqrt{39} \) is 2.
(v) The order of the surd \( \sqrt[3]{18} \) is 3. The order of a surd is determined by the index of its radical sign.
In simple words: The order of a surd is just the small number sitting on the root sign. If there is no number written, it is a square root, which always has an order of 2.

🎯 Exam Tip: Remember that a standard square root symbol \( \sqrt{x} \) has an implicit order of 2. Do not write 1 or leave it blank in your final answer.

Question 2. State which of the following are surds Justify. [2 Marks each]
(i) \( \sqrt[3]{51} \)
(ii) \( \sqrt[4]{16} \)
(iii) \( \sqrt[5]{81} \)
(iv) \( \sqrt{256} \)
(v) \( \sqrt[3]{64} \)
(vi) \( \sqrt{\frac{22}{7}} \)
Answer:
(i) \( \sqrt[3]{51} \) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \( \sqrt[3]{51} \) is irrational.
(ii) \( \sqrt[4]{16} \) is not a surd because \( \sqrt[4]{16} = (16)^{\frac{1}{4}} = (2^4)^{\frac{1}{4}} = 2 \), which is not an irrational number.
(iii) \( \sqrt[5]{81} \) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \( \sqrt[5]{81} \) is irrational.
(iv) \( \sqrt{256} \) is not a surd because \( \sqrt{256} = (256)^{\frac{1}{2}} = (16^2)^{\frac{1}{2}} = 16 \), which is not an irrational number.
(v) \( \sqrt[3]{64} \) is not a surd because \( \sqrt[3]{64} = (64)^{\frac{1}{3}} = (4^3)^{\frac{1}{3}} = 4 \), which is not an irrational number.
(vi) \( \sqrt{\frac{22}{7}} \) is a surd because \( \frac{22}{7} \) is a positive rational number, 2 is a positive integer greater than 1 and \( \sqrt{\frac{22}{7}} \) is irrational. Understanding these conditions helps us easily distinguish surds from other real numbers.
In simple words: A surd is a root of a positive number that cannot be simplified to a whole number or a simple fraction. If the root simplifies perfectly into a normal number, it is not a surd.

🎯 Exam Tip: To score full marks, clearly state all three conditions for a surd: the radicand must be positive and rational, the order must be an integer greater than 1, and the final value must be irrational.

Question 3. Classify the given pair of surds into like surds and unlike surds. [2 Marks each]

Question 1. Classify the given pair of surds into like surds and unlike surds.
(i) \( \sqrt{52}, 5\sqrt{13} \)
(ii) \( \sqrt{68}, 5\sqrt{3} \)
(iii) \( 4\sqrt{18}, 7\sqrt{2} \)
(iv) \( 19\sqrt{12}, 6\sqrt{3} \)
(v) \( 5\sqrt{22}, 7\sqrt{33} \)
(vi) \( 5\sqrt{5}, \sqrt{75} \)
Answer:
If the order of the surds and the radicands are same, then the surds are like surds.

(i) \( \sqrt{52}, 5\sqrt{13} \)
\( \sqrt{52} = \sqrt{4 \times 13} \)
\( = \sqrt{4} \times \sqrt{13} \)
\( = 2\sqrt{13} \)
Here, the order of \( 2\sqrt{13} \) and \( 5\sqrt{13} \) is same and their radicands are also same.
\( \implies \sqrt{52} \) and \( 5\sqrt{13} \) are like surds.

(ii) \( \sqrt{68}, 5\sqrt{3} \)
\( \sqrt{68} = \sqrt{4 \times 17} \)
\( = \sqrt{4} \times \sqrt{17} \)
\( = 2\sqrt{17} \)
Here, the order of \( 2\sqrt{17} \) and \( 5\sqrt{3} \) is same but their radicands are not.
\( \implies \sqrt{68} \) and \( 5\sqrt{3} \) are unlike surds.

(iii) \( 4\sqrt{18}, 7\sqrt{2} \)
\( 4\sqrt{18} = 4 \times \sqrt{9 \times 2} \)
\( = 4 \times \sqrt{9} \times \sqrt{2} \)
\( = 4 \times 3\sqrt{2} \)
\( = 12\sqrt{2} \)
Here, the order of \( 12\sqrt{2} \) and \( 7\sqrt{2} \) is same and their radicands are also same.
\( \implies 4\sqrt{18} \) and \( 7\sqrt{2} \) are like surds.

(iv) \( 19\sqrt{12}, 6\sqrt{3} \)
\( 19\sqrt{12} = 19 \times \sqrt{4 \times 3} \)
\( = 19 \times \sqrt{4} \times \sqrt{3} \)
\( = 19 \times 2\sqrt{3} \)
\( = 38\sqrt{3} \)
Here, the order of \( 38\sqrt{3} \) and \( 6\sqrt{3} \) is same and their radicands are also same.
\( \implies 19\sqrt{12} \) and \( 6\sqrt{3} \) are like surds.

(v) \( 5\sqrt{22}, 7\sqrt{33} \)
These surds cannot be simplified further. Here, the order of both surds is same but their radicands are not.
\( \implies 5\sqrt{22} \) and \( 7\sqrt{33} \) are unlike surds.

(vi) \( 5\sqrt{5}, \sqrt{75} \)
\( \sqrt{75} = \sqrt{25 \times 3} \)
\( = 5\sqrt{3} \)
Here, the order of \( 5\sqrt{5} \) and \( 5\sqrt{3} \) is same but their radicands are not.
\( \implies 5\sqrt{5} \) and \( \sqrt{75} \) are unlike surds.
In simple words: To find if two surds are alike, we first simplify them as much as possible. If they have the same root number (radicand) and the same root index (order), they are like surds; otherwise, they are unlike surds.

🎯 Exam Tip: Always simplify the surds to their simplest form before comparing their radicands. Clearly state whether the order and radicands are equal to justify your classification.

Question 4. Simplify the following surds.
i. \( \sqrt{27} \)
ii. \( \sqrt{50} \)
iii. \( \sqrt{250} \)
iv. \( \sqrt{112} \)
v. \( \sqrt{168} \)
Answer:
i. \( \sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3} \)
ii. \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \)
iii. \( \sqrt{250} = \sqrt{25 \times 10} = \sqrt{25} \times \sqrt{10} = 5\sqrt{10} \)
iv. \( \sqrt{112} = \sqrt{16 \times 7} = \sqrt{16} \times \sqrt{7} = 4\sqrt{7} \)
v. \( \sqrt{168} = \sqrt{4 \times 42} = \sqrt{4} \times \sqrt{42} = 2\sqrt{42} \)
This simplification helps in performing arithmetic operations like addition and subtraction on surds more easily.
In simple words: To simplify a surd, we find the largest perfect square factor of the number inside the root, take its square root outside, and leave the remaining non-square factor inside.

🎯 Exam Tip: Always look for the largest perfect square factor (like 9, 16, 25) to simplify the surd completely in one step.

Question 5. Compare the following pair of surds.
i. \( 7\sqrt{2}, 5\sqrt{3} \)
ii. \( \sqrt{247}, \sqrt{274} \)
iii. \( 2\sqrt{7}, \sqrt{28} \)
iv. \( 5\sqrt{5}, 7\sqrt{2} \)
v. \( 4\sqrt{42}, 9\sqrt{2} \)
vi. \( 5\sqrt{3}, 9 \)
vii. \( 7, 2\sqrt{5} \)
Answer:
i. \( 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98} \)
\( 5\sqrt{3} = \sqrt{25 \times 3} = \sqrt{75} \)
Since \( 98 > 75 \),
\( \implies \sqrt{98} > \sqrt{75} \)
\( \implies 7\sqrt{2} > 5\sqrt{3} \)

ii. Since \( 247 < 274 \),
\( \implies \sqrt{247} < \sqrt{274} \)

iii. \( 2\sqrt{7} = \sqrt{4 \times 7} = \sqrt{28} \)
Since \( 28 = 28 \),
\( \implies \sqrt{28} = \sqrt{28} \)
\( \implies 2\sqrt{7} = \sqrt{28} \)

iv. \( 5\sqrt{5} = \sqrt{25 \times 5} = \sqrt{125} \)
\( 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98} \)
Since \( 125 > 98 \),
\( \implies \sqrt{125} > \sqrt{98} \)
\( \implies 5\sqrt{5} > 7\sqrt{2} \)

v. \( 4\sqrt{42} = \sqrt{16 \times 42} = \sqrt{672} \)
\( 9\sqrt{2} = \sqrt{81 \times 2} = \sqrt{162} \)
Since \( 672 > 162 \),
\( \implies \sqrt{672} > \sqrt{162} \)
\( \implies 4\sqrt{42} > 9\sqrt{2} \)

vi. \( 5\sqrt{3} = \sqrt{25 \times 3} = \sqrt{75} \)
\( 9 = \sqrt{81} \)
Since \( 75 < 81 \),
\( \implies \sqrt{75} < \sqrt{81} \)
\( \implies 5\sqrt{3} < 9 \)

vii. \( 7 = \sqrt{49} \)
\( 2\sqrt{5} = \sqrt{4 \times 5} = \sqrt{20} \)
Since \( 49 > 20 \),
\( \implies \sqrt{49} > \sqrt{20} \)
\( \implies 7 > 2\sqrt{5} \)
Comparing surds is a fundamental step in arranging real numbers in ascending or descending order.
In simple words: To compare two surds, we square both numbers or write them completely inside the square root. Then, we simply compare the numbers inside the roots to see which one is larger.

🎯 Exam Tip: Always convert both numbers into pure surds (entirely under the radical sign) before comparing their values to avoid calculation errors.

Question 1. Compare the following pairs of surds:
(i) \( 7\sqrt{2}, 5\sqrt{3} \)
(ii) \( \sqrt{247}, \sqrt{274} \)
(iii) \( 2\sqrt{7}, \sqrt{28} \)
(iv) \( 5\sqrt{5}, 7\sqrt{2} \)
Answer:
(i) \( 7\sqrt{2} = \sqrt{49} \times \sqrt{2} = \sqrt{98} \)
\( 5\sqrt{3} = \sqrt{25} \times \sqrt{3} = \sqrt{75} \)
Since, \( 98 > 75 \)

\( \therefore \sqrt{98} > \sqrt{75} \)

\( \therefore 7\sqrt{2} > 5\sqrt{3} \)

(ii) Since, \( 247 < 274 \)

\( \dots \sqrt{247} < \sqrt{274} \)

(iii) \( 2\sqrt{7} = \sqrt{4} \times \sqrt{7} = \sqrt{28} \)
Since, \( 28 = 28 \)

\( \therefore \sqrt{28} = \sqrt{28} \)

\( \dots 2\sqrt{7} = \sqrt{28} \)

(iv) \( 5\sqrt{5} = \sqrt{25} \times \sqrt{5} = \sqrt{125} \)
\( 7\sqrt{2} = \sqrt{49} \times \sqrt{2} = \sqrt{98} \)
Since, \( 125 > 98 \)

\( \therefore \sqrt{125} > \sqrt{98} \)

\( \therefore 5\sqrt{5} > 7\sqrt{2} \)
In simple words: To compare surds, we square the number outside the root and multiply it with the number inside to make it a single square root. Then, we simply compare the numbers under the square root sign to see which one is larger.

🎯 Exam Tip: Always convert mixed surds into pure surds before comparing. Clearly show the step where you compare the radicands (the numbers inside the square root) to score full marks.

Question 5. Compare the following pairs of surds.
(v) \( 4\sqrt{42}, 9\sqrt{2} \)
(vi) \( 5\sqrt{3}, 9 \)
(vii) \( 7, 2\sqrt{5} \)
Answer:
(v) \( 4\sqrt{42} = \sqrt{16} \times \sqrt{42} = \sqrt{672} \)
\( 9\sqrt{2} = \sqrt{81} \times \sqrt{2} = \sqrt{162} \)
Since, \( 672 > 162 \)
\( \implies \sqrt{672} > \sqrt{162} \)
\( \implies 4\sqrt{42} > 9\sqrt{2} \)

(vi) \( 5\sqrt{3} = \sqrt{25} \times \sqrt{3} = \sqrt{75} \)
\( 9 = \sqrt{81} \)
Since, \( 75 < 81 \)
\( \implies \sqrt{75} < \sqrt{81} \)
\( \implies 5\sqrt{3} < 9 \)

(vii) \( 7 = \sqrt{49} \)
\( 2\sqrt{5} = \sqrt{4} \times \sqrt{5} = \sqrt{20} \)
Since, \( 49 > 20 \)
\( \implies \sqrt{49} > \sqrt{20} \)
\( \implies 7 > 2\sqrt{5} \)
Comparing surds becomes much easier when we convert them into pure surds under a single radical sign.
In simple words: To compare two surds, we write them completely inside the square root by squaring the outside number. The surd with the larger number inside the square root has the greater value.

🎯 Exam Tip: To compare surds quickly, convert mixed surds into pure surds by taking the outside coefficient inside the square root as its square.

Question 6. Simplify.
(i) \( 5\sqrt{3} + 8\sqrt{3} \)
(ii) \( 9\sqrt{5} - 4\sqrt{5} + \sqrt{125} \)
(iii) \( 7\sqrt{48} - \sqrt{27} - \sqrt{3} \)
(iv) \( \sqrt{7} - \frac{3}{5}\sqrt{7} + 2\sqrt{7} \)
Answer:
(i) \( 5\sqrt{3} + 8\sqrt{3} \)
\( = (5 + 8)\sqrt{3} \)
\( = 13\sqrt{3} \)

(ii) \( 9\sqrt{5} - 4\sqrt{5} + \sqrt{125} \)
\( = 9\sqrt{5} - 4\sqrt{5} + \sqrt{25 \times 5} \)
\( = 9\sqrt{5} - 4\sqrt{5} + 5\sqrt{5} \)
\( = (9 - 4 + 5)\sqrt{5} \)
\( = 10\sqrt{5} \)

(iii) \( 7\sqrt{48} - \sqrt{27} - \sqrt{3} \)
\( = 7\sqrt{16 \times 3} - \sqrt{9 \times 3} - \sqrt{3} \)
\( = 7(4\sqrt{3}) - 3\sqrt{3} - \sqrt{3} \)
\( = 28\sqrt{3} - 3\sqrt{3} - \sqrt{3} \)
\( = (28 - 3 - 1)\sqrt{3} \)
\( = 24\sqrt{3} \)

(iv) \( \sqrt{7} - \frac{3}{5}\sqrt{7} + 2\sqrt{7} \)
\( = \left(1 - \frac{3}{5} + 2\right)\sqrt{7} \)
\( = \left(3 - \frac{3}{5}\right)\sqrt{7} \)
\( = \left(\frac{15 - 3}{5}\right)\sqrt{7} \)
\( = \frac{12}{5}\sqrt{7} \)
Simplifying surds is highly similar to combining like terms in algebraic expressions.
In simple words: We can add or subtract surds only when they are like surds, meaning they have the same number inside the square root. We simplify the numbers inside first, then add or subtract their coefficients.

🎯 Exam Tip: Always simplify the numbers inside the square roots to their simplest form first to find like surds before adding or subtracting.

Question 6. Simplify the following surds:
(i) \( 5\sqrt{3} + 8\sqrt{3} \)
(ii) \( 9\sqrt{5} - 4\sqrt{5} + \sqrt{125} \)
(iii) \( 7\sqrt{48} - \sqrt{27} - \sqrt{3} \)
(iv) \( \sqrt{7} - \frac{3}{5}\sqrt{7} + 2\sqrt{7} \)
Answer:
(i) \( 5\sqrt{3} + 8\sqrt{3} \)
\( = (5+8)\sqrt{3} \)
\( = 13\sqrt{3} \)
\( \therefore 5\sqrt{3} + 8\sqrt{3} = 13\sqrt{3} \)

(ii) \( 9\sqrt{5} - 4\sqrt{5} + \sqrt{125} \)
\( = 9\sqrt{5} - 4\sqrt{5} + \sqrt{25 \times 5} \)
\( = 9\sqrt{5} - 4\sqrt{5} + \sqrt{25} \times \sqrt{5} \)
\( = 9\sqrt{5} - 4\sqrt{5} + 5\sqrt{5} \)
\( = (9 - 4 + 5)\sqrt{5} \)
\( = 10\sqrt{5} \)
\( \therefore 9\sqrt{5} - 4\sqrt{5} + \sqrt{125} = 10\sqrt{5} \)

(iii) \( 7\sqrt{48} - \sqrt{27} - \sqrt{3} \)
\( = 7\sqrt{16 \times 3} - \sqrt{9 \times 3} - \sqrt{3} \)
\( = 7 \times \sqrt{16} \times \sqrt{3} - \sqrt{9} \times \sqrt{3} - \sqrt{3} \)
\( = 7 \times 4\sqrt{3} - 3\sqrt{3} - \sqrt{3} \)
\( = 28\sqrt{3} - 3\sqrt{3} - \sqrt{3} \)
\( = (28 - 3 - 1)\sqrt{3} \)
\( = 24\sqrt{3} \)
\( \therefore 7\sqrt{48} - \sqrt{27} - \sqrt{3} = 24\sqrt{3} \)

(iv) \( \sqrt{7} - \frac{3}{5}\sqrt{7} + 2\sqrt{7} \)
\( = \left(1 - \frac{3}{5} + 2\right)\sqrt{7} \)
\( = \left(3 - \frac{3}{5}\right)\sqrt{7} \)
\( = \left(\frac{15 - 3}{5}\right)\sqrt{7} \)
\( = \frac{12\sqrt{7}}{5} \)
\( \therefore \sqrt{7} - \frac{3}{5}\sqrt{7} + 2\sqrt{7} = \frac{12\sqrt{7}}{5} \)
In simple words: To add or subtract surds, we first simplify each term so they have the same number inside the square root, and then we combine their outside coefficients just like normal numbers.

🎯 Exam Tip: Always look for perfect square factors (like 9, 16, 25) inside the square root to simplify the surds before performing addition or subtraction.

Question 7. Multiply and write the answer in the simplest form.
(i) \( 3\sqrt{12} \times \sqrt{18} \)
(ii) \( 3\sqrt{12} \times 7\sqrt{15} \)
(iii) \( 3\sqrt{8} \times \sqrt{5} \)
(iv) \( 5\sqrt{8} \times 2\sqrt{8} \)
Answer:
(i) \( 3\sqrt{12} \times \sqrt{18} \)
\( = 3\sqrt{4 \times 3} \times \sqrt{9 \times 2} \)
\( = 3 \times 2\sqrt{3} \times 3\sqrt{2} \)
\( = 6\sqrt{3} \times 3\sqrt{2} \)
\( = (6 \times 3) \times \sqrt{3 \times 2} \)
\( = 18\sqrt{6} \)

(ii) \( 3\sqrt{12} \times 7\sqrt{15} \)
\( = 3\sqrt{4 \times 3} \times 7\sqrt{15} \)
\( = 3 \times 2\sqrt{3} \times 7\sqrt{15} \)
\( = 6\sqrt{3} \times 7\sqrt{15} \)
\( = (6 \times 7) \times \sqrt{3 \times 15} \)
\( = 42\sqrt{45} \)
\( = 42\sqrt{9 \times 5} \)
\( = 42 \times 3\sqrt{5} \)
\( = 126\sqrt{5} \)

(iii) \( 3\sqrt{8} \times \sqrt{5} \)
\( = 3\sqrt{4 \times 2} \times \sqrt{5} \)
\( = 3 \times 2\sqrt{2} \times \sqrt{5} \)
\( = 6\sqrt{2} \times \sqrt{5} \)
\( = 6\sqrt{2 \times 5} \)
\( = 6\sqrt{10} \)

(iv) \( 5\sqrt{8} \times 2\sqrt{8} \)
\( = (5 \times 2) \times (\sqrt{8} \times \sqrt{8}) \)
\( = 10 \times 8 \)
\( = 80 \)
In simple words: To multiply surds, multiply the numbers outside the square roots together, and multiply the numbers inside the square roots together, then simplify the final radical.

🎯 Exam Tip: Remember that multiplying a square root by itself, like \( \sqrt{a} \times \sqrt{a} \), simply equals \( a \), which makes calculations much faster.

Question 7. Multiply and write the answer in the simplest form.
(i) \( 3\sqrt{12} \times \sqrt{18} \)
(ii) \( 3\sqrt{12} \times 7\sqrt{15} \)
(iii) \( 3\sqrt{8} \times \sqrt{5} \)
(iv) \( 5\sqrt{8} \times 2\sqrt{8} \)
Answer:
(i) \( 3\sqrt{12} \times \sqrt{18} = 3 \times \sqrt{4 \times 3} \times \sqrt{9 \times 2} \)
\( = 3 \times 2\sqrt{3} \times 3\sqrt{2} \)
\( = 3 \times 2 \times 3 \times \sqrt{3} \times \sqrt{2} \)
\( = 18\sqrt{6} \)
\( \therefore 3\sqrt{12} \times \sqrt{18} = 18\sqrt{6} \)

(ii) \( 3\sqrt{12} \times 7\sqrt{15} = 3 \times \sqrt{4 \times 3} \times 7 \times \sqrt{5 \times 3} \)
\( = 3 \times 2\sqrt{3} \times 7\sqrt{5} \times \sqrt{3} \)
\( = 3 \times 2 \times 7 \times \sqrt{3} \times \sqrt{3} \times \sqrt{5} \)
\( = 42 \times 3 \times \sqrt{5} \)
\( = 126\sqrt{5} \)
\( \therefore 3\sqrt{12} \times 7\sqrt{15} = 126\sqrt{5} \)

(iii) \( 3\sqrt{8} \times \sqrt{5} = 3 \times \sqrt{4 \times 2} \times \sqrt{5} \)
\( = 3 \times 2\sqrt{2} \times \sqrt{5} \)
\( = 6\sqrt{10} \)
\( \therefore 3\sqrt{8} \times \sqrt{5} = 6\sqrt{10} \)

(iv) \( 5\sqrt{8} \times 2\sqrt{8} = 5 \times 2 \times \sqrt{8} \times \sqrt{8} \)
\( = 5 \times 2 \times 8 \)
\( = 80 \)
\( \therefore 5\sqrt{8} \times 2\sqrt{8} = 80 \)
Simplifying surds before multiplying makes the calculation much easier and prevents dealing with very large numbers under the radical sign.
In simple words: To multiply square root numbers, first simplify each square root by pulling out perfect squares. Then, multiply the outside numbers together and the inside numbers together to get your final simplified answer.

🎯 Exam Tip: Always look for perfect square factors (like 4, 9, 16, 25) inside the square root first to simplify the surds before performing multiplication.

Question 8. Divide and write form.
(i) \( \sqrt{98} \div \sqrt{2} \)
(ii) \( \sqrt{125} \div \sqrt{50} \)
(iii) \( \sqrt{54} \div \sqrt{27} \)
(iv) \( \sqrt{310} \div \sqrt{5} \)
Answer:
(i) \( \sqrt{98} \div \sqrt{2} = \sqrt{\frac{98}{2}} \)
\( = \sqrt{49} \)
\( = 7 \)

(ii) \( \sqrt{125} \div \sqrt{50} = \sqrt{\frac{125}{50}} \)
\( = \sqrt{\frac{25 \times 5}{25 \times 2}} \)
\( = \sqrt{\frac{5}{2}} \)

(iii) \( \sqrt{54} \div \sqrt{27} = \sqrt{\frac{54}{27}} \)
\( = \sqrt{2} \)

(iv) \( \sqrt{310} \div \sqrt{5} = \sqrt{\frac{310}{5}} \)
\( = \sqrt{62} \)
When dividing surds of the same order, we can combine them under a single radical sign to simplify the fraction directly.
In simple words: To divide square roots, you can put both numbers under one single square root sign, divide them like a normal fraction, and then simplify the final root.

🎯 Exam Tip: Remember that \( \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \). Combining them under a single square root often allows you to simplify the fraction easily before taking the root.

Question 8. Simplify the following:
(i) \( \frac{\sqrt{98}}{\sqrt{2}} \)
(ii) \( \frac{\sqrt{125}}{\sqrt{50}} \)
(iii) \( \frac{\sqrt{54}}{\sqrt{27}} \)
(iv) \( \frac{\sqrt{310}}{\sqrt{5}} \)
Answer:
(i) \( \frac{\sqrt{98}}{\sqrt{2}} = \sqrt{\frac{98}{2}} = \sqrt{49} = 7 \)
(ii) \( \frac{\sqrt{125}}{\sqrt{50}} = \sqrt{\frac{125}{50}} = \sqrt{\frac{25 \times 5}{25 \times 2}} = \sqrt{\frac{5}{2}} \)
(iii) \( \frac{\sqrt{54}}{\sqrt{27}} = \sqrt{\frac{54}{27}} = \sqrt{2} \)
(iv) \( \frac{\sqrt{310}}{\sqrt{5}} = \sqrt{\frac{310}{5}} = \sqrt{\frac{5 \times 62}{5}} = \sqrt{62} \)
These simplifications are performed by dividing the radicands under a single radical sign.
In simple words: To divide square roots, you can put both numbers under one square root sign, simplify the fraction inside, and then find the square root of the result.

🎯 Exam Tip: Always look for common factors in the numerator and denominator under the radical to simplify the expression before taking the square root.

Question 9. Rationalize the denominator.
(i) \( \frac{3}{\sqrt{5}} \)
(ii) \( \frac{1}{\sqrt{14}} \)
(iii) \( \frac{5}{\sqrt{7}} \)
(iv) \( \frac{6}{9\sqrt{3}} \)
(v) \( \frac{11}{\sqrt{3}} \)
Answer:
(i) \( \frac{3}{\sqrt{5}} = \frac{3 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{3\sqrt{5}}{5} \)
(ii) \( \frac{1}{\sqrt{14}} = \frac{1 \times \sqrt{14}}{\sqrt{14} \times \sqrt{14}} = \frac{\sqrt{14}}{14} \)
(iii) \( \frac{5}{\sqrt{7}} = \frac{5 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{5\sqrt{7}}{7} \)
(iv) \( \frac{6}{9\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2 \times \sqrt{3}}{3\sqrt{3} \times \sqrt{3}} = \frac{2\sqrt{3}}{3 \times 3} = \frac{2\sqrt{3}}{9} \)
(v) \( \frac{11}{\sqrt{3}} = \frac{11 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{11\sqrt{3}}{3} \)
Rationalizing the denominator helps in simplifying algebraic fractions by eliminating irrational numbers from the divisor.
In simple words: To rationalize a denominator, multiply both the top and the bottom of the fraction by the square root that is in the denominator so that the bottom becomes a whole number.

🎯 Exam Tip: Remember to simplify any common factors between the numerator and the rationalized denominator to get the final answer in its simplest form.

Question 1. Rationalize the denominator of the following:
(i) \( \frac{3}{\sqrt{5}} \)
(ii) \( \frac{1}{\sqrt{14}} \)
(iii) \( \frac{5}{\sqrt{7}} \)
Answer:
(i) \( \frac{3}{\sqrt{5}} \)
\( = \frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \) ...[Multiplying the numerator and denominator by \( \sqrt{5} \)]
\( = \frac{3 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{3\sqrt{5}}{5} \)
\( \therefore \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \)

(ii) \( \frac{1}{\sqrt{14}} \)
\( = \frac{1}{\sqrt{14}} \times \frac{\sqrt{14}}{\sqrt{14}} \) ...[Multiplying the numerator and denominator by \( \sqrt{14} \)]
\( = \frac{1 \times \sqrt{14}}{\sqrt{14} \times \sqrt{14}} = \frac{\sqrt{14}}{14} \)
\( \therefore \frac{1}{\sqrt{14}} = \frac{\sqrt{14}}{14} \)

(iii) \( \frac{5}{\sqrt{7}} \)
\( = \frac{5}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} \) ...[Multiplying the numerator and denominator by \( \sqrt{7} \)]
\( = \frac{5 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{5\sqrt{7}}{7} \)
\( \therefore \frac{5}{\sqrt{7}} = \frac{5\sqrt{7}}{7} \)
This mathematical process helps simplify expressions by removing radical signs from the denominator.
In simple words: To rationalize a denominator, we multiply both the top and bottom of the fraction by the square root that is in the denominator. This turns the bottom number into a regular whole number without changing the value of the fraction.

🎯 Exam Tip: Always multiply both the numerator and denominator by the exact same radical term to keep the fraction balanced, and simplify the final fraction if possible.

Question iv. Simplify: \( \frac{6}{9\sqrt{3}} \)
Answer:
\( \frac{6}{9\sqrt{3}} = \frac{6}{9\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) [Multiplying the numerator and denominator by \( \sqrt{3} \)]
\( = \frac{6 \times \sqrt{3}}{9\sqrt{3} \times \sqrt{3}} \)
\( = \frac{6\sqrt{3}}{9 \times 3} = \frac{2\sqrt{3}}{9} \)

\( \therefore \frac{6}{9\sqrt{3}} = \frac{2\sqrt{3}}{9} \)
In simple words: To remove the square root from the bottom of the fraction, we multiply both the top and bottom by \( \sqrt{3} \) and then simplify the numbers.

🎯 Exam Tip: Always simplify the final fraction to its lowest terms to ensure you get full marks.

Question v. Simplify: \( \frac{11}{\sqrt{3}} \)
Answer:
\( \frac{11}{\sqrt{3}} = \frac{11}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) [Multiplying the numerator and denominator by \( \sqrt{3} \)]
\( = \frac{11 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{11\sqrt{3}}{3} \)

\( \therefore \frac{11}{\sqrt{3}} = \frac{11\sqrt{3}}{3} \)
In simple words: We multiply the top and bottom by \( \sqrt{3} \) to make the denominator a rational number without any square root.

🎯 Exam Tip: Rationalizing the denominator is a standard step in algebra; make sure to show the multiplication step clearly.

Question 1. \( \sqrt{9 + 16} \text{ ? } \sqrt{9} + \sqrt{16} \) (Textbook pg. no. 28)
Answer:
\( \sqrt{9+16} = \sqrt{25} = 5 \)
\( \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \)

\( \therefore \sqrt{9+16} \neq \sqrt{9} + \sqrt{16} \)
In simple words: Adding numbers inside a square root first gives a different result than taking their square roots individually and then adding them.

🎯 Exam Tip: Remember that square roots do not distribute over addition, so \( \sqrt{a+b} \) is not equal to \( \sqrt{a} + \sqrt{b} \).

Question 2. \( \sqrt{100 + 36} \text{ ? } \sqrt{100} + \sqrt{36} \) (Textbook pg. no. 28)
Answer:
\( \sqrt{100+36} = \sqrt{136} = 2\sqrt{34} \)
\( \sqrt{100} + \sqrt{36} = 10 + 6 = 16 \)

\( \therefore \sqrt{100+36} \neq \sqrt{100} + \sqrt{36} \)
From the above examples, we can conclude that:
\( \sqrt{a+b} \neq \sqrt{a} + \sqrt{b} \)
In simple words: Just like the previous example, this shows that you cannot split a square root across an addition sign.

🎯 Exam Tip: State the general rule \( \sqrt{a+b} \neq \sqrt{a} + \sqrt{b} \) clearly at the end of your proof to show a complete understanding.

Question 3. Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)
Answer: To complete the chart on page 34, follow each mathematical operation (such as addition, subtraction, multiplication, or division of surds) in the direction of the arrows step-by-step to find the final simplified values.
In simple words: Follow the arrows in your textbook chart and perform the math operations one by one to fill in the empty boxes.

🎯 Exam Tip: Carefully perform operations on surds, keeping like terms together when adding or subtracting.

Question (II). Complete the mathematical flowchart by performing the operations indicated along the arrows:
• Start with \( \sqrt{3} \)
• Multiply by 2 to get \( 2\sqrt{3} \)
• Multiply by \( \sqrt{8} \) to get \( 4\sqrt{6} \)
• Add \( 10\sqrt{6} \) to get [ Box 1 ]
• Divide by \( \sqrt{3} \) to get [ Box 2 ]
• Multiply by \( \sqrt{2} \) to get [ Box 3 ]
• Add 3 to get 31
• Multiply by \( \sqrt{5} \) to get [ Box 4 ]
• Multiply by \( 3\sqrt{5} \) to get 465
• Divide by 5 to get [ Box 5 ] (End)
Answer:
The completed values for the boxes in the flowchart are calculated as follows:
• Box 1: \( 4\sqrt{6} + 10\sqrt{6} = 14\sqrt{6} \)
• Box 2: \( 14\sqrt{6} \div \sqrt{3} = 14\sqrt{2} \)
• Box 3: \( 14\sqrt{2} \times \sqrt{2} = 14 \times 2 = 28 \)
• Box 4: \( 31 \times \sqrt{5} = 31\sqrt{5} \)
• Box 5 (End): \( 465 \div 5 = 93 \)
In simple words: Follow each arrow step-by-step, performing basic operations like addition, multiplication, and division on surds to fill in the missing numbers.

🎯 Exam Tip: When multiplying surds like \( \sqrt{2} \times \sqrt{2} \), remember it simplifies to the rational number 2, which makes further calculations much easier.

Question 4. There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)
(The numbers on the card sheet are: \( 12 \), \( 2\sqrt{5} \), \( -\frac{3}{4} \), \( -9\sqrt{5} \), \( 9\sqrt{2} \), \( -11 \), \( 3\sqrt{11} \), \( 5\sqrt{7} \), \( -3\sqrt{2} \))
Answer:
Using the real numbers from the card sheet, we can construct and solve the following examples:

1. Addition Examples:
(i) \( 9\sqrt{2} + (-3\sqrt{2}) = 9\sqrt{2} - 3\sqrt{2} = 6\sqrt{2} \)
(ii) \( 2\sqrt{5} + (-9\sqrt{5}) = 2\sqrt{5} - 9\sqrt{5} = -7\sqrt{5} \)

2. Subtraction Examples:
(i) \( 9\sqrt{2} - (-3\sqrt{2}) = 9\sqrt{2} + 3\sqrt{2} = 12\sqrt{2} \)
(ii) \( 2\sqrt{5} - (-9\sqrt{5}) = 2\sqrt{5} + 9\sqrt{5} = 11\sqrt{5} \)

3. Multiplication Examples:
(i) \( 9\sqrt{2} \times (-3\sqrt{2}) = [9 \times (-3)] \times [\sqrt{2} \times \sqrt{2}] = -27 \times 2 = -54 \)
(ii) \( 2\sqrt{5} \times (-9\sqrt{5}) = [2 \times (-9)] \times [\sqrt{5} \times \sqrt{5}] = -18 \times 5 = -90 \)

4. Division Examples:
(i) \( 9\sqrt{2} \div (-3\sqrt{2}) = \frac{9\sqrt{2}}{-3\sqrt{2}} = \frac{9}{-3} = -3 \)
(ii) \( -9\sqrt{5} \div 2\sqrt{5} = \frac{-9\sqrt{5}}{2\sqrt{5}} = -\frac{9}{2} \)
In simple words: We choose pairs of similar surds (like those with \( \sqrt{2} \) or \( \sqrt{5} \)) from the cards to make adding, subtracting, multiplying, and dividing them straightforward and easy to solve.

🎯 Exam Tip: To make calculations simple, always choose like surds (surds with the same radicand) when constructing your own examples for addition and subtraction.

Question. Simplify the following expressions:
(i) \( 9\sqrt{2} + (-3\sqrt{2}) \)
(ii) \( 12 - 2\sqrt{5} \)
(iii) \( 2\sqrt{5} \times 3\sqrt{11} \)
(iv) \( \frac{2\sqrt{5}}{9\sqrt{2}} \)
Answer:
(i) \( 9\sqrt{2} + (-3\sqrt{2}) = 9\sqrt{2} - 3\sqrt{2} = 6\sqrt{2} \)
(ii) \( 12 - 2\sqrt{5} = 2(6 - \sqrt{5}) \)
(iii) \( 2\sqrt{5} \times 3\sqrt{11} = 6\sqrt{55} \)
(iv) \( \frac{2\sqrt{5}}{9\sqrt{2}} = \frac{2\sqrt{5}}{9\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{10}}{9 \times 2} = \frac{\sqrt{10}}{9} \)
In simple words: To simplify these expressions, we combine similar square roots, factor out common numbers, multiply the outside numbers and inside numbers separately, or multiply the top and bottom by the square root to clear it from the bottom.

🎯 Exam Tip: When rationalizing the denominator as in sub-part (iv), always multiply both the numerator and denominator by the radical term to eliminate the square root from the bottom of the fraction.