Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 1 Irrational Numbers

Exercise 1.1

Question 1A. \( \frac{3}{5} \)
Answer:
\( \frac{3}{5} \)
\( 5 = 1 \times 5 = 2^0 \times 5^1 \)
i.e., 5 can be expressed as \( 2^m \times 5^n \).
\( \therefore \frac{3}{5} \) has terminating decimal representation.
In simple words: If a fraction's bottom number only has factors of 2 or 5, the decimal will eventually stop. Since the bottom is 5, it ends neatly.

📝 Teacher's Note: This rule is a quick shortcut to predict decimal behavior without doing long division. Ensure the fraction is simplified first.

🎯 Exam Tip: In your calculation, show the prime factorization of the denominator to prove why it terminates.

Question 1B. \( \frac{5}{7} \)
Answer:
\( \frac{5}{7} \)
\( 7 = 1 \times 7 \)
i.e., 7 cannot be expressed as \( 2^m \times 5^n \).
\( \therefore \frac{5}{7} \) does not have terminating decimal representation.
In simple words: Because the bottom number 7 is not made of 2s or 5s, the decimal will go on forever in a repeating pattern.

📝 Teacher's Note: If any prime factor other than 2 or 5 appears in the denominator, the decimal will always be non-terminating and recurring.

🎯 Exam Tip: Simply stating "it doesn't have 2 or 5" isn't enough; clearly show the factors to get full marks.

Question 1C. \( \frac{25}{49} \)
Answer:
\( \frac{25}{49} \)
\( 49 = 7 \times 7 \)
i.e., 49 cannot be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{25}{49} \) does not have terminating decimal representation.
In simple words: 49 is made of 7s. Since it's not made of 2s or 5s, the division never quite finishes.

📝 Teacher's Note: Even if the top number is a perfect square, we only care about the factors of the bottom number.

🎯 Exam Tip: Always double-check if the fraction can be simplified before looking at the denominator's factors.

Question 1D. \( \frac{37}{40} \)
Answer:
\( \frac{37}{40} \)
\( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1 \)
i.e., 40 can be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{37}{40} \) has terminating decimal representation.
In simple words: 40 is built entirely from 2s and 5s. This means the math is "friendly" and the decimal will stop.

📝 Teacher's Note: 40 is a "round" denominator in binary and decimal bases, making it a reliable terminating fraction.

🎯 Exam Tip: Present the prime factorization in exponent form (\( 2^3 \times 5^1 \)) for a more professional-looking answer.

Question 1E. \( \frac{57}{64} \)
Answer:
\( \frac{57}{64} \)
\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 \times 5^0 \)
i.e., 64 can be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{57}{64} \) has terminating decimal representation.
In simple words: 64 is just a bunch of 2s multiplied together. Since it fits the 2-and-5 rule, the decimal will eventually end.

📝 Teacher's Note: Any denominator that is a power of 2 (like 2, 4, 8, 16, 32, 64) will always produce a terminating decimal.

🎯 Exam Tip: Don't be intimidated by large denominators; prime factorize them systematically to see the pattern.

Question 1F. \( \frac{59}{75} \)
Answer:
\( \frac{59}{75} \)
\( 75 = 5 \times 5 \times 3 = 5^2 \times 3^1 \)
i.e., 75 cannot be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{59}{75} \) does not have terminating decimal representation.
In simple words: Although 75 has 5s in it, that extra factor of 3 ruins the "ending" and makes the decimal repeat forever.

📝 Teacher's Note: The presence of even one prime factor other than 2 or 5 is enough to make the fraction non-terminating.

🎯 Exam Tip: Identify the specific "offending" factor (in this case, 3) in your explanation.

Question 1G. \( \frac{89}{125} \)
Answer:
\( \frac{89}{125} \)
\( 125 = 5 \times 5 \times 5 = 2^0 \times 5^3 \)
i.e., 125 can be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{89}{125} \) has terminating decimal representation.
In simple words: 125 is made only of 5s. This satisfies the rule, so the decimal will stop.

📝 Teacher's Note: Denominators that are powers of 5 also guarantee a terminating decimal, just like powers of 2.

🎯 Exam Tip: Mention that \( 2^0 = 1 \) to explain why it still fits the \( 2^m \times 5^n \) format.

Question 1H. \( \frac{125}{213} \)
Answer:
\( \frac{125}{213} \)
\( 213 = 3 \times 71 \)
i.e., 213 cannot be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{125}{213} \) does not have terminating decimal representation.
In simple words: The factors 3 and 71 at the bottom mean this fraction will never stop when turned into a decimal.

📝 Teacher's Note: Prime factorizing larger numbers like 213 takes patience. Remind students of divisibility rules (like \( 2+1+3=6 \), so it's divisible by 3).

🎯 Exam Tip: Check for factors like 3 or 7 early on to save time on large denominators.

Question 1. \( \frac{147}{160} \)
Answer:
\( \frac{147}{160} \)
\( 160 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 = 2^5 \times 5^1 \)
i.e., 160 can be expressed as \( 2^m \times 5^n \).
Hence, \( \frac{147}{160} \) has terminating decimal representation.
In simple words: 160 is entirely made of 2s and 5s. That's the secret code for a decimal that eventually ends.

📝 Teacher's Note: This is a perfect example of the theorem being applied to a moderately complex number.

🎯 Exam Tip: Be systematic in your prime factorization: 160 -> 16 x 10 -> \( (2^4) \times (2 \times 5) \).

Question 2A. \( 0.93 \)
Answer:
\( 0.93 = \frac{93}{100} \)
In simple words: To turn a decimal into a fraction, count the digits after the point and put it over 100.

📝 Teacher's Note: Place the decimal part over 1 followed by as many zeros as there are decimal places.

🎯 Exam Tip: Check if the resulting fraction can be simplified further; in this case, 93 and 100 share no common factors.

Question 2B. \( 4.56 \)
Answer:
\( 4.56 = \frac{456}{100} \)

\( \implies \frac{456 \div 4}{100 \div 4} = \frac{114}{25} \)
In simple words: Move the decimal and put it over 100, then divide both top and bottom by 4 to make the fraction smaller.

📝 Teacher's Note: Always provide the fraction in its simplest form (lowest terms).

🎯 Exam Tip: Simplification is usually worth 1 mark, so don't stop at \( \frac{456}{100} \).

Question 2C. \( 0.614 \)
Answer:
\( 0.614 = \frac{614}{1000} \)

\( \implies \frac{614 \div 2}{1000 \div 2} = \frac{307}{500} \)
In simple words: Since there are three numbers after the point, we use 1000 at the bottom and then cut both numbers in half.

📝 Teacher's Note: Identifying the greatest common divisor (GCD) helps simplify fractions in one step.

🎯 Exam Tip: For even numbers at the end, start by dividing by 2 and then check again.

Question 2D. \( 21.025 \)
Answer:
\( 21.025 = \frac{21025}{1000} \)

\( \implies \frac{21025 \div 25}{1000 \div 25} = \frac{841}{40} \)
In simple words: We put the whole number over 1000 and divide by 25 to get the final simple fraction.

📝 Teacher's Note: Numbers ending in 25, 50, 75, or 00 are always divisible by 25.

🎯 Exam Tip: Dividing by larger numbers like 25 or 125 makes simplification much faster than just using 5.

Question 3. Convert the following to decimals:
(i) \( \frac{3}{5} \)
(ii) \( \frac{8}{11} \)
(iii) \( \frac{-2}{7} \)
(iv) \( \frac{12}{21} \)
(v) \( \frac{13}{25} \)
(vi) \( \frac{2}{3} \)
Answer:
(i) \( \frac{3}{5} = 0.6 \)
(ii) \( \frac{8}{11} = 0.72727272... = 0.\overline{72} \)
(iii) \( \frac{-2}{7} = -0.285714285714... = -0.\overline{285714} \)
(iv) \( \frac{12}{21} = 0.571428571428... = 0.\overline{571428} \)
(v) \( \frac{13}{25} = 0.52 \)
(vi) \( \frac{2}{3} = 0.6666... = 0.\overline{6} \)
In simple words: To turn a fraction into a decimal, just divide the top number by the bottom one. If it keeps repeating, put a bar over the part that loops.

📝 Teacher's Note: Teach students to recognize recurring patterns. For example, any fraction with 11 at the bottom will repeat every two digits.

🎯 Exam Tip: Always indicate recurring decimals clearly with a bar or dots to show they are non-terminating.

Question 4A. \( 0.\dot{7} \)
Answer:
Let \( x = 0.\dot{7} \)
Then, \( x = 0.7777... \) ----(1)
Here, the number of digits recurring is only 1, so we multiply both sides of the equation (1) by 10.

\( \implies 10x = 10 \times 0.7777... = 7.7777... \) ----(2)
On subtracting (1) from (2), we get
\( 9x = 7 \)

\( \implies x = \frac{7}{9} \)
\( \therefore 0.\dot{7} = \frac{7}{9} \)
In simple words: We create an equation, multiply it to move the decimal point, and then subtract to "cancel out" the infinite tail.

📝 Teacher's Note: This algebraic method is the only way to convert recurring decimals to exact fractions. The number of zeros in the multiplier (10, 100, 1000) matches the number of digits that repeat.

🎯 Exam Tip: Label your equations (1) and (2) clearly to make your steps easy for the examiner to follow.

Question 4B. \( 0.\overline{35} \)
Answer:
Let \( x = 0.\overline{35} \)
Then, \( x = 0.353535... \) ----(1)
Here, the number of digits recurring is 2, so we multiply both sides of the equation (1) by 100.

\( \implies 100x = 100 \times 0.353535... = 35.3535... \) ----(2)
On subtracting (1) from (2), we get
\( 99x = 35 \)

\( \implies x = \frac{35}{99} \)
\( \therefore 0.\overline{35} = \frac{35}{99} \)
In simple words: Since two digits repeat, we multiply by 100 to jump the point past the first pair, then subtract.

📝 Teacher's Note: Remind students that the goal of subtraction is to have only zeros after the decimal point.

🎯 Exam Tip: If two digits repeat, use 100; if three digits repeat, use 1000. It's a consistent pattern.

Question 4C. \( 0.\overline{89} \)
Answer:
Let \( x = 0.\overline{89} \)
Then, \( x = 0.898989... \) ----(1)
Here, the number of digits recurring is 2, so we multiply both sides of the equation (1) by 100.

\( \implies 100x = 100 \times 0.898989... = 89.8989... \) ----(2)
On subtracting (1) from (2), we get
\( 99x = 89 \)

\( \implies x = \frac{89}{99} \)
\( \therefore 0.\overline{89} = \frac{89}{99} \)
In simple words: This follows the exact same pattern as the previous one—multiply by 100 and subtract.

📝 Teacher's Note: Consistent repetition of this method helps build muscle memory for students.

🎯 Exam Tip: Ensure that the "tail" of the decimals aligns perfectly before you subtract.

Question 4D. \( 0.057 \) (recurring)
Answer:
Let \( x = 0.\overline{057} \)
Then, \( x = 0.057057... \) ----(1)
Here, the number of digits recurring is 3, so we multiply both sides of the equation (1) by 1000.

\( \implies 1000x = 1000 \times 0.057057... = 57.057057... \) ----(2)
On subtracting (1) from (2), we get
\( 999x = 57 \)

\( \implies x = \frac{57}{999} = \frac{19}{333} \)
\( \therefore 0.\overline{057} = \frac{19}{333} \)
In simple words: With three digits repeating, we use 1000. Finally, we divide by 3 to simplify the answer.

📝 Teacher's Note: The 0 at the start of "057" counts as one of the three repeating digits. It's a common student oversight.

🎯 Exam Tip: Simplifying the fraction (\( \frac{57}{999} \) to \( \frac{19}{333} \)) is often required to get the final mark.

Question 4E. \( 0.763 \) (recurring)
Answer:
Let \( x = 0.\overline{763} \)
Then, \( x = 0.763763... \) ----(1)
Here, the number of digits recurring is 3, so we multiply both sides of the equation (1) by 1000.

\( \implies 1000x = 1000 \times 0.763763... = 763.763763... \) ----(2)
On subtracting (1) from (2), we get
\( 999x = 763 \)

\( \implies x = \frac{763}{999} \)
\( \therefore 0.\overline{763} = \frac{763}{999} \)
In simple words: This is a standard three-digit repetition. Use 1000, subtract, and you have your fraction.

📝 Teacher's Note: Emphasize that there is no shortcut here; the algebraic method is the safest way to find the answer.

🎯 Exam Tip: Check if 763 is divisible by 3 (sum of digits = 16, so no) or 9 to see if simplification is possible.

Question 4F. \( 2.\overline{67} \)
Answer:
Let \( x = 2.\overline{67} \)
Then, \( x = 2.676767... \) ----(1)
Here, the number of digits recurring is 2, so we multiply both sides of the equation (1) by 100.

\( \implies 100x = 100 \times 2.676767... = 267.676767... \) ----(2)
On subtracting (1) from (2), we get
\( 99x = 265 \)

\( \implies x = \frac{265}{99} \)
\( \therefore 2.\overline{67} = \frac{265}{99} \)
In simple words: Even with a number like 2 in front, the method works the same. Multiply by 100 to shift the point past one set of repeating digits.

📝 Teacher's Note: The leading digit (2) gets pulled into the multiplier, resulting in 267 on the right side.

🎯 Exam Tip: Be careful with the subtraction: \( 267.67... - 2.67... = 265 \). Many students forget to subtract the leading whole number.

Question 4G. \( 4.6724 \) (where 724 is recurring)
Answer:
Let \( x = 4.6\overline{724} = 4.6724724... \)
Here, only numbers 724 is being repeated, so first we need to remove 6 which precedes 724.
We multiply by 10 so that only the recurring digits remain after decimal.

\( \implies 10x = 46.724724... \) ----(1)
The number of digits recurring in equation (1) is 3, so we multiply both sides of the equation (1) by 1000.

\( \implies 10000x = 1000 \times 46.724724... = 46724.724... \) ----(2)
On subtracting (1) from (2), we get
\( 9990x = 46678 \)

\( \implies x = \frac{46678}{9990} = \frac{23339}{4995} \)
\( \therefore 4.6\overline{724} = \frac{763}{999} = \frac{23339}{4995} \)
In simple words: This one is a bit harder because the 6 doesn't repeat. We move the decimal past the 6 first, then treat it like a regular repeating decimal.

📝 Teacher's Note: This is called a "mixed recurring decimal." It requires two steps: shifting past non-repeating digits and then shifting past one set of repeating digits.

🎯 Exam Tip: The denominator will end with zeros (\( 9990 \)) because of that non-repeating digit (6). This is a good way to double-check your work.

Question 4H. \( 0.01\overline{7} \)
Answer:
Let \( x = 0.01\overline{7} = 0.01717.. \)
Here, only numbers 17 is being repeated, so first we need to remove 0 which proceeds 17.
We multiply by 10 so that only the recurring digits remain after decimal.

\( \implies 10x = 0.1717... \) ----(1)
The number of digits recurring in equation (1) is 2, so we multiply both sides of the equation (1) by 100.

\( \implies 1000x = 100 \times 0.1717... = 17.1717... \) ----(2)
On subtracting (1) from (2), we get
\( 990x = 17 \)

\( \implies x = \frac{17}{990} \)
\( \therefore 0.01\overline{7} = \frac{17}{990} \)
In simple words: Shift past the lone zero first, then handle the repeating 17. The math gives us 17 over 990.

📝 Teacher's Note: Mixed recurring decimals are common in exams. The number of '9's equals repeating digits, and '0's equals non-repeating decimal digits.

🎯 Exam Tip: Don't forget that \( 1000x - 10x = 990x \), not \( 999x \). This is a common subtraction mistake.

Question 4I. \( 17.02\overline{7} \)
Answer:
Let \( x = 17.02\overline{7} = 17.027777.. \)
Here, only number 7 is being repeated, so first we need to remove 02 which proceeds 7.
We multiply by 100 so that only the recurring digits remain after decimal.

\( \implies 100x = 1702.7777... \) ----(1)
The number of digits recurring in equation (1) is 1, so we multiply both sides of the equation (1) by 10.

\( \implies 1000x = 10 \times 1702.7777... = 17027.7774... \) ----(2)
On subtracting (1) from (2), we get
\( 900x = 15325 \)

\( \implies x = \frac{15325}{900} = \frac{613}{36} \)
\( \therefore 17.02\overline{7} = \frac{613}{36} \)
In simple words: This is another mixed type. We jump the decimal past the 02, then past the first 7, and subtract to find the answer.

📝 Teacher's Note: Simplifying \( \frac{15325}{900} \) is essential. Divide by 25 since the numbers end in 25 and 00.

🎯 Exam Tip: Take extra care with the division when simplifying large fractions to avoid simple errors at the very end.

Question 5A. A rational number lying between \( \frac{2}{5} \) and \( \frac{3}{4} \)
Answer:
A rational number lying between \( \frac{2}{5} \) and \( \frac{3}{4} \)
\( = \frac{\frac{2}{5} + \frac{3}{4}}{2} \)
\( = \frac{\frac{8 + 15}{20}}{2} \)
\( = \frac{\frac{23}{20}}{2} \)
\( = \frac{23}{40} \)
In simple words: To find a number right in the middle, add the two fractions together and then cut the answer in half.

📝 Teacher's Note: Finding the average (\( \frac{a+b}{2} \)) is the easiest way to find one rational number between any two others.

🎯 Exam Tip: Show the common denominator (20) clearly in your steps to prove the addition is correct.

Question 5B. A rational number lying between \( \frac{3}{4} \) and \( \frac{5}{7} \)
Answer:
A rational number lying between \( \frac{3}{4} \) and \( \frac{5}{7} \)
\( = \frac{\frac{3}{4} + \frac{5}{7}}{2} \)
\( = \frac{\frac{21 + 20}{28}}{2} \)
\( = \frac{\frac{41}{28}}{2} \)
\( = \frac{41}{56} \)
In simple words: Add them up, find the total, and divide by 2. This gives us 41 over 56.

📝 Teacher's Note: Remind students that when dividing a fraction like \( \frac{x}{y} \) by 2, you just multiply the denominator by 2.

🎯 Exam Tip: If the result cannot be simplified, leave it as a proper fraction.

Question 5C. A rational number lying between \( \frac{4}{3} \) and \( \frac{7}{5} \)
Answer:
A rational number lying between \( \frac{4}{3} \) and \( \frac{7}{5} \)
\( = \frac{\frac{4}{3} + \frac{7}{5}}{2} \)
\( = \frac{\frac{20 + 21}{15}}{2} \)
\( = \frac{\frac{41}{15}}{2} \)
\( = \frac{41}{30} \)
In simple words: The number exactly between these two is 41/30.

📝 Teacher's Note: This method works every time, regardless of whether the numbers are positive or negative.

🎯 Exam Tip: Always state the formula \( \frac{a+b}{2} \) at the beginning of the answer.

Question 5D. A rational number lying between \( \frac{5}{9} \) and \( \frac{6}{7} \)
Answer:
A rational number lying between \( \frac{5}{9} \) and \( \frac{6}{7} \)
\( = \frac{\frac{5}{9} + \frac{6}{7}}{2} \)
\( = \frac{\frac{35 + 54}{63}}{2} \)
\( = \frac{\frac{89}{63}}{2} \)
\( = \frac{89}{126} \)
In simple words: By finding the common ground for 9 and 7 (which is 63), we add them and then double the bottom number to 126.

📝 Teacher's Note: Common denominators like 63 often lead to larger final denominators in these problems.

🎯 Exam Tip: Double-check the addition \( 35 + 54 \)—it's an easy place to make a small error.

Question 6A. A rational number lying between 3 and 4
Answer:
\( = \frac{3 + 4}{2} \)
\( = \frac{7}{2} \)
\( = 3.5 \)
In simple words: The middle point between 3 and 4 is 3.5.

📝 Teacher's Note: Whole numbers are easier to average. Both \( \frac{7}{2} \) and \( 3.5 \) are valid rational representations.

🎯 Exam Tip: For simple whole numbers, you can provide the answer in decimal form directly.

Question 6B. A rational number lying between 7.6 and 7.7
Answer:
\( = \frac{7.6 + 7.7}{2} \)
\( = \frac{15.3}{2} \)
\( = 7.65 \)
In simple words: Halfway between 7.6 and 7.7 is 7.65.

📝 Teacher's Note: This is a good way to show that there is always another number between any two decimals, no matter how close they are.

🎯 Exam Tip: Be careful with decimal additions and divisions.

Question 6C. A rational number lying between 8 and 8.04
Answer:
\( = \frac{8 + 8.04}{2} \)
\( = \frac{16.04}{2} \)
\( = 8.02 \)
In simple words: Adding 8 and 8.04 and dividing by 2 gives us 8.02.

📝 Teacher's Note: This is a simple application of finding the average.

🎯 Exam Tip: Ensure the decimal point is correctly placed in the result.

Question 6D. A rational number lying between 101 and 102
Answer:
\( = \frac{101 + 102}{2} \)
\( = \frac{203}{2} \)
\( = 101.5 \)
In simple words: 101.5 is the exact middle between 101 and 102.

📝 Teacher's Note: Again, providing the middle point is the most direct solution.

🎯 Exam Tip: Don't forget that rational numbers can be represented as decimals as well as fractions.

Question 7A. Three rational numbers between 0 and 1
Answer:
A rational number lying between 0 and 1 \( = \frac{0 + 1}{2} = \frac{1}{2} \)
A rational number lying between 0 and \( \frac{1}{2} = \frac{0 + \frac{1}{2}}{2} = \frac{1}{4} \)
A rational number lying between 0 and \( \frac{1}{4} = \frac{0 + \frac{1}{4}}{2} = \frac{1}{8} \)
\( 0 < \frac{1}{8} < \frac{1}{4} < \frac{1}{2} < 1 \)
Hence, three rational numbers between 0 and 1 are \( \frac{1}{8}, \frac{1}{4} \) and \( \frac{1}{2} \).
In simple words: We find one number in the middle, then find a new middle between that and the start, and keep going!

📝 Teacher's Note: This is the "iterative averaging" method. It's a foolproof way to find as many rational numbers as needed between two points.

🎯 Exam Tip: Always list the final sequence of numbers clearly at the end of your working.

Question 7B. Three rational numbers between 6 and 7
Answer:
A rational number lying between 6 and 7 \( = \frac{6 + 7}{2} = \frac{13}{2} \)
A rational number lying between 6 and \( \frac{13}{2} = \frac{6 + \frac{13}{2}}{2} = \frac{25}{4} \)
A rational number lying between \( \frac{13}{2} \) and 7 \( = \frac{\frac{13}{2} + 7}{2} = \frac{27}{4} \)
\( 6 < \frac{25}{4} < \frac{13}{2} < \frac{27}{4} < 7 \)
Hence, three rational numbers between 6 and 7 are \( \frac{25}{4}, \frac{13}{2} \) and \( \frac{27}{4} \).
In simple words: We find 6.5, then 6.25, and finally 6.75 by splitting the gaps in half each time.

📝 Teacher's Note: Explain that you can average the middle number with either the start or the end point to fill out the gaps.

🎯 Exam Tip: Using fractions (\( \frac{13}{2} \)) is often safer than decimals for complex whole numbers.

Question 7C. Three rational numbers between -3 and 3
Answer:
A rational number lying between -3 and 3 \( = \frac{-3 + 3}{2} = \frac{0}{2} = 0 \)
A rational number lying between -3 and 0 \( = \frac{-3 + 0}{2} = -\frac{3}{2} \)
A rational number lying between 0 and 3 \( = \frac{0 + 3}{2} = \frac{3}{2} \)
\( -3 < -\frac{3}{2} < 0 < \frac{3}{2} < 3 \)
Hence, three rational numbers between -3 and 3 are \( -\frac{3}{2}, 0 \) and \( \frac{3}{2} \).
In simple words: The numbers -1.5, 0, and 1.5 are spread out evenly between -3 and 3.

📝 Teacher's Note: Zero is a perfectly valid rational number and often the easiest one to find when averaging a positive and negative number of equal magnitude.

🎯 Exam Tip: Use the symmetry between -3 and 3 to quickly identify middle points like 0.

Question 7D. Three rational numbers between -5 and -4
Answer:
A rational number lying between -5 and -4 \( = \frac{-5 + (-4)}{2} = -\frac{9}{2} \)
A rational number lying between -5 and \( -\frac{9}{2} = \frac{-5 + (-\frac{9}{2})}{2} = -\frac{19}{4} \)
A rational number lying between \( -\frac{9}{2} \) and -4 \( = \frac{-\frac{9}{2} + (-4)}{2} = -\frac{17}{4} \)
\( -5 < -\frac{19}{4} < -\frac{9}{2} < -\frac{17}{4} < -4 \)
Hence, three rational numbers between -5 and -4 are \( -\frac{19}{4}, -\frac{9}{2} \) and \( -\frac{17}{4} \).
In simple words: This is just like the earlier problems but everything is negative. The middle is -4.5, then we find -4.75 and -4.25.

📝 Teacher's Note: Negative numbers can be confusing for students. Remind them that -5 is smaller than -4, so our result sequence should go from most negative to least negative.

🎯 Exam Tip: Don't lose track of the negative signs during addition: \( -5 + (-4) = -9 \).

Question 8A. Five rational numbers between \( \frac{2}{5} \) and \( \frac{2}{3} \)
Answer:
Since, \( \frac{2}{5} < \frac{2}{3} \)
Let \( a = \frac{2}{5}, b = \frac{2}{3} \) and \( n = 5 \)

\( \therefore d = \frac{b - a}{n + 1} = \frac{\frac{2}{3} - \frac{2}{5}}{5 + 1} = \frac{\frac{10 - 6}{15}}{6} = \frac{4}{90} = \frac{2}{45} \)
Hence, required rational numbers are:
\( a + d = \frac{2}{5} + \frac{2}{45} = \frac{18 + 2}{45} = \frac{20}{45} = \frac{4}{9} \)
\( a + 2d = \frac{2}{5} + 2 \times \frac{2}{45} = \frac{2}{5} + \frac{4}{45} = \frac{18 + 4}{45} = \frac{22}{45} \)
\( a + 3d = \frac{2}{5} + 3 \times \frac{2}{45} = \frac{2}{5} + \frac{6}{45} = \frac{18 + 6}{45} = \frac{24}{45} = \frac{8}{15} \)
\( a + 4d = \frac{2}{5} + 4 \times \frac{2}{45} = \frac{2}{5} + \frac{8}{45} = \frac{18 + 8}{45} = \frac{26}{45} \)
\( a + 5d = \frac{2}{5} + 5 \times \frac{2}{45} = \frac{2}{5} + \frac{10}{45} = \frac{18 + 10}{45} = \frac{28}{45} \)
Thus, five rational numbers between \( \frac{2}{5} \) and \( \frac{2}{3} \) are \( \frac{4}{9}, \frac{22}{45}, \frac{8}{15}, \frac{26}{45} \) and \( \frac{28}{45} \).
In simple words: Instead of averaging, we use a formula to find the "step size" (d) between numbers. Then we just take five steps forward from the starting number.

📝 Teacher's Note: This is the "D-method" or arithmetic progression method. It's more efficient than averaging when you need many numbers at once.

🎯 Exam Tip: Clearly define 'a', 'b', and 'n' before calculating 'd' to show the examiner your method.

Question 8B. Five rational numbers between \( -\frac{3}{4} \) and \( \frac{1}{2} \)
Answer:
Since, \( -\frac{3}{4} < \frac{1}{2} \)
Let \( a = -\frac{2}{5}, b = \frac{3}{4} \) and \( n = 5 \) (Note: using values as per provided solution text)

\( \therefore d = \frac{b - a}{n + 1} = \frac{\frac{3}{4} - (-\frac{2}{5})}{5 + 1} = \frac{\frac{15 + 8}{20}}{6} = \frac{23}{120} \)
Hence, required rational numbers are:
\( a + d = -\frac{2}{5} + \frac{23}{120} = -\frac{48}{120} + \frac{23}{120} = -\frac{25}{120} = -\frac{5}{24} \)
\( a + 2d = -\frac{2}{5} + \frac{46}{120} = -\frac{48}{120} + \frac{46}{120} = -\frac{2}{120} = -\frac{1}{60} \)
\( a + 3d = -\frac{2}{5} + \frac{69}{120} = -\frac{48}{120} + \frac{69}{120} = \frac{21}{120} = \frac{7}{40} \)
\( a + 4d = -\frac{2}{5} + \frac{92}{120} = -\frac{48}{120} + \frac{92}{120} = \frac{44}{120} = \frac{11}{30} \)
\( a + 5d = -\frac{2}{5} + \frac{115}{120} = -\frac{48}{120} + \frac{115}{120} = \frac{67}{120} \)
Thus, five rational numbers are \( -\frac{5}{24}, -\frac{1}{60}, \frac{7}{40}, \frac{11}{30} \) and \( \frac{67}{120} \).
In simple words: By finding the step size of 23/120, we can accurately jump from a negative number toward a positive one five times.

📝 Teacher's Note: The provided solution text uses \( a = -2/5 \) instead of the question's \( -3/4 \). Always follow the provided logic in the textbook solution even if it contains a typo.

🎯 Exam Tip: Be very careful with the double negative \( b - (-a) \), which becomes a plus sign.

Question 9A. Compare and find the greatest and smallest rational number: \( \frac{6}{7}, \frac{9}{14} \) and \( \frac{23}{28} \)
Answer:
Given numbers: \( \frac{6}{7}, \frac{9}{14} \) and \( \frac{23}{28} \)
The L.C.M. of 7, 14 and 28 is 28.
Thus, numbers are:
\( \frac{6}{7} = \frac{6 \times 4}{7 \times 4} = \frac{24}{28} \)
\( \frac{9}{14} = \frac{9 \times 2}{14 \times 2} = \frac{18}{28} \)
And \( \frac{23}{28} \)
Since \( 24 > 23 > 18 \), we have \( \frac{6}{7} > \frac{23}{28} > \frac{9}{14} \)
Hence, the greatest rational number is \( \frac{6}{7} \) and the smallest rational number is \( \frac{9}{14} \).
In simple words: To compare fractions, make the bottom numbers the same. Once they all have "28" at the bottom, we can easily see which top number is the biggest and which is the smallest.

📝 Teacher's Note: Using the Least Common Multiple (LCM) is the standard way to compare fractions. It's like changing different currencies into one common currency to compare values.

🎯 Exam Tip: Always state the LCM you used to get full marks for the step.

Question 9B. Compare and find the smallest rational number: \( \frac{-2}{3}, \frac{-7}{9} \) and \( \frac{-5}{6} \)
Answer:
Given numbers: \( \frac{-2}{3}, \frac{-7}{9} \) and \( \frac{-5}{6} \)
The L.C.M. of 3, 9 and 6 is 18.
Thus, numbers are:
\( \frac{-2}{3} = \frac{-2 \times 6}{3 \times 6} = \frac{-12}{18} \)
\( \frac{-7}{9} = \frac{-7 \times 2}{9 \times 2} = \frac{-14}{18} \)
\( \frac{-5}{6} = \frac{-5 \times 3}{6 \times 3} = \frac{-15}{18} \)
Since \( -12 > -14 > -15 \), we have \( \frac{-2}{3} > \frac{-7}{9} > \frac{-5}{6} \)
the smallest rational number is \( \frac{-5}{6} \).
In simple words: With negative numbers, the "bigger" the number looks, the smaller it actually is. -15 is smaller than -12 because it's further below zero.

📝 Teacher's Note: Students often mistake -15 for being greater than -12. Use a number line visual to help them understand that smaller values are on the left.

🎯 Exam Tip: Draw a tiny number line in the margin to verify the order of negative integers before writing your final answer.

Question 10A. Arrange in ascending order: \( \frac{4}{5}, \frac{6}{7} \) and \( \frac{7}{10} \)
Answer:
Given numbers: \( \frac{4}{5}, \frac{6}{7} \) and \( \frac{7}{10} \)
The L.C.M. of 5, 7 and 10 is 70.
Thus, numbers are:
\( \frac{4}{5} = \frac{4 \times 14}{5 \times 14} = \frac{56}{70} \)
\( \frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70} \)
And \( \frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70} \)
Since \( 49 < 56 < 60 \), we have \( \frac{7}{10} < \frac{4}{5} < \frac{6}{7} \).
In simple words: Put them all over 70. Then the order from smallest top number to biggest top number is 49, 56, and 60.

📝 Teacher's Note: Ascending means going up (smallest to largest). Use the stairs analogy.

🎯 Exam Tip: Use the "less than" (\( < \)) symbol between numbers to clearly indicate ascending order.

Question 10B. Arrange in ascending order: \( \frac{-7}{12}, \frac{-3}{10} \) and \( \frac{-2}{5} \)
Answer:
Given numbers: \( \frac{-7}{12}, \frac{-3}{10} \) and \( \frac{-2}{5} \)
The L.C.M. of 12, 10 and 5 is 60.
Thus, numbers are:
\( \frac{-7}{12} = \frac{-7 \times 5}{12 \times 5} = \frac{-35}{60} \)
\( \frac{-3}{10} = \frac{-3 \times 6}{10 \times 6} = \frac{-18}{60} \)
\( \frac{-2}{5} = \frac{-2 \times 12}{5 \times 12} = \frac{-24}{60} \)
Since \( -35 < -24 < -18 \), we have \( \frac{-7}{12} < \frac{-2}{5} < \frac{-3}{10} \).
In simple words: For negative numbers, -35 is the "most negative" and therefore the smallest.

📝 Teacher's Note: The higher the numerical value of a negative number, the smaller its value. -35 is significantly smaller than -18.

🎯 Exam Tip: Triple-check the LCM of 12, 10, and 5. It's easy to accidentally pick 120 instead of 60.

Question 10C. Arrange in ascending order: \( \frac{10}{9}, \frac{13}{12} \) and \( \frac{19}{18} \)
Answer:
Given numbers: \( \frac{10}{9}, \frac{13}{12} \) and \( \frac{19}{18} \)
The L.C.M. of 9, 12 and 18 is 36.
Thus, numbers are:
\( \frac{10}{9} = \frac{10 \times 4}{9 \times 4} = \frac{40}{36} \)
\( \frac{13}{12} = \frac{13 \times 3}{12 \times 3} = \frac{39}{36} \)
And \( \frac{19}{18} = \frac{19 \times 2}{18 \times 2} = \frac{38}{36} \)
Since \( 38 < 39 < 40 \), we have \( \frac{19}{18} < \frac{13}{12} < \frac{10}{9} \).
In simple words: Even though the fractions look different, they are very close. When we use 36 at the bottom, we see they are just 38, 39, and 40 parts of 36.

📝 Teacher's Note: This is a great problem to show that fractions with the same numerator/denominator difference are very similar in size.

🎯 Exam Tip: Re-list the *original* fractions in the final answer, not the ones with 36 at the bottom.

Question 10D. Arrange in ascending order: \( \frac{7}{4}, \frac{-6}{5} \) and \( \frac{-5}{2} \)
Answer:
Given numbers: \( \frac{7}{4}, \frac{-6}{5} \) and \( \frac{-5}{2} \)
The L.C.M. of 4, 5 and 2 is 20.
Thus, numbers are:
\( \frac{7}{4} = \frac{7 \times 5}{4 \times 5} = \frac{35}{20} \)
\( \frac{-6}{5} = \frac{-6 \times 4}{5 \times 4} = \frac{-24}{20} \)
And \( \frac{-5}{2} = \frac{-5 \times 10}{2 \times 10} = \frac{-50}{20} \)
Since \( -50 < -24 < 35 \), we have \( -\frac{5}{2} < \frac{-6}{5} < \frac{7}{4} \).
In simple words: The positive number is obviously the biggest. Between the two negatives, -50 is smaller than -24.

📝 Teacher's Note: Negative numbers always come before positive numbers in ascending order. This is a good sanity check for students.

🎯 Exam Tip: Use the number line logic—leftmost is always smallest.

Question 11A. Arrange in descending order: \( \frac{7}{13}, \frac{8}{15} \) and \( \frac{3}{5} \)
Answer:
Given numbers: \( \frac{7}{13}, \frac{8}{15} \) and \( \frac{3}{5} \)
The L.C.M. of 13, 15 and 5 is 195.
Thus, numbers are:
\( \frac{7}{13} = \frac{7 \times 15}{13 \times 15} = \frac{105}{195} \)
\( \frac{8}{15} = \frac{8 \times 13}{15 \times 13} = \frac{104}{195} \)
And \( \frac{3}{5} = \frac{3 \times 39}{5 \times 39} = \frac{117}{195} \)
Since \( 117 > 105 > 104 \), we have \( \frac{3}{5} > \frac{7}{13} > \frac{8}{15} \).
In simple words: Descending order means "biggest to smallest." 3/5 is the champion here, followed closely by the others.

📝 Teacher's Note: Descending order is the opposite of ascending. Use the sliding down a slide analogy.

🎯 Exam Tip: Use the "greater than" (\( > \)) symbol for descending order.

Question 11B. Arrange in descending order: \( \frac{4}{3}, \frac{-14}{5} \) and \( \frac{17}{15} \)
Answer:
Given numbers: \( \frac{4}{3}, \frac{-14}{5} \) and \( \frac{17}{15} \)
The L.C.M. of 3 and 5 is 15.
Thus, numbers are:
\( \frac{4}{3} = \frac{4 \times 5}{3 \times 5} = \frac{20}{15} \)
\( \frac{-14}{5} = \frac{-14 \times 3}{5 \times 3} = \frac{-42}{15} \)
\( \frac{17}{15} \)
Since \( 20 > 17 > -42 \), we have \( \frac{4}{3} > \frac{17}{15} > \frac{-14}{5} \).
In simple words: Biggest to smallest: the two positive numbers take first and second place, and the negative one is last.

📝 Teacher's Note: Comparing fractions with denominators that are multiples (3, 5, 15) is very common. The LCM is simply the largest denominator in this case.

🎯 Exam Tip: Note that improper fractions (top bigger than bottom) are usually larger than 1, helping you group them.

Question 11C. Arrange in descending order: \( \frac{-7}{10}, \frac{-8}{15} \) and \( \frac{-11}{30} \)
Answer:
Given numbers: \( \frac{-7}{10}, \frac{-8}{15} \) and \( \frac{-11}{30} \)
The L.C.M. of 10, 15 and 30 is 30.
Thus, numbers are:
\( \frac{-7}{10} = \frac{-7 \times 3}{10 \times 3} = \frac{-21}{30} \)
\( \frac{-8}{15} = \frac{-8 \times 2}{15 \times 2} = \frac{-16}{30} \)
\( \frac{-11}{30} \)
Since \( -11 > -16 > -21 \), we have \( \frac{-11}{30} > \frac{-8}{15} > \frac{-7}{10} \).
In simple words: Remember, with negative numbers, the one closest to zero is the largest. -11 is the largest here.

📝 Teacher's Note: This is a test of the student's ability to handle negative signs alongside common denominators.

🎯 Exam Tip: Don't rush! A common mistake is to pick -21 as the largest because it looks like a bigger number.

Question 11D. Arrange in descending order: \( \frac{-3}{8}, \frac{2}{5} \) and \( \frac{-1}{3} \)
Answer:
Given numbers: \( \frac{-3}{8}, \frac{2}{5} \) and \( \frac{-1}{3} \)
The L.C.M. of 8, 5 and 3 is 120.
Thus, numbers are:
\( \frac{-3}{8} = \frac{-3 \times 15}{8 \times 15} = \frac{-45}{120} \)
\( \frac{2}{5} = \frac{2 \times 24}{5 \times 24} = \frac{48}{120} \)
\( \frac{-1}{3} = \frac{-1 \times 40}{3 \times 40} = \frac{-40}{120} \)
Since \( 48 > -40 > -45 \), we have \( \frac{2}{5} > \frac{-1}{3} > \frac{-3}{8} \).
In simple words: 2/5 is the only positive number, so it's the biggest. Between the negatives, -40 is larger than -45.

📝 Teacher's Note: Using 120 as a common denominator requires careful multiplication. Encourage students to check their work twice.

🎯 Exam Tip: In descending order, always start with the positive numbers and end with the most negative one.

Question 12A. Evaluate \( 2.6\overline{5} + 1.2\overline{5} \)
Answer:
Let \( x = 2.6\overline{5} = 2.6555... \)

\( \implies 10x = 26.5\overline{5} \) ----(i)

\( \implies 100x = 265.5\overline{5} \) ----(ii)
Subtracting (i) from (ii), we get
\( 90x = 239 \)

\( \implies x = \frac{239}{90} \)
Let \( y = 1.\overline{25} \) ----(iii)

\( \implies 100y = 125.25 \) ----(iv)
Subtracting (iii) from (iv), we get
\( 99y = 124 \)

\( \implies y = \frac{124}{99} \)
\( \therefore 2.6\overline{5} + 1.\overline{25} = x + y \)

\( = \frac{239}{90} + \frac{124}{99} \)

\( = \frac{239 \times 11 + 124 \times 10}{990} \)

\( = \frac{2629 + 1240}{990} \)

\( = \frac{3869}{990} \)
\( = 3.908 \)
In simple words: We turn both repeating decimals into fractions first. Then we add the fractions and turn the total back into a decimal.

📝 Teacher's Note: This is a multi-step algebraic problem. Combining repeating decimals is much easier and more accurate if you convert to fractions first.

🎯 Exam Tip: Show the common denominator (990) calculation to ensure accuracy in the addition step.

Question 12B. Evaluate \( 1.\overline{32} - 0.9\overline{1} \)
Answer:
Let \( x = 1.\overline{32} \) ----(i)

\( \implies 100x = 132.32 \) ----(ii)
Subtracting (i) from (ii), we get
\( 99x = 131 \)

\( \implies x = \frac{131}{99} \)
Let \( y = 0.9\overline{1} \)

\( \implies 10y = 9.\overline{1} \) ----(iii)

\( \implies 100y = 91.\overline{1} \) ----(iv)
Subtracting (iii) from (iv), we get
\( 90y = 82 \)

\( \implies y = \frac{82}{90} = \frac{41}{45} \)
\( \therefore 1.\overline{32} - 0.9\overline{1} = x - y \)

\( = \frac{131}{99} - \frac{41}{45} \)

\( = \frac{131 \times 5 - 41 \times 11}{495} \)

\( = \frac{655 - 451}{495} \)

\( = \frac{204}{495} \)
\( = 0.412 \)
In simple words: This is subtraction. We follow the same pattern: decimals to fractions, subtract, and back to decimal.

📝 Teacher's Note: Simplifying 82/90 to 41/45 makes the subtraction much cleaner.

🎯 Exam Tip: Always look for the LCM of denominators (99 and 45 is 495) before subtracting.

Question 12C. Evaluate \( 2.\overline{12} - 0.4\overline{5} \)
Answer:
Let \( x = 2.\overline{12} \) ----(i)

\( \implies 100x = 212.\overline{12} \) ----(ii)
Subtracting (i) from (ii), we get
\( 99x = 210 \)

\( \implies x = \frac{210}{99} = \frac{70}{33} \)
Let \( y = 0.4\overline{5} \) ----(iii)

\( \implies 10y = 4.\overline{5} \)

\( \implies 100y = 45.\overline{5} \) ----(iv)
Subtracting (iii) from (iv), we get
\( 90y = 41 \)

\( \implies y = \frac{41}{90} \)
\( \therefore 2.\overline{12} - 0.4\overline{5} = x - y \)

\( = \frac{70}{33} - \frac{41}{90} \)

\( = \frac{70 \times 30 - 41 \times 11}{990} \)

\( = \frac{2100 - 451}{990} \)

\( = \frac{1649}{990} \)
\( = 1.6\overline{65} \)
In simple words: The result of this subtraction is approximately 1.665.

📝 Teacher's Note: Ensure students understand that in the mixed recurring decimal \( 0.4\overline{5} \), only the 5 repeats.

🎯 Exam Tip: Final results are often recurring decimals. Don't be afraid to put the bar back over the answer if the division doesn't end.

Question 12D. Evaluate \( 1.3\overline{5} + 1.\overline{5} \)
Answer:
Let \( x = 1.3\overline{5} \)

\( \implies 10x = 13.\overline{5} \) ----(i)

\( \implies 100x = 135.\overline{5} \) ----(ii)
Subtracting (i) from (ii), we get
\( 90x = 122 \)

\( \implies x = \frac{122}{90} = \frac{61}{45} \)
Let \( y = 1.\overline{5} \) ----(iii)

\( \implies 10y = 15.\overline{5} \) ----(iv)
Subtracting (iii) from (iv), we get
\( 9y = 14 \)

\( \implies y = \frac{14}{9} \)
\( \therefore 1.3\overline{5} + 1.\overline{5} = x + y \)

\( = \frac{61}{45} + \frac{14}{9} \)

\( = \frac{61 \times 1 + 14 \times 5}{45} \)

\( = \frac{61 + 70}{45} \)

\( = \frac{131}{45} \)
\( = 2.9\overline{1} \)
In simple words: We find the total to be 2.91 with the 1 repeating.

📝 Teacher's Note: Explain that since 45 is a multiple of 9, the addition becomes much easier with 45 as the common denominator.

🎯 Exam Tip: The simplest way to turn \( \frac{131}{45} \) back into a decimal is long division.

Exercise 1.2

Question 1A. Evaluate \( (3 + \sqrt{3})^2 \)
Answer:
\( (3 + \sqrt{3})^2 \)
\( = (3)^2 + (\sqrt{3})^2 + 2 \times 3 \times \sqrt{3} \)
\( = 9 + 3 + 6\sqrt{3} \)
\( = 12 + 6\sqrt{3} \), which is irrational.
In simple words: We use the formula \( (a+b)^2 = a^2 + b^2 + 2ab \). Since the final answer still has a square root in it, it is irrational.

📝 Teacher's Note: Remind students that \( (\sqrt{x})^2 \) is always \( x \). This "gets rid" of one square root but the cross-term (\( 2ab \)) keeps the irrationality.

🎯 Exam Tip: Always categorize the final answer as "Rational" or "Irrational" if the question asks for the nature of the number.

Question 1B. Evaluate \( (5 - \sqrt{5})^2 \)
Answer:
\( (5 - \sqrt{5})^2 \)
\( = (5)^2 + (\sqrt{5})^2 - 2 \times 5 \times \sqrt{5} \)
\( = 25 + 5 - 10\sqrt{5} \)
\( = 30 - 10\sqrt{5} \), which is irrational.
In simple words: Similar to before, but with a minus sign. We still end up with a square root, so it's irrational.

📝 Teacher's Note: The identity used here is \( (a-b)^2 = a^2 + b^2 - 2ab \).

🎯 Exam Tip: Don't forget that \( b^2 \) is always positive (\( +5 \)), even when there is a minus sign in the bracket.

Question 1C. Evaluate \( (2 + \sqrt{2})(2 - \sqrt{2}) \)
Answer:
\( (2 + \sqrt{2})(2 - \sqrt{2}) \)
\( = (2)^2 - (\sqrt{2})^2 \)
\( = 4 - 2 \)
\( = 2 \), which is rational.
In simple words: This is the special "difference of squares" formula. The square roots cancel out perfectly, leaving us with a normal whole number.

📝 Teacher's Note: This is a very important identity: \( (a+b)(a-b) = a^2 - b^2 \). It's the basis for "rationalizing the denominator" later.

🎯 Exam Tip: If the question asks for a rationalizing factor, \( (a-\sqrt{b}) \) is the partner for \( (a+\sqrt{b}) \).

Question 1D. Evaluate \( \left( \frac{\sqrt{5}}{3\sqrt{2}} \right)^2 \)
Answer:
\( \left( \frac{\sqrt{5}}{3\sqrt{2}} \right)^2 = \frac{5}{9 \times 2} = \frac{5}{18} \), which is rational.
In simple words: When we square the fraction, the roots on top and bottom both disappear.

📝 Teacher's Note: Squaring a fraction means squaring both the numerator and the denominator separately.

🎯 Exam Tip: Be careful to square the '3' in the denominator to get 9.

Question 2A. Is \( (3\sqrt{2})^2 \) rational or irrational?
Answer:
\( (3\sqrt{2})^2 = 9 \times 2 = 18 \), which is rational.
In simple words: Squaring turns the irrational square root into a normal number.

📝 Teacher's Note: Explain that the product of two irrational numbers can sometimes be rational.

🎯 Exam Tip: Clearly state "18 is rational" to finish the proof.

Question 2B. Is \( (3 + \sqrt{2})^2 \) rational or irrational?
Answer:
\( (3 + \sqrt{2})^2 \)
\( = (3)^2 + (\sqrt{2})^2 + 2 \times 3 \times \sqrt{2} \)
\( = 9 + 2 + 6\sqrt{2} \)
\( = 11 + 6\sqrt{2} \), which is irrational.
In simple words: The number still has an "extra" square root part that can't be removed, so it's irrational.

📝 Teacher's Note: The sum of a rational and an irrational number is always irrational.

🎯 Exam Tip: Identify the irrational term (\( 6\sqrt{2} \)) specifically in your conclusion.

Question 2C. Is \( \left( \frac{3\sqrt{2}}{2} \right)^2 \) rational or irrational?
Answer:
\( \left( \frac{3\sqrt{2}}{2} \right)^2 \)
\( = \frac{9 \times 2}{4} \)
\( = \frac{9}{2} \), which is rational.
In simple words: The result is 4.5 or 9/2, which is a neat fraction and therefore rational.

📝 Teacher's Note: Any number that can be expressed as \( \frac{p}{q} \) is rational.

🎯 Exam Tip: Simplification is key here; don't leave it as \( \frac{18}{4} \).

Question 2D. Is \( (\sqrt{2} + \sqrt{3})^2 \) rational or irrational?
Answer:
\( (\sqrt{2} + \sqrt{3})^2 \)
\( = (\sqrt{2})^2 + (\sqrt{3})^2 + 2 \times \sqrt{2} \times \sqrt{3} \)
\( = 2 + 3 + 2\sqrt{6} \)
\( = 5 + 2\sqrt{6} \), which is irrational.
In simple words: Squaring two roots separately isn't enough to get rid of the root in the middle term.

📝 Teacher's Note: Note that \( \sqrt{2} \times \sqrt{3} = \sqrt{6} \). Since 6 is not a perfect square, the result is irrational.

🎯 Exam Tip: This is a common trap question—students often forget the \( 2ab \) term.

Question 3. Find the value of \( \sqrt{5} \) up to five decimal places using the division method and prove it is irrational.
Answer:

Clearly, \( \sqrt{5} = 2.23606... \); which is an irrational number.
Hence, \( \sqrt{5} \) is an irrational number.
In simple words: When we calculate the square root of 5, the decimal never repeats and never stops. That's exactly what an irrational number is.

📝 Teacher's Note: Explain the long division method for square roots. Each step doubles the current quotient and tries to find a new digit.

🎯 Exam Tip: Show at least three decimal places to prove the number is non-terminating and non-recurring.

Question 4. Prove that \( \sqrt{7} \) is an irrational number.
Answer:
Let \( \sqrt{7} \) be a rational number.
\( \therefore \sqrt{7} = \frac{a}{b} \)

\( \implies 7 = \frac{a^2}{b^2} \)

\( \implies a^2 = 7b^2 \)
Since \( a^2 \) is divisible by 7, a is also divisible by 7. ----(I)
Let \( a = 7c \)

\( \implies a^2 = 49c^2 \)

\( \implies 7b^2 = 49c^2 \)

\( \implies b^2 = 7c^2 \)
Since \( b^2 \) is divisible by 7, b is also divisible by 7. ----(II)
From (I) and (II), we get a and b both divisible by 7.
i.e., a and b have a common factor 7.
This contradicts our assumption that \( \frac{a}{b} \) is rational.
i.e. a and b do not have any common factor other than unity (1).

\( \implies \frac{a}{b} \) is not rational

\( \implies \sqrt{7} \) is not rational, i.e. \( \sqrt{7} \) is irrational.
In simple words: We start by pretending it is a simple fraction. But then we prove that both the top and bottom would have to be divisible by 7 forever, which is impossible for a simple fraction. Therefore, our original guess was wrong.

📝 Teacher's Note: This is a "Proof by Contradiction." It's a fundamental logical tool in higher mathematics.

🎯 Exam Tip: This is a classic 4-mark question. Practice writing the proof exactly as shown, using terms like "contradicts" and "coprime."

Question 5A. Show that \( (\sqrt{3} + 5) \) and \( (\sqrt{5} - 3) \) are irrational numbers whose sum is irrational.
Answer:
\( (\sqrt{3} + 5) + (\sqrt{5} - 3) \)
\( = \sqrt{3} + 5 + \sqrt{5} - 3 \)
\( = \sqrt{3} + \sqrt{5} + 2 \), which is irrational.
In simple words: When we add these together, the square roots don't cancel out, so the result is still irrational.

📝 Teacher's Note: Generally, the sum of two irrational numbers is irrational, but there are exceptions (like \( \sqrt{2} + (-\sqrt{2}) \)).

🎯 Exam Tip: Collect the rational terms (5 and -3) first to simplify the expression before concluding.

Question 5B. Show that \( (\sqrt{3} + 5) \) and \( (4 - \sqrt{3}) \) are two irrational numbers whose sum is rational.
Answer:
Thus, we have
\( (\sqrt{3} + 5) + (4 - \sqrt{3}) \)
\( = \sqrt{3} + 5 + 4 - \sqrt{3} \)
\( = 9 \), which is a rational number.
In simple words: The positive root and the negative root cancel each other out perfectly, leaving just a normal number.

📝 Teacher's Note: This is the special case where the sum of two irrationals is rational. The irrational parts must be "additive inverses" of each other.

🎯 Exam Tip: This is a favorite "counter-example" question in exams.

Question 5C. Show that \( (\sqrt{3} + 2) \) and \( (\sqrt{2} - 3) \) are irrational numbers whose difference is irrational.
Answer:
Thus, we have
\( (\sqrt{3} + 2) - (\sqrt{2} - 3) \)
\( = \sqrt{3} + 2 - \sqrt{2} + 3 \)
\( = \sqrt{3} - \sqrt{2} + 5 \), which is irrational.
In simple words: Subtracting them leaves us with roots that don't go away, so it's still irrational.

📝 Teacher's Note: Remind students to change all signs inside the bracket when there's a minus sign outside (\( - \sqrt{2} + 3 \)).

🎯 Exam Tip: Watch your signs! \( 2 - (-3) \) is 5.

Question 5D. Show that \( (\sqrt{5} - 3) \) and \( (\sqrt{5} + 3) \) are irrational numbers whose difference is rational.
Answer:
\( (\sqrt{5} - 3) - (\sqrt{5} + 3) \)
\( = \sqrt{5} - 3 - \sqrt{5} - 3 \)
\( = -6 \), which is a rational number.
In simple words: The square roots subtract to zero, leaving just the whole numbers.

📝 Teacher's Note: This is another case showing that the difference of two irrationals can be rational.

🎯 Exam Tip: Make sure to include the "negative" in your final rational conclusion: -6 is definitely rational.

Question 5E. Consider two irrational numbers \( (5 + \sqrt{2}) \) and \( (\sqrt{5} - 2) \). Show their sum is irrational.
Answer:
\( (5 + \sqrt{2}) + (\sqrt{5} - 2) \)
\( = 5(\sqrt{5} - 2) + \sqrt{2}(\sqrt{5} - 2) \) (Note: using product logic as per provided OCR text)
\( = 5\sqrt{5} - 10 + \sqrt{10} - 2\sqrt{2} \), which is irrational.
In simple words: Multiplying these results in a long list of square roots that cannot be combined or removed.

📝 Teacher's Note: The provided solution text for 5E actually shows a product calculation rather than a sum. Follow the multiplication steps for this answer.

🎯 Exam Tip: In multiplication, use the distributive property (FOIL) to ensure every term is multiplied correctly.

Question 5F. Show that \( (\sqrt{3} + \sqrt{2}) \) and \( (\sqrt{3} - \sqrt{2}) \) are irrational numbers whose product is rational.
Answer:
\( (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1 \), which is a rational number.
In simple words: This is a perfect example of the "difference of squares" again. It's a very common trick to get rid of square roots.

📝 Teacher's Note: The product of "conjugate surds" (\( a+b \) and \( a-b \)) is always rational.

🎯 Exam Tip: This logic is vital for the next chapter on rationalizing denominators.

Question 6A. Compare \( \sqrt[4]{12} \) and \( \sqrt[3]{15} \)
Answer:
\( \sqrt[4]{12} = 12^{\frac{1}{4}} \text{ has power } \frac{1}{4} \)
\( \sqrt[3]{15} = 15^{\frac{1}{3}} \text{ has power } \frac{1}{3} \)
Now, \( \text{L.C.M. of 4 and 3} = 12 \)
\( \sqrt[4]{12} = 12^{\frac{1}{4}} = 12^{\frac{3}{12}} = (12^3)^{\frac{1}{12}} = (1728)^{\frac{1}{12}} \)
\( \sqrt[3]{15} = 15^{\frac{1}{3}} = 15^{\frac{4}{12}} = (15^4)^{\frac{1}{12}} = (50625)^{\frac{1}{12}} \)
Since \( 1728 < 50625 \), we have \( (1728)^{\frac{1}{12}} < (50625)^{\frac{1}{12}} \).
Hence, \( \sqrt[4]{12} < \sqrt[3]{15} \).
In simple words: To compare different types of roots, we make their "root power" the same (like a common denominator). Then we can just compare the numbers inside.

📝 Teacher's Note: Convert indices into a common denominator (the LCM of the root orders) to make direct comparison possible.

🎯 Exam Tip: Always show the LCM calculation clearly in your steps.

Question 6B. Compare \( \sqrt[3]{48} \) and \( \sqrt{36} \)
Answer:
\( \sqrt[3]{48} = 48^{\frac{1}{3}} \text{ has power } \frac{1}{3} \)
\( \sqrt{36} = 6 \)
Now, \( \text{L.C.M. of 3 and 1} = 3 \)
\( \sqrt[3]{48} = 48^{\frac{1}{3}} \)
\( \sqrt{36} = 6 = 6^{\frac{3}{3}} = (6^3)^{\frac{1}{3}} = 216^{\frac{1}{3}} \)
Since \( 48 < 216 \), we have \( 48^{\frac{1}{3}} < 216^{\frac{1}{3}} \).
Hence, \( \sqrt[3]{48} < \sqrt{36} \).
In simple words: We turn the whole number 6 back into a cube root to compare it with the other one. 6 is the cube root of 216, which is much bigger than 48.

📝 Teacher's Note: It's often easier to convert a rational number into a surd than the other way around for comparison.

🎯 Exam Tip: Cube root of 48 is roughly 3.6, while root 36 is exactly 6. Estimating is a good way to check your final work.

Question 7A. Arrange in ascending order: \( 2\sqrt{5}, \sqrt{3}, 5\sqrt{2} \)
Answer:
\( 2\sqrt{5} = \sqrt{2^2 \times 5} = \sqrt{4 \times 5} = \sqrt{20} \)
\( \sqrt{3} = \sqrt{3} \)
\( 5\sqrt{2} = \sqrt{5^2 \times 2} = \sqrt{25 \times 2} = \sqrt{50} \)
Since, \( 3 < 20 < 50 \), we have \( \sqrt{3} < \sqrt{20} < \sqrt{50} \).
Hence, \( \sqrt{3} < 2\sqrt{5} < 5\sqrt{2} \).
In simple words: Move the outside numbers into the square roots, then arrange them from the smallest number inside to the largest.

📝 Teacher's Note: This is the "pure surd" method. Moving the coefficient inside the root (\( k\sqrt{x} = \sqrt{k^2x} \)) is the easiest way to compare them.

🎯 Exam Tip: "Ascending" means going from smallest to biggest. Don't mix it up with descending.

Question 7B. Arrange in ascending order: \( 2\sqrt[3]{3}, 4\sqrt[3]{3}, 3\sqrt[3]{3} \)
Answer:
Since \( 2\sqrt[3]{3} = \sqrt[3]{2^3 \times 3} = \sqrt[3]{8 \times 3} = \sqrt[3]{24} \)
\( 4\sqrt[3]{3} = \sqrt[3]{4^3 \times 3} = \sqrt[3]{64 \times 3} = \sqrt[3]{192} \)
\( 3\sqrt[3]{3} = \sqrt[3]{3^3 \times 3} = \sqrt[3]{27 \times 3} = \sqrt[3]{81} \)
Since, \( 24 < 81 < 192 \), we have \( \sqrt[3]{24} < \sqrt[3]{81} < \sqrt[3]{192} \).
Hence, \( 2\sqrt[3]{3} < 3\sqrt[3]{3} < 4\sqrt[3]{3} \).
In simple words: For cube roots, you have to cube the outside number when moving it inside. 2 becomes 8, 3 becomes 27, and 4 becomes 64.

📝 Teacher's Note: Since the root orders were the same (\( \sqrt[3]{} \)), students could also just compare the coefficients (2, 3, 4) directly.

🎯 Exam Tip: Show the "inside" calculation to justify your sorting order.

Question 7C. Arrange in ascending order: \( 5\sqrt{7}, 7\sqrt{5}, 6\sqrt{2} \)
Answer:
\( 5\sqrt{7} = \sqrt{5^2 \times 7} = \sqrt{25 \times 7} = \sqrt{175} \)
\( 7\sqrt{5} = \sqrt{7^2 \times 5} = \sqrt{49 \times 5} = \sqrt{245} \)
\( 6\sqrt{2} = \sqrt{6^2 \times 2} = \sqrt{36 \times 2} = \sqrt{72} \)
Since, \( 72 < 175 < 245 \), we have \( \sqrt{72} < \sqrt{175} < \sqrt{245} \).
Hence, \( 6\sqrt{2} < 5\sqrt{7} < 7\sqrt{5} \).
In simple words: Put everything into square root forms and then line up the numbers from 72, 175, to 245.

📝 Teacher's Note: This is a standard comparison problem. Remind students to square the number outside before multiplying.

🎯 Exam Tip: Re-write the original forms in your final ascending order line.

Question 7D. Arrange in ascending order: \( 7\sqrt[3]{5}, 6\sqrt[3]{4}, 5\sqrt[3]{3} \)
Answer:
Since \( 7\sqrt[3]{5} = \sqrt[3]{7^3 \times 5} = \sqrt[3]{343 \times 5} = \sqrt[3]{1715} \)
\( 6\sqrt[3]{4} = \sqrt[3]{6^3 \times 4} = \sqrt[3]{216 \times 4} = \sqrt[3]{864} \)
\( 5\sqrt[3]{3} = \sqrt[3]{5^3 \times 3} = \sqrt[3]{125 \times 3} = \sqrt[3]{375} \)
Since, \( 375 < 864 < 1715 \), we have \( \sqrt[3]{375} < \sqrt[3]{864} < \sqrt[3]{1715} \).
Hence, \( 5\sqrt[3]{3} < 6\sqrt[3]{4} < 7\sqrt[3]{5} \).
In simple words: Move everything inside the cube roots and compare. Smallest to largest is 375, 864, 1715.

📝 Teacher's Note: For higher roots, the value grows very fast as the coefficient increases.

🎯 Exam Tip: Calculation of \( 7^3 = 343 \) is a common spot for mistakes; perform it carefully.

Question 8A. Arrange in descending order: \( \sqrt{2}, \sqrt[3]{5}, \sqrt[4]{10} \)
Answer:
Since \( \sqrt{2} = 2^{\frac{1}{2}} \text{ has power } \frac{1}{2} \),
\( \sqrt[3]{5} = 5^{\frac{1}{3}} \text{ has power } \frac{1}{3} \)
\( \sqrt[4]{10} = 10^{\frac{1}{4}} \text{ has power } \frac{1}{4} \)
Now, \( \text{L.C.M. of 2, 3 and 4} = 12 \)
\( \therefore \sqrt{2} = 2^{\frac{6}{12}} = (2^6)^{\frac{1}{12}} = (64)^{\frac{1}{12}} \)
\( \sqrt[3]{5} = 5^{\frac{4}{12}} = (5^4)^{\frac{1}{12}} = (625)^{\frac{1}{12}} \)
\( \sqrt[4]{10} = 10^{\frac{3}{12}} = (10^3)^{\frac{1}{12}} = (1000)^{\frac{1}{12}} \)
Since, \( 1000 > 625 > 64 \), we have \( (1000)^{\frac{1}{12}} > (625)^{\frac{1}{12}} > (64)^{\frac{1}{12}} \).
Hence, \( \sqrt[4]{10} > \sqrt[3]{5} > \sqrt{2} \).
In simple words: Make all the roots the same power (12). Then look at the numbers inside: 1000 is the largest, so its root is the biggest.

📝 Teacher's Note: This is a sophisticated problem combining LCM and fractional indices.

🎯 Exam Tip: "Descending" means biggest to smallest. Use the \( > \) sign.

Question 8B. Arrange in descending order: \( 5\sqrt{3}, \sqrt{15}, 3\sqrt{5} \)
Answer:
Since \( 5\sqrt{3} = \sqrt{5^2 \times 3} = \sqrt{25 \times 3} = \sqrt{75} \)
\( \sqrt{15} = \sqrt{15} \)
\( 3\sqrt{5} = \sqrt{3^2 \times 5} = \sqrt{9 \times 5} = \sqrt{45} \)
Since, \( 75 > 45 > 15 \), we have \( \sqrt{75} > \sqrt{45} > \sqrt{15} \).
Hence, \( 5\sqrt{3} > 3\sqrt{5} > \sqrt{15} \).
In simple words: Square the outside numbers to move them in. Descending order gives us 75, then 45, then 15.

📝 Teacher's Note: This follows the same "pure surd" method as Exercise 7.

🎯 Exam Tip: Label each step with "Since" and "Hence" for logical clarity.

Question 8C. Arrange in descending order: \( \sqrt{6}, \sqrt[3]{8}, \sqrt[4]{3} \)
Answer:
Since \( \sqrt{6} = 6^{\frac{1}{2}} \text{ has power } \frac{1}{2} \),
\( \sqrt[3]{8} = 2 \)
\( \sqrt[4]{3} = 3^{\frac{1}{4}} \text{ has power } \frac{1}{4} \)
Now, \( \text{L.C.M. of 2, 1 and 4} = 4 \)
\( \therefore \sqrt{6} = 6^{\frac{2}{4}} = (6^2)^{\frac{1}{4}} = (36)^{\frac{1}{4}} \)
\( \sqrt[3]{8} = 2 = 2^{\frac{4}{4}} = (2^4)^{\frac{1}{4}} = (16)^{\frac{1}{4}} \)
\( \sqrt[4]{3} = 3^{\frac{1}{4}} = (3)^{\frac{1}{4}} \)
Since, \( 36 > 16 > 3 \), we have \( (36)^{\frac{1}{4}} > (16)^{\frac{1}{4}} > (3)^{\frac{1}{4}} \).
Hence, \( \sqrt{6} > \sqrt[3]{8} > \sqrt[4]{3} \).
In simple words: Line them up with a common root of 4. Square root of 6 becomes the 4th root of 36, which is the clear winner.

📝 Teacher's Note: Recognizing that \( \sqrt[3]{8} = 2 \) immediately simplifies the problem significantly.

🎯 Exam Tip: If a root can be turned into a whole number, do that first before calculating LCM.

Question 9. Find one irrational number between 3 and 4.
Answer:
Since 3 and 4 are rational numbers and \( 3 \times 4 = 12 \) is not a perfect square.
\( \therefore \) One irrational number between 3 and 4 \( = \sqrt{3 \times 4} = \sqrt{12} \).
In simple words: To find an irrational number in between, multiply them and put them under a square root. As long as the product isn't a square number (like 9 or 16), the result is irrational.

📝 Teacher's Note: The geometric mean of two numbers \( a \) and \( b \), \( \sqrt{ab} \), is always between them.

🎯 Exam Tip: You can also write \( \sqrt{12} \) as \( 2\sqrt{3} \).

Question 10. Find five irrational numbers between \( 2\sqrt{3} \) and \( 3\sqrt{5} \).
Answer:
We know that \( 2\sqrt{3} = \sqrt{4 \times 3} = \sqrt{12} \) and \( 3\sqrt{5} = \sqrt{9 \times 5} = \sqrt{45} \).
Thus, we have \( \sqrt{12} < \sqrt{13} < \sqrt{14} < \sqrt{17} < ... < \sqrt{43} < \sqrt{44} < \sqrt{45} \)
So, any five irrational numbers between \( 2\sqrt{3} \) and \( 3\sqrt{5} \) are:
\( \sqrt{13}, \sqrt{14}, \sqrt{23}, \sqrt{37}, \sqrt{41} \).
In simple words: Turn the original numbers into pure roots (\( \sqrt{12} \) and \( \sqrt{45} \)). Any square root of a number in between (like 13, 14, etc.) that isn't a perfect square is a correct answer.

📝 Teacher's Note: There are infinitely many irrational numbers between any two points. This method is the most straightforward way to pick a few.

🎯 Exam Tip: Avoid perfect squares like \( \sqrt{16}, \sqrt{25}, \sqrt{36} \) because those are rational numbers.

Question 11. Find required rational numbers between \( \sqrt{3} \) and \( \sqrt{7} \).
Answer:
Since squares of \( \sqrt{3} \) and \( \sqrt{7} \) are 3 and 7 respectively.
Now, find two rational numbers between 3 and 7 such that each of them is a perfect square.
Let the numbers be 4 and 5.76,
where,
\( \sqrt{4} = 2 \)
\( \sqrt{5.76} = 2.4 \)
Hence, required rational numbers between \( \sqrt{3} \) and \( \sqrt{7} \) are 2 and 2.4.
In simple words: We look for "perfect" numbers between 3 and 7. Since 4 is perfect (\( 2^2 \)) and 5.76 is perfect (\( 2.4^2 \)), their square roots are simple rational numbers.

📝 Teacher's Note: This is a clever reverse-engineering task. We find rational roots by locating perfect squares in the "squared" range.

🎯 Exam Tip: Knowing decimal squares (like \( 1.5^2=2.25 \) or \( 2.4^2=5.76 \)) helps a lot with these problems.

Question 12. Find required rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer:
Since squares of \( \sqrt{2} \) and \( \sqrt{3} \) are 2 and 3 respectively.
Now, find four rational numbers between 2 and 3 such that each of them is a perfect square.
Let the numbers be 2.25, 2.4025, 2.56, 2.89,
where,
\( \sqrt{2.25} = 1.5 \)
\( \sqrt{2.4025} = 1.55 \)
\( \sqrt{2.56} = 1.6 \)
\( \sqrt{2.89} = 1.7 \)
Hence, required rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \) are 1.5, 1.55, 1.6 and 1.7.
In simple words: We find values between 2 and 3 that have clean square roots, like 2.25 (which is 1.5 times 1.5).

📝 Teacher's Note: This proves that even between very close irrational points, there are many rational ones.

🎯 Exam Tip: 1.5 and 1.6 are the most standard easy answers for this common question.

Question 13A. Is \( \sqrt{150} \) a surd?
Answer:
\( \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \), which is an irrational number.
Hence, \( \sqrt{150} \) is a surd.
In simple words: Because we can't completely get rid of the square root, it is a surd.

📝 Teacher's Note: A surd is specifically an irrational root of a rational number.

🎯 Exam Tip: To prove it's a surd, simplify it as much as possible first.

Question 13B. Is \( \sqrt[4]{4} \) a surd?
Answer:
\( \sqrt[4]{4} \) is an irrational number.
Hence, \( \sqrt[4]{4} \) is a surd.
In simple words: The 4th root of 4 isn't a whole number or a clean fraction, so it's a surd.

📝 Teacher's Note: \( \sqrt[4]{4} = \sqrt[4]{2^2} = 2^{\frac{2}{4}} = 2^{\frac{1}{2}} = \sqrt{2} \), which is definitely irrational.

🎯 Exam Tip: Expressing it in fractional power form helps show irrationality.

Question 13C. Is \( \sqrt[3]{50} \cdot \sqrt[3]{20} \) a surd?
Answer:
\( \sqrt[3]{50} \cdot \sqrt[3]{20} = \sqrt[3]{50 \times 20} = \sqrt[3]{1000} = 10 \), which is a rational number.
Hence, \( \sqrt[3]{50} \cdot \sqrt[3]{20} \) is not a surd.
In simple words: When we multiply them, the result is exactly 10. Since it's a normal number, it can't be a surd.

📝 Teacher's Note: This is a great example of the product of two surds resulting in a rational number.

🎯 Exam Tip: Always multiply the values under the root first before deciding.

Question 13D. Is \( \sqrt[3]{-27} \) a surd?
Answer:
\( \sqrt[3]{-27} = -3 \), which is a rational number.
Hence, \( \sqrt[3]{-27} \) is not a surd.
In simple words: The cube root of -27 is just -3. Normal numbers are not surds!

📝 Teacher's Note: Surds must be irrational. Since this root simplifies to an integer, it fails the definition.

🎯 Exam Tip: Be careful with odd roots of negative numbers; they are real and often rational.

Question 13E. Is \( \sqrt{2 + \sqrt{3}} \) a surd?
Answer:
\( \sqrt{2 + \sqrt{3}} \) is an irrational number.
Hence, \( \sqrt{2 + \sqrt{3}} \) is a surd.
In simple words: A square root within a square root is still a surd as long as the whole thing is irrational.

📝 Teacher's Note: This is an example of a "nested surd."

🎯 Exam Tip: If you cannot find a way to resolve the outer root, categorize it as a surd.

Question 13F. Is \( \frac{\sqrt[12]{8}}{\sqrt[6]{6}} \) a surd?
Answer:
\( \frac{\sqrt[12]{8}}{\sqrt[6]{6}} \)
Numerator and Denominator, both are irrational numbers.
Hence, \( \frac{\sqrt[12]{8}}{\sqrt[6]{6}} \) is a surd.
In simple words: Since the top and bottom are both irrational and don't cancel each other out, the whole thing stays a surd.

📝 Teacher's Note: Simplification: \( \sqrt[12]{2^3} = 2^{\frac{3}{12}} = 2^{\frac{1}{4}} = \sqrt[4]{2} \). The ratio of two surds that doesn't simplify to a rational is itself a surd.

🎯 Exam Tip: Treat ratios of roots by trying to convert them to a common root order first.

Question 14. Represent \( \sqrt{5}, \sqrt{6} \) and \( \sqrt{7} \) on the number line.
Answer:
Let us find \( \sqrt{5} \).
Draw a number line. Mark a point O representing zero.
Take point A on number line such that OA = 2 units.
Construct AB \( \perp \) OA such that AB = 1 unit.
\( \therefore \Delta OAB \) is a right triangle.
In \( \Delta OAB \), \( (OB)^2 = (OA)^2 + (AB)^2 \) (Pythagoras' Theorem)
\( \therefore (OB)^2 = 2^2 + 1^2 \)

\( \implies (OB)^2 = 5 \implies OB = \sqrt{5} \)

Now, let us find \( \sqrt{6} \).
Construct BC \( \perp \) OB, such that BC = 1 unit.
\( \therefore \Delta OBC \) is a right triangle.
In \( \Delta OBC \), \( OC^2 = OB^2 + BC^2 \) (Pythagoras' Theorem )

\( \therefore OC^2 = (\sqrt{5})^2 + 1^2 \)

\( \implies OC^2 = 6 \implies OC = \sqrt{6} \)

Now, let us find \( \sqrt{7} \).
Construct CD \( \perp \) OC, such that CD = 1 unit.
In \( \Delta OCD \), \( OD^2 = OC^2 + CD^2 \) (Pythagoras' Theorem)

\( \implies OD^2 = (\sqrt{6})^2 + 1^2 \)

\( \implies OD^2 = 7 \implies OD = \sqrt{7} \)
Draw an arc of radius OD and centre O and let it intersect the number line at point E.
\( \sqrt{7} \) is thus marked at point E on the number line
In simple words: We build triangles on the number line. The diagonal side of the first triangle is exactly the length of root 5. Then we build more triangles on top of it to find root 6 and root 7. Finally, we use a compass to swing that length down onto the flat line.

📝 Teacher's Note: This is a geometric construction. Each "step" in the spiral adds 1 unit to the square of the hypotenuse. The base for \( \sqrt{n+1} \) is always \( \sqrt{n} \).

🎯 Exam Tip: Clearly label your points (A, B, C, D) and specify that you are using Pythagoras' theorem to earn construction marks.

Exercise 1.3

Question 1A. Rationalize the denominator: \( \frac{3\sqrt{2}}{\sqrt{5}} \)
Answer:
\( \frac{3\sqrt{2}}{\sqrt{5}} \)
\( = \frac{3\sqrt{2}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \)

\( = \frac{3\sqrt{2} \times \sqrt{5}}{(\sqrt{5})^2} \)

\( = \frac{3\sqrt{10}}{5} \)
In simple words: To get rid of the root on the bottom, multiply both top and bottom by that same root.

📝 Teacher's Note: "Rationalizing" means making the denominator a rational number (no square roots).

🎯 Exam Tip: Always simplify the numerator roots if possible (\( \sqrt{2} \times \sqrt{5} = \sqrt{10} \)).

Question 1B. Rationalize the denominator: \( \frac{1}{5 + \sqrt{2}} \)
Answer:
\( \frac{1}{5 + \sqrt{2}} \)
\( = \frac{1}{5 + \sqrt{2}} \times \frac{5 - \sqrt{2}}{5 - \sqrt{2}} \)

\( = \frac{5 - \sqrt{2}}{(5)^2 - (\sqrt{2})^2} \)

\( = \frac{5 - \sqrt{2}}{25 - 2} \)

\( = \frac{5 - \sqrt{2}}{23} \)
In simple words: Use the opposite sign (\( 5-\sqrt{2} \)) to multiply. This makes the square roots at the bottom disappear.

📝 Teacher's Note: The term \( 5-\sqrt{2} \) is the "conjugate" of \( 5+\sqrt{2} \).

🎯 Exam Tip: Use the \( (a+b)(a-b) = a^2 - b^2 \) identity every time you have a two-term denominator.

Question 1C. Rationalize the denominator: \( \frac{1}{\sqrt{3} + \sqrt{2}} \)
Answer:
\( \frac{1}{\sqrt{3} + \sqrt{2}} \)
\( = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \)

\( = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \)

\( = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} \)

\( = \frac{\sqrt{3} - \sqrt{2}}{1} \)
\( = \sqrt{3} - \sqrt{2} \)
In simple words: Multiplying by the conjugate (\( \sqrt{3}-\sqrt{2} \)) clears the denominator completely because \( 3-2 \) is just 1.

📝 Teacher's Note: This is a very common textbook problem where the denominator simplifies to 1.

🎯 Exam Tip: Don't leave your final answer with a denominator of 1; write the numerator alone.

Question 1D. Rationalize the denominator: \( \frac{2}{3 + \sqrt{7}} \)
Answer:
\( \frac{2}{3 + \sqrt{7}} \)
\( = \frac{2}{3 + \sqrt{7}} \times \frac{3 - \sqrt{7}}{3 - \sqrt{7}} \)

\( = \frac{2(3 - \sqrt{7})}{(3)^2 - (\sqrt{7})^2} \)

\( = \frac{2(3 - \sqrt{7})}{9 - 7} \)

\( = \frac{2(3 - \sqrt{7})}{2} \)
\( = 3 - \sqrt{7} \)
In simple words: The '2' at the top cancels out the '2' we got at the bottom after rationalizing.

📝 Teacher's Note: Advise students NOT to multiply out the numerator bracket early. Often it will cancel out with the simplified denominator.

🎯 Exam Tip: Simplify the fraction by cancelling common factors before writing the final result.

Question 1E. Rationalize the denominator: \( \frac{5}{\sqrt{7} - \sqrt{2}} \)
Answer:
\( \frac{5}{\sqrt{7} - \sqrt{2}} \)
\( = \frac{5}{\sqrt{7} - \sqrt{2}} \times \frac{\sqrt{7} + \sqrt{2}}{\sqrt{7} + \sqrt{2}} \)

\( = \frac{5(\sqrt{7} + \sqrt{2})}{(\sqrt{7})^2 - (\sqrt{2})^2} \)

\( = \frac{5(\sqrt{7} + \sqrt{2})}{7 - 2} \)

\( = \frac{5(\sqrt{7} + \sqrt{2})}{5} \)
\( = \sqrt{7} + \sqrt{2} \)
In simple words: The denominator becomes 5, which cancels the 5 on top, leaving a clean answer.

📝 Teacher's Note: Another example where waiting to distribute the numerator coefficient saves time.

🎯 Exam Tip: Always look for the cancellation step (\( 5/5 \)) to simplify your final result.

Question 1F. Rationalize the denominator: \( \frac{42}{2\sqrt{3} + 3\sqrt{2}} \)
Answer:
\( \frac{42}{2\sqrt{3} + 3\sqrt{2}} \)
\( = \frac{42}{2\sqrt{3} + 3\sqrt{2}} \times \frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}} \)

\( = \frac{42(2\sqrt{3} - 3\sqrt{2})}{(2\sqrt{3})^2 - (3\sqrt{2})^2} \)

\( = \frac{84\sqrt{3} - 126\sqrt{2}}{12 - 18} \)

\( = \frac{84\sqrt{3} - 126\sqrt{2}}{-6} \)
\( = -14\sqrt{3} + 21\sqrt{2} \)
\( = 21\sqrt{2} - 14\sqrt{3} \)
\( = 7(3\sqrt{2} - 2\sqrt{3}) \)
In simple words: This one has numbers outside the roots. We square both parts (\( 2^2 \times 3 \) and \( 3^2 \times 2 \)) to find the bottom value, then divide everything by -6.

📝 Teacher's Note: Calculation of \( (2\sqrt{3})^2 = 4 \times 3 = 12 \). This is a common error point.

🎯 Exam Tip: Factor out common terms at the end (like 7) to show the most elegant version of the answer.

Question 1G. Rationalize the denominator: \( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
Answer:
\( \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
\( = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \)

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

\( = \frac{(\sqrt{3})^2 + 2 \times \sqrt{3} \times 1 + (1)^2}{3 - 1} \)

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

\( = \frac{4 + 2\sqrt{3}}{2} \)
\( = 2 + \sqrt{3} \)
In simple words: Since both the top and bottom use root 3 and 1, we end up squaring the top part while rationalizing the bottom.

📝 Teacher's Note: This problem requires both the \( (a+b)(a-b) \) identity for the bottom and \( (a+b)^2 \) for the top.

🎯 Exam Tip: In the final step, divide BOTH 4 and \( 2\sqrt{3} \) by 2. Don't leave it as \( 2 + 2\sqrt{3} \).

Question 1H. Rationalize the denominator: \( \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \)
Answer:
\( \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \)
\( = \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)

\( = \frac{\sqrt{5} \times \sqrt{3} - \sqrt{7} \times \sqrt{3}}{(\sqrt{3})^2} \)

\( = \frac{\sqrt{15} - \sqrt{21}}{3} \)
In simple words: Just multiply the top two roots by the root 3 from the bottom.

📝 Teacher's Note: This is a simple case where the denominator has only one term. Multiply by it directly.

🎯 Exam Tip: Ensure that \( \sqrt{5} \times \sqrt{3} \) is correctly combined into \( \sqrt{15} \).

Question 1I. Rationalize the denominator: \( \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \)
Answer:
\( \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \)
\( = \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} \)

\( = \frac{3(2 - \sqrt{2}) - \sqrt{3}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2} \)

\( = \frac{6 - 3\sqrt{2} - 2\sqrt{3} + \sqrt{6}}{4 - 2} \)

\( = \frac{6 - 3\sqrt{2} - 2\sqrt{3} + \sqrt{6}}{2} \)
In simple words: This requires full multiplication for the top part (FOIL), creating four different terms.

📝 Teacher's Note: When the roots are different (\( \sqrt{2} \) and \( \sqrt{3} \)), they cannot be simplified further. Let the terms stay separate.

🎯 Exam Tip: Double-check the sign of the last term: \( (-\sqrt{3}) \times (-\sqrt{2}) = +\sqrt{6} \).

Question 1H. Rationalize the denominator: \( \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \)
Answer:
\( \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \)
\( = \frac{\sqrt{5} - \sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)

\( \implies \frac{\sqrt{5} \times \sqrt{3} - \sqrt{7} \times \sqrt{3}}{(\sqrt{3})^2} \)

\( \implies \frac{\sqrt{15} - \sqrt{21}}{3} \)
In simple words: To remove the square root from the bottom of a fraction, multiply both the top and the bottom by that same square root.

📝 Teacher's Note: Explain to students that rationalizing the denominator makes the expression easier to work with in complex calculations. Remind them that \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \).

🎯 Exam Tip: Always check if the resulting surds in the numerator can be simplified further before finalizing your answer.

Question 1I. Rationalize the denominator: \( \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \)
Answer:
\( \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \)
\( = \frac{3 - \sqrt{3}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} \)

\( \implies \frac{3(2 - \sqrt{2}) - \sqrt{3}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2} \)

\( \implies \frac{6 - 3\sqrt{2} - 2\sqrt{3} + \sqrt{6}}{4 - 2} \)

\( \implies \frac{6 - 3\sqrt{2} - 2\sqrt{3} + \sqrt{6}}{2} \)
In simple words: When the bottom has two parts, multiply by the "opposite" version (change the plus to a minus) to cancel out the roots.

📝 Teacher's Note: This process uses the identity \( (a+b)(a-b) = a^2 - b^2 \). Make sure students distribute the multiplication in the numerator correctly using the FOIL method.

🎯 Exam Tip: Be very careful with signs; a common mistake is getting the sign of the last term in the numerator (\( -\sqrt{3} \times -\sqrt{2} = +\sqrt{6} \)) wrong.

Question 2. Simplify the following by rationalizing the denominator:
(i) \( \frac{5 + \sqrt{6}}{5 - \sqrt{6}} \)
(ii) \( \frac{4 + \sqrt{8}}{4 - \sqrt{8}} \)
(iii) \( \frac{\sqrt{15} + 3}{\sqrt{15} - 3} \)
(iv) \( \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}} \)
(v) \( \frac{3\sqrt{5} + \sqrt{7}}{3\sqrt{5} - \sqrt{7}} \)
(vi) \( \frac{2\sqrt{3} - \sqrt{6}}{2\sqrt{3} + \sqrt{6}} \)
(vii) \( \frac{5\sqrt{3} - \sqrt{15}}{5\sqrt{3} + \sqrt{15}} \)
(viii) \( \frac{2\sqrt{6} - \sqrt{5}}{3\sqrt{5} - 2\sqrt{6}} \)
(ix) \( \frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \)
(x) \( \frac{\sqrt{12} + \sqrt{18}}{\sqrt{75} - \sqrt{50}} \)
Answer:
(i) \( \frac{5 + \sqrt{6}}{5 - \sqrt{6}} = \frac{5 + \sqrt{6}}{5 - \sqrt{6}} \times \frac{5 + \sqrt{6}}{5 + \sqrt{6}} \)

\( \implies \frac{(5 + \sqrt{6})^2}{(5)^2 - (\sqrt{6})^2} = \frac{25 + 6 + 10\sqrt{6}}{25 - 6} \)

\( \implies \frac{31 + 10\sqrt{6}}{19} \)

(ii) \( \frac{4 + \sqrt{8}}{4 - \sqrt{8}} = \frac{4 + \sqrt{8}}{4 - \sqrt{8}} \times \frac{4 + \sqrt{8}}{4 + \sqrt{8}} \)

\( \implies \frac{(4 + \sqrt{8})^2}{(4)^2 - (\sqrt{8})^2} = \frac{16 + 8 + 8\sqrt{8}}{16 - 8} \)

\( \implies \frac{24 + 8\sqrt{8}}{8} = 3 + \sqrt{8} \)

(iii) \( \frac{\sqrt{15} + 3}{\sqrt{15} - 3} = \frac{\sqrt{15} + 3}{\sqrt{15} - 3} \times \frac{\sqrt{15} + 3}{\sqrt{15} + 3} \)

\( \implies \frac{(\sqrt{15} + 3)^2}{(\sqrt{15})^2 - (3)^2} = \frac{15 + 9 + 6\sqrt{15}}{15 - 9} \)

\( \implies \frac{24 + 6\sqrt{15}}{6} = 4 + \sqrt{15} \)

(iv) \( \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} - \sqrt{5}} \)

\( \implies \frac{(\sqrt{7} - \sqrt{5})^2}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{7 + 5 - 2\sqrt{35}}{7 - 5} = \frac{12 - 2\sqrt{35}}{2} \)

\( \implies 6 - \sqrt{35} \)

(v) \( \frac{3\sqrt{5} + \sqrt{7}}{3\sqrt{5} - \sqrt{7}} = \frac{3\sqrt{5} + \sqrt{7}}{3\sqrt{5} - \sqrt{7}} \times \frac{3\sqrt{5} + \sqrt{7}}{3\sqrt{5} + \sqrt{7}} \)

\( \implies \frac{(3\sqrt{5} + \sqrt{7})^2}{(3\sqrt{5})^2 - (\sqrt{7})^2} = \frac{45 + 7 + 6\sqrt{35}}{45 - 7} \)

\( \implies \frac{52 + 6\sqrt{35}}{38} = \frac{26 + 3\sqrt{35}}{19} \)

(vi) \( \frac{2\sqrt{3} - \sqrt{6}}{2\sqrt{3} + \sqrt{6}} = \frac{2\sqrt{3} - \sqrt{6}}{2\sqrt{3} + \sqrt{6}} \times \frac{2\sqrt{3} - \sqrt{6}}{2\sqrt{3} - \sqrt{6}} \)

\( \implies \frac{(2\sqrt{3} - \sqrt{6})^2}{(2\sqrt{3})^2 - (\sqrt{6})^2} = \frac{12 + 6 - 4\sqrt{18}}{12 - 6} \)

\( \implies \frac{18 - 4\sqrt{18}}{6} = \frac{9 - 2\sqrt{18}}{3} = \frac{9 - 6\sqrt{2}}{3} = 3 - 2\sqrt{2} \)

(vii) \( \frac{5\sqrt{3} - \sqrt{15}}{5\sqrt{3} + \sqrt{15}} = \frac{5\sqrt{3} - \sqrt{15}}{5\sqrt{3} + \sqrt{15}} \times \frac{5\sqrt{3} - \sqrt{15}}{5\sqrt{3} - \sqrt{15}} \)

\( \implies \frac{(5\sqrt{3} - \sqrt{15})^2}{(5\sqrt{3})^2 - (\sqrt{15})^2} = \frac{75 + 15 - 10\sqrt{45}}{75 - 15} \)

\( \implies \frac{90 - 10\sqrt{45}}{60} = \frac{9 - \sqrt{45}}{6} = \frac{9 - 3\sqrt{5}}{6} = \frac{3 - \sqrt{5}}{2} \)

(viii) \( \frac{2\sqrt{6} - \sqrt{5}}{3\sqrt{5} - 2\sqrt{6}} = \frac{2\sqrt{6} - \sqrt{5}}{3\sqrt{5} - 2\sqrt{6}} \times \frac{3\sqrt{5} + 2\sqrt{6}}{3\sqrt{5} + 2\sqrt{6}} \)

\( \implies \frac{6\sqrt{30} + 24 - 15 - 2\sqrt{30}}{(3\sqrt{5})^2 - (2\sqrt{6})^2} \)

\( \implies \frac{6\sqrt{30} + 9 - 2\sqrt{30}}{45 - 24} = \frac{4\sqrt{30} + 9}{21} \)

(ix) \( \frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} = \frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \times \frac{\sqrt{48} - \sqrt{18}}{\sqrt{48} - \sqrt{18}} \)

\( \implies \frac{7\sqrt{144} - 7\sqrt{54} - 5\sqrt{96} + 5\sqrt{36}}{(\sqrt{48})^2 - (\sqrt{18})^2} \)

\( \implies \frac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{48 - 18} \)

\( \implies \frac{114 - 41\sqrt{6}}{30} \)

(x) \( \frac{\sqrt{12} + \sqrt{18}}{\sqrt{75} - \sqrt{50}} = \frac{\sqrt{12} + \sqrt{18}}{\sqrt{75} - \sqrt{50}} \times \frac{\sqrt{75} + \sqrt{50}}{\sqrt{75} + \sqrt{50}} \)

\( \implies \frac{(2\sqrt{3} + 3\sqrt{2})(5\sqrt{3} + 5\sqrt{2})}{(\sqrt{75})^2 - (\sqrt{50})^2} \)

\( \implies \frac{30 + 10\sqrt{6} + 15\sqrt{6} + 30}{75 - 50} \)

\( \implies \frac{60 + 25\sqrt{6}}{25} = \frac{12 + 5\sqrt{6}}{5} \)
In simple words: To simplify complex fractions with square roots, we use the formula \( (a+b)(a-b) = a^2 - b^2 \) for the bottom part and multiply carefully on top to make the answer as clean as possible.

📝 Teacher's Note: Remind students to simplify surds before multiplying if possible (e.g., \( \sqrt{12} = 2\sqrt{3} \)). This makes the numbers smaller and the calculation easier.

🎯 Exam Tip: If the question asks for a simplified form, always divide out common factors in the final fraction to get full marks.

Question 3. Evaluate the following:
(i) \( \frac{3}{5 - \sqrt{3}} + \frac{2}{5 + \sqrt{3}} \)
(ii) \( \frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}} \)
(iii) \( \frac{\sqrt{5} - 2}{\sqrt{5} + 2} - \frac{\sqrt{5} + 2}{\sqrt{5} - 2} \)
(iv) \( \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} - \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} \)
(v) \( \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \)
Answer:
(i) \( \frac{3}{5 - \sqrt{3}} + \frac{2}{5 + \sqrt{3}} = \frac{3(5 + \sqrt{3}) + 2(5 - \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})} \)

\( \implies \frac{15 + 3\sqrt{3} + 10 - 2\sqrt{3}}{(5)^2 - (\sqrt{3})^2} = \frac{25 + \sqrt{3}}{25 - 3} \)

\( \implies \frac{25 + \sqrt{3}}{22} \)

(ii) \( \frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}} = \frac{(4 + \sqrt{5})^2 + (4 - \sqrt{5})^2}{(4 - \sqrt{5})(4 + \sqrt{5})} \)

\( \implies \frac{16 + 5 + 8\sqrt{5} + 16 + 5 - 8\sqrt{5}}{16 - 5} \)

\( \implies \frac{42}{11} \)

(iii) \( \frac{\sqrt{5} - 2}{\sqrt{5} + 2} - \frac{\sqrt{5} + 2}{\sqrt{5} - 2} = \frac{(\sqrt{5} - 2)^2 - (\sqrt{5} + 2)^2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} \)

\( \implies \frac{5 + 4 - 4\sqrt{5} - 5 - 4 - 4\sqrt{5}}{(\sqrt{5})^2 - (2)^2} \)

\( \implies \frac{-8\sqrt{5}}{5 - 4} = -8\sqrt{5} \)

(iv) \( \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} - \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} = \frac{(\sqrt{7} - \sqrt{3})^2 - (\sqrt{7} + \sqrt{3})^2}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} \)

\( \implies \frac{7 + 3 - 2\sqrt{21} - 7 - 3 - 2\sqrt{21}}{(\sqrt{7})^2 - (\sqrt{3})^2} \)

\( \implies \frac{-4\sqrt{21}}{7 - 3} = -\sqrt{21} \)

(v) \( \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \)

\( \implies \frac{5 + 3 + 2\sqrt{15} + 5 + 3 - 2\sqrt{15}}{5 - 3} \)

\( \implies \frac{16}{2} = 8 \)
In simple words: To add or subtract these fractions, we find a common denominator. Because the denominators are "conjugates" (like twins with different signs), the bottom part becomes a nice simple number.

📝 Teacher's Note: Using common denominators is faster than rationalizing each term separately. Show students how the middle "2ab" terms often cancel out in addition or subtract in subtraction.

🎯 Exam Tip: When subtracting, always put the second numerator in brackets to avoid missing a sign change (\( -(a+b) = -a - b \)).

Question 4. Simplify each of the following:
(i) \( \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} \)
(ii) \( \frac{3\sqrt{2}}{\sqrt{6} - \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} + \frac{2\sqrt{3}}{\sqrt{6} + 2} \)
(iii) \( \frac{6}{2\sqrt{3} - \sqrt{6}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} - \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} \)
(iv) \( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \)
(v) \( \frac{4\sqrt{3}}{2 - \sqrt{2}} - \frac{30}{4\sqrt{3} - 3\sqrt{2}} - \frac{3\sqrt{2}}{3 + 2\sqrt{3}} \)
Answer:
(i) Rationalizing the denominator of each term, we have
\( \frac{\sqrt{6}(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} + \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{(\sqrt{6} + \sqrt{3})(\sqrt{6} - \sqrt{3})} - \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})} \)

\( \implies \frac{\sqrt{12} - \sqrt{18}}{2 - 3} + \frac{3\sqrt{12} - 3\sqrt{6}}{6 - 3} - \frac{4\sqrt{18} - 4\sqrt{6}}{6 - 2} \)

\( \implies \frac{\sqrt{12} - \sqrt{18}}{-1} + \frac{3\sqrt{12} - 3\sqrt{6}}{3} - \frac{4\sqrt{18} - 4\sqrt{6}}{4} \)

\( \implies \sqrt{18} - \sqrt{12} + \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6} \)

\( \implies 0 \)

(ii) Rationalizing the denominator of each term, we have
\( \frac{3\sqrt{2}(\sqrt{6} + \sqrt{3})}{(\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3})} - \frac{4\sqrt{3}(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} + \frac{2\sqrt{3}(\sqrt{6} - 2)}{(\sqrt{6} + 2)(\sqrt{6} - 2)} \)

\( \implies \frac{3\sqrt{12} + 3\sqrt{6}}{6 - 3} - \frac{4\sqrt{18} + 4\sqrt{6}}{6 - 2} + \frac{2\sqrt{18} - 4\sqrt{3}}{6 - 4} \)

\( \implies \frac{3\sqrt{12} + 3\sqrt{6}}{3} - \frac{4\sqrt{18} + 4\sqrt{6}}{4} + \frac{2\sqrt{18} - 4\sqrt{3}}{2} \)

\( \implies \sqrt{12} + \sqrt{6} - \sqrt{18} - \sqrt{6} + \sqrt{18} - 2\sqrt{3} \)

\( \implies \sqrt{12} - 2\sqrt{3} = 2\sqrt{3} - 2\sqrt{3} = 0 \)

(iii) Rationalizing the denominator of each term, we have
\( \frac{6(2\sqrt{3} + \sqrt{6})}{(2\sqrt{3} - \sqrt{6})(2\sqrt{3} + \sqrt{6})} + \frac{\sqrt{6}(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} - \frac{4\sqrt{3}(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} \)

\( \implies \frac{12\sqrt{3} + 6\sqrt{6}}{12 - 6} + \frac{\sqrt{18} - \sqrt{12}}{3 - 2} - \frac{4\sqrt{18} + 4\sqrt{6}}{6 - 2} \)

\( \implies \frac{12\sqrt{3} + 6\sqrt{6}}{6} + \frac{\sqrt{18} - \sqrt{12}}{1} - \frac{4\sqrt{18} + 4\sqrt{6}}{4} \)

\( \implies 2\sqrt{3} + \sqrt{6} + \sqrt{18} - \sqrt{12} - \sqrt{18} - \sqrt{6} \)

\( \implies 2\sqrt{3} - \sqrt{12} = 2\sqrt{3} - 2\sqrt{3} = 0 \)

(iv) Rationalizing the denominator of each term, we have
\( \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10} + \sqrt{3})(\sqrt{10} - \sqrt{3})} - \frac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})} - \frac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{(\sqrt{15} + 3\sqrt{2})(\sqrt{15} - 3\sqrt{2})} \)

\( \implies \frac{7\sqrt{30} - 21}{10 - 3} - \frac{2\sqrt{30} - 10}{6 - 5} - \frac{3\sqrt{30} - 18}{15 - 18} \)

\( \implies \frac{7\sqrt{30} - 21}{7} - \frac{2\sqrt{30} - 10}{1} - \frac{3\sqrt{30} - 18}{-3} \)

\( \implies \sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6 \)

\( \implies -1 \)

(v) Rationalizing the denominator of each term, we have
\( \frac{4\sqrt{3}(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} - \frac{30(4\sqrt{3} + 3\sqrt{2})}{(4\sqrt{3} - 3\sqrt{2})(4\sqrt{3} + 3\sqrt{2})} - \frac{3\sqrt{2}(3 - 2\sqrt{3})}{(3 + 2\sqrt{3})(3 - 2\sqrt{3})} \)

\( \implies \frac{8\sqrt{3} + 4\sqrt{6}}{4 - 2} - \frac{120\sqrt{3} + 90\sqrt{2}}{48 - 18} - \frac{9\sqrt{2} - 6\sqrt{6}}{9 - 12} \)

\( \implies \frac{8\sqrt{3} + 4\sqrt{6}}{2} - \frac{120\sqrt{3} + 90\sqrt{2}}{30} + \frac{9\sqrt{2} - 6\sqrt{6}}{3} \)

\( \implies 4\sqrt{3} + 2\sqrt{6} - 4\sqrt{3} - 3\sqrt{2} + 3\sqrt{2} - 2\sqrt{6} \)

\( \implies 0 \)
In simple words: This is a long problem where we handle three separate parts one by one. After cleaning up all the square roots, the final total surprisingly turns out to be zero or a small simple number.

📝 Teacher's Note: This is an exhaustive problem. Encourage students to rationalize each term separately on different lines to avoid clutter and small mistakes. If the final answer is zero, it's usually a sign that they followed the steps correctly.

🎯 Exam Tip: For expressions like this, 1 mark is usually awarded for each correctly rationalized term. Show every simplification clearly.

Question 5. Evaluate \( \frac{\sqrt{2.5} - \sqrt{0.75}}{\sqrt{2.5} + \sqrt{0.75}} \) and find p and q such that the result is \( p + q\sqrt{30} \).
Answer:
\( \frac{\sqrt{2.5} - \sqrt{0.75}}{\sqrt{2.5} + \sqrt{0.75}} = \frac{\sqrt{2.5} - \sqrt{0.75}}{\sqrt{2.5} + \sqrt{0.75}} \times \frac{\sqrt{2.5} - \sqrt{0.75}}{\sqrt{2.5} - \sqrt{0.75}} \)

\( \implies \frac{(\sqrt{2.5} - \sqrt{0.75})^2}{(\sqrt{2.5})^2 - (\sqrt{0.75})^2} \)

\( \implies \frac{2.5 - 2 \times \sqrt{2.5} \times \sqrt{0.75} + 0.75}{2.5 - 0.75} \)

\( \implies \frac{3.25 - 2 \times \sqrt{0.25 \times 10} \times \sqrt{0.25 \times 3}}{1.75} \)

\( \implies \frac{3.25 - 2 \times 0.25\sqrt{30}}{1.75} \)

\( \implies \frac{3.25 - 0.5\sqrt{30}}{1.75} \)

\( \implies \frac{3.25}{1.75} - \frac{0.5}{1.75}\sqrt{30} \)

\( \implies \frac{325}{175} - \frac{50}{175}\sqrt{30} \)

\( \implies \frac{13}{7} - \frac{2}{7}\sqrt{30} = \frac{13}{7} + \left( -\frac{2}{7} \right)\sqrt{30} \)

\( \implies p = \frac{13}{7} \text{ and } q = -\frac{2}{7} \)
In simple words: We simplify the messy square roots step by step until they look like a simple math pattern. Then we just match the numbers up to find the values of p and q.

📝 Teacher's Note: This problem involves "comparison of coefficients." Students must first rationalize and simplify the entire LHS before they can determine the values of p and q.

🎯 Exam Tip: To avoid decimal confusion, multiply the numerator and denominator by 100 before simplifying the fraction (\( \frac{3.25}{1.75} = \frac{325}{175} \)).

Question 6A. Find a and b if \( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b\sqrt{3} \)
Answer:
\( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)

\( \implies \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} \)

\( \implies \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \)

\( \implies 2 + (-1)\sqrt{3} = a + b\sqrt{3} \)
Hence, a = 2 and b = -1.
In simple words: Rationalizing gives us \( 2 - \sqrt{3} \). By looking at the pattern, we see that 'a' is 2 and 'b' is the negative 1 hiding in front of the root.

📝 Teacher's Note: Remind students that if there is no number before the root, the coefficient is either 1 or -1.

🎯 Exam Tip: Always state the values of a and b clearly at the end of your derivation.

Question 6B. Find a and b if \( \frac{3 + \sqrt{7}}{3 - \sqrt{7}} = a + b\sqrt{7} \)
Answer:
\( \frac{3 + \sqrt{7}}{3 - \sqrt{7}} = \frac{3 + \sqrt{7}}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}} \)

\( \implies \frac{(3 + \sqrt{7})^2}{(3)^2 - (\sqrt{7})^2} = \frac{9 + 6\sqrt{7} + 7}{9 - 7} \)

\( \implies \frac{16 + 6\sqrt{7}}{2} = 8 + 3\sqrt{7} \)

\( \implies a + b\sqrt{7} \)
Hence, a = 8 and b = 3.
In simple words: We find that a is 8 and b is 3 after rationalizing.

📝 Teacher's Note: This is a standard comparison problem. Distributing the division by 2 is the most important final step.

🎯 Exam Tip: Ensure that both parts of the numerator are divided by the denominator.

Question 6C. Find a and b if \( \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a + b\sqrt{3} \)
Answer:
\( \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{5(7 - 4\sqrt{3}) + 2\sqrt{3}(7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \)

\( \implies \frac{35 - 20\sqrt{3} + 14\sqrt{3} - 24}{49 - 48} = \frac{11 - 6\sqrt{3}}{1} \)

\( \implies 11 + (-6)\sqrt{3} = a + b\sqrt{3} \)
Hence, a = 11 and b = -6.
In simple words: Multiplying the terms carefully gives us 11 minus 6-root-3. So a is 11 and b is -6.

📝 Teacher's Note: Note the denominator calculation: \( (4\sqrt{3})^2 = 16 \times 3 = 48 \). This is where students often make arithmetic mistakes.

🎯 Exam Tip: Always rewrite the final expression with a plus sign for comparison, like \( 11 + (-6)\sqrt{3} \), to correctly identify a negative 'b' value.

Question 6D. Find a and b if \( \frac{1}{\sqrt{5} - \sqrt{3}} = a\sqrt{5} - b\sqrt{3} \)
Answer:
\( \frac{1}{\sqrt{5} - \sqrt{3}} = \frac{1}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} \)

\( \implies \frac{\sqrt{5} + \sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \frac{\sqrt{5} + \sqrt{3}}{2} \)

\( \implies \frac{1}{2}\sqrt{5} + \frac{1}{2}\sqrt{3} = \frac{1}{2}\sqrt{5} - \left( -\frac{1}{2} \right)\sqrt{3} \)

\( \implies a\sqrt{5} - b\sqrt{3} \)
Hence, a = \( \frac{1}{2} \) and b = \( -\frac{1}{2} \).
In simple words: The final fraction is half of root 5 plus half of root 3. So a is 1/2 and b is negative half.

📝 Teacher's Note: Pay attention to the specific format asked in the question (\( a\sqrt{5} - b\sqrt{3} \)). Since our answer has a plus, 'b' must be negative.

🎯 Exam Tip: Check the target format carefully before choosing the sign for 'b'.

Question 6E. Find a and b if \( \frac{\sqrt{3} - 2}{\sqrt{3} + 2} = a\sqrt{3} - b \)
Answer:
\( \frac{\sqrt{3} - 2}{\sqrt{3} + 2} \times \frac{\sqrt{3} - 2}{\sqrt{3} - 2} = \frac{(\sqrt{3} - 2)^2}{(\sqrt{3})^2 - (2)^2} \)

\( \implies \frac{3 - 2\sqrt{3} \times 2 + 4}{3 - 4} = \frac{7 - 4\sqrt{3}}{-1} \)

\( \implies -(7 - 4\sqrt{3}) = -7 + 4\sqrt{3} \)

\( \implies 4\sqrt{3} - 7 = 4\sqrt{3} - (-7) \)

\( \implies a\sqrt{3} + b \)
Hence, a = 4 and b = -7.
In simple words: After rationalizing, we get 4-root-3 minus 7. This matches up to give a=4 and b=-7.

📝 Teacher's Note: The negative 1 in the denominator flips the signs of the numerator. This is a very common place for mistakes.

🎯 Exam Tip: Be sure to re-order the terms to match the required format (\( a\sqrt{3} \) first, then \( b \)).

Question 6F. Find a and b if \( \frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} = a - b\sqrt{77} \)
Answer:
\( \frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} \times \frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} - \sqrt{7}} = \frac{(\sqrt{11} - \sqrt{7})^2}{(\sqrt{11})^2 - (\sqrt{7})^2} \)

\( \implies \frac{11 + 7 - 2\sqrt{77}}{11 - 7} = \frac{18 - 2\sqrt{77}}{4} \)

\( \implies \frac{18}{4} - \frac{2}{4}\sqrt{77} = \frac{9}{2} - \frac{1}{2}\sqrt{77} \)

\( \implies a - b\sqrt{77} \)
Hence, a = \( \frac{9}{2} \) and b = \( \frac{1}{2} \).
In simple words: The result is 4.5 minus half of root 77. This gives a=9/2 and b=1/2.

📝 Teacher's Note: Note that \( \sqrt{11} \times \sqrt{7} = \sqrt{77} \). The target format already has a minus sign, so 'b' itself is positive.

🎯 Exam Tip: Always reduce fractions like 18/4 to their lowest terms (9/2) for full marks.

Question 6G. Find a and b if \( \frac{7\sqrt{3} - 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} = a - b\sqrt{6} \)
Answer:
\( \frac{7\sqrt{3} - 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} \times \frac{4\sqrt{3} - 3\sqrt{2}}{4\sqrt{3} - 3\sqrt{2}} = \frac{7\sqrt{3}(4\sqrt{3} - 3\sqrt{2}) - 5\sqrt{2}(4\sqrt{3} - 3\sqrt{2})}{(4\sqrt{3})^2 - (3\sqrt{2})^2} \)

\( \implies \frac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{48 - 18} = \frac{114 - 41\sqrt{6}}{30} \)

\( \implies \frac{114}{30} - \frac{41}{30}\sqrt{6} = \frac{19}{5} - \frac{41}{30}\sqrt{6} \)

\( \implies a - b\sqrt{6} \)
Hence, a = \( \frac{19}{5} \) and b = \( \frac{41}{30} \).
In simple words: After full multiplication, we find a=11/3 and b=41/30.

📝 Teacher's Note: This is a high-level rationalization problem. Accuracy in calculating \( (4\sqrt{3})^2 = 48 \) and \( (3\sqrt{2})^2 = 18 \) is paramount.

🎯 Exam Tip: Simplify the constant fraction (114/30) by dividing both parts by 6 to reach the most correct answer.

Question 6H. Find a and b if \( \frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt{6} \)
Answer:
\( \frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} \times \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} = \frac{(\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2})^2 - (2\sqrt{3})^2} \)

\( \implies \frac{\sqrt{2}(3\sqrt{2} + 2\sqrt{3}) + \sqrt{3}(3\sqrt{2} + 2\sqrt{3})}{(9 \times 2) - (4 \times 3)} = \frac{(3 \times 2 + 2\sqrt{6}) + (3\sqrt{6} + 2 \times 3)}{18 - 12} \)

\( \implies \frac{6 + 2\sqrt{6} + 3\sqrt{6} + 6}{6} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6} \)

\( \implies 2 - \left( -\frac{5}{6} \right)\sqrt{6} = a - b\sqrt{6} \)
Hence, a = 2 and b = \( -\frac{5}{6} \).
In simple words: After solving, we get 2 plus five-sixths root 6. Since the target format has a minus, b must be the negative version of our fraction.

📝 Teacher's Note: This requires two sign-flips in logic. Be patient and show the step where plus becomes "minus of minus."

🎯 Exam Tip: If your result has a '+' but the question shows '-', the value for 'b' MUST be negative.

Question 6I. Find a and b if \( \frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + b\sqrt{5} \)
Answer:
\( \frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = \frac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{7^2 - (\sqrt{5})^2} \)

\( \implies \frac{7^2 + 2 \times 7 \times \sqrt{5} + (\sqrt{5})^2 - (7^2 - 2 \times 7 \times \sqrt{5} + (\sqrt{5})^2)}{49 - 5} \)

\( \implies \frac{49 + 14\sqrt{5} + 5 - 49 + 14\sqrt{5} - 5}{44} = \frac{28\sqrt{5}}{44} = \frac{7\sqrt{5}}{11} \)

\( \implies 0 + \frac{7}{11}\sqrt{5} = a + b\sqrt{5} \)
Hence, a = 0 and b = \( \frac{7}{11} \).
In simple words: Since all the whole numbers cancel each other out, a is 0. The remaining root part gives b=7/11.

📝 Teacher's Note: This is an application of the identity \( (x+y)^2 - (x-y)^2 = 4xy \). It's a much faster way to solve the numerator.

🎯 Exam Tip: Don't forget that if no rational part remains, the value for 'a' is zero.

Question 6J. Find a and b if \( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3} \)
Answer:
\( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} \)

\( \implies \frac{(\sqrt{3})^2 - 2\sqrt{3} + 1 + (\sqrt{3})^2 + 2\sqrt{3} + 1}{3 - 1} \)

\( \implies \frac{3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1}{2} = \frac{8}{2} = 4 \)

\( \implies 4 + 0\sqrt{3} = a + b\sqrt{3} \)
Hence, a = 4 and b = 0.
In simple words: The roots cancel out completely this time, leaving only a whole number. This means b must be 0.

📝 Teacher's Note: This identity is \( (x+y)^2 + (x-y)^2 = 2(x^2 + y^2) \). Use this to simplify the numerator instantly.

🎯 Exam Tip: If the result is a rational number like 4, clearly state that the coefficient of the irrational part (b) is 0.

Question 7. (i) Given \( x = 7 + 4\sqrt{3} \), find \( \sqrt{x} + \frac{1}{\sqrt{x}} \)
Answer:
\( (\sqrt{x} + \frac{1}{\sqrt{x}})^2 = x + \frac{1}{x} + 2 \) ----(1)
We will first find out \( x + \frac{1}{x} \)
\( x + \frac{1}{x} = (7 + 4\sqrt{3}) + \frac{1}{(7 + 4\sqrt{3})} \)

\( \implies \frac{(7 + 4\sqrt{3})^2 + 1}{(7 + 4\sqrt{3})} = \frac{49 + 48 + 56\sqrt{3} + 1}{(7 + 4\sqrt{3})} \)

\( \implies \frac{98 + 56\sqrt{3}}{7 + 4\sqrt{3}} = \frac{14(7 + 4\sqrt{3})}{(7 + 4\sqrt{3})} = 14 \)
Substituting in (1), we get
\( (\sqrt{x} + \frac{1}{\sqrt{x}})^2 = 14 + 2 = 16 \)

\( \therefore \sqrt{x} + \frac{1}{\sqrt{x}} = 4 \)
In simple words: We square the target first to find its value, which turns out to be 16. Then, we take the square root of 16 to get the final answer of 4.

📝 Teacher's Note: This is a brilliant trick. Squaring the required term simplifies the problem because the cross-term \( 2 \cdot a \cdot b \) becomes just \( 2 \cdot 1 = 2 \).

🎯 Exam Tip: Remember to take the square root at the very end to finish the problem.

Question 7. (ii) Find \( x^2 + \frac{1}{x^2} \) if \( x = 7 + 4\sqrt{3} \).
Answer:
\( (x^2 + \frac{1}{x^2}) = (x + \frac{1}{x})^2 - 2 \) ----(1)
From the previous part, we know \( x + \frac{1}{x} = 14 \)
Substituting in (1):
\( x^2 + \frac{1}{x^2} = (14)^2 - 2 = 196 - 2 = 194 \)

\( \therefore x^2 + \frac{1}{x^2} = 194 \)
In simple words: Since we know that x + 1/x is 14, we just square 14 and subtract 2 to find the answer.

📝 Teacher's Note: If a problem has multiple sub-parts, use the results of earlier parts to solve the later ones faster.

🎯 Exam Tip: The identity \( x^2 + y^2 = (x+y)^2 - 2xy \) is essential for these types of questions.

Question 7. (iii) Find \( x^3 + \frac{1}{x^3} \) if \( x = 7 + 4\sqrt{3} \).
Answer:
\( (x^3 + \frac{1}{x^3}) = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \) ----(1)
From previous work, \( x + \frac{1}{x} = 14 \)
Substituting in (1):
\( x^3 + \frac{1}{x^3} = (14)^3 - 3 \times 14 = 2744 - 42 = 2702 \)

\( \therefore x^3 + \frac{1}{x^3} = 2702 \)
In simple words: We use the cubed version of our formula with the same number 14.

📝 Teacher's Note: For higher powers like cubes, ensure students can calculate \( 14^3 \) accurately by doing \( 14 \times 14 \times 14 \).

🎯 Exam Tip: Memorizing cubes up to 15 or 20 is a major time-saver in mathematics exams.

Question 7. (iv) Find \( x + \frac{1}{x} \) given \( x = 7 + 4\sqrt{3} \).
Answer:
\( x = 7 + 4\sqrt{3} \)
\( \frac{1}{x} = \frac{1}{7 + 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{7^2 - (4\sqrt{3})^2} = \frac{7 - 4\sqrt{3}}{49 - 48} = 7 - 4\sqrt{3} \)
\( x + \frac{1}{x} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \)
Hence, \( (x + \frac{1}{x})^2 = (14)^2 = 196 \)
In simple words: 1/x is just the twin of x with a minus sign. Adding them together wipes out the roots and leaves us with 14.

📝 Teacher's Note: This sub-part actually proves the groundwork for parts (i), (ii), and (iii). If \( a^2 - b^2 = 1 \), then \( \frac{1}{a+b} = a-b \).

🎯 Exam Tip: This "Difference of 1" shortcut is extremely common in entrance exams and competitive tests.

Question 8. Given \( x = 4 - \sqrt{15} \), evaluate the following:
(i) \( \frac{1}{x} \)
(ii) \( x + \frac{1}{x} \)
(iii) \( x^2 + \frac{1}{x^2} \)
(iv) \( x^3 + \frac{1}{x^3} \)
(v) \( \left( x + \frac{1}{x} \right)^2 \)
Answer:
(i) \( \frac{1}{x} = \frac{1}{4 - \sqrt{15}} = \frac{1}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} = \frac{4 + \sqrt{15}}{16 - 15} = 4 + \sqrt{15} \)

(ii) \( x + \frac{1}{x} = (4 - \sqrt{15}) + (4 + \sqrt{15}) = 8 \)

(iii) \( x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 8^2 - 2 = 64 - 2 = 62 \)

(iv) \( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = 8^3 - 3(8) = 512 - 24 = 488 \)

(v) \( \left( x + \frac{1}{x} \right)^2 = (8)^2 = 64 \)
In simple words: This is a chain of logic. We find 1/x first, then their sum (8), then use that sum to find squares and cubes easily.

📝 Teacher's Note: This mirrors Question 7. It's excellent reinforcement for the algebraic sum-of-reciprocals concept.

🎯 Exam Tip: These problems are modular. If you get the sum in part (ii) wrong, every subsequent part will be wrong. Check your addition twice!

Question 9. Evaluate \( x^2 - y^2 \) if \( x = \frac{2 + \sqrt{5}}{2 - \sqrt{5}} \) and \( y = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} \)
Answer:
\( x = \frac{2 + \sqrt{5}}{2 - \sqrt{5}} = \frac{(2 + \sqrt{5})^2}{2^2 - (\sqrt{5})^2} = \frac{4 + 5 + 4\sqrt{5}}{-1} = -9 - 4\sqrt{5} \)
\( y = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} = \frac{(2 - \sqrt{5})^2}{2^2 - (\sqrt{5})^2} = \frac{4 + 5 - 4\sqrt{5}}{-1} = -9 + 4\sqrt{5} \)
\( \therefore x^2 - y^2 = (x+y)(x-y) \)
\( = (-9 - 4\sqrt{5} - 9 + 4\sqrt{5}) \times (-9 - 4\sqrt{5} + 9 - 4\sqrt{5}) \)
\( = (-18) \times (-8\sqrt{5}) = 144\sqrt{5} \)
In simple words: We rationalize x and y first. Then we add them and subtract them, and multiply those two results together to find the final answer.

📝 Teacher's Note: Use the identity \( x^2 - y^2 = (x+y)(x-y) \). It's always faster than squaring both messy expressions separately.

🎯 Exam Tip: Keep track of the negative sign from the denominator (\( 4 - 5 = -1 \)). It flips the entire numerator.

Question 10. Evaluate the following given \( x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \) and \( y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \):
(i) \( x^2 + y^2 \)
(ii) \( x^3 + y^3 \)
(iii) \( x^2 - y^2 + xy \)
Answer:
\( (x+y) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2 + (\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}}{2} = \frac{8}{2} = 4 \)
\( xy = 1 \) (as they are reciprocals)
(i) \( x^2 + y^2 = (x + y)^2 - 2xy = 16 - 2 = 14 \)
(ii) \( x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 64 - 3(4) = 64 - 12 = 52 \)
(iii) \( (x - y) = \frac{(\sqrt{3} + 1)^2 - (\sqrt{3} - 1)^2}{3 - 1} = \frac{4\sqrt{3}}{2} = 2\sqrt{3} \)
\( x^2 - y^2 + xy = (x + y)(x - y) + xy = 4 \times 2\sqrt{3} + 1 = 8\sqrt{3} + 1 \)
In simple words: This problem is all about the sum and product of x and y. Once you know their sum is 4 and their product is 1, all parts become easy.

📝 Teacher's Note: This is a very common exam pattern. Master the sum (\( x+y \)) and product (\( xy \)) first, then use identities.

🎯 Exam Tip: Notice that for these specific values, \( xy = 1 \). This always happens when x and y are conjugates of each other in the numerator/denominator.

Question 11. Evaluate (i) \( x^2 + y^2 \) and (ii) \( x^3 + y^3 \) if \( x = \frac{1}{3 - 2\sqrt{2}} \) and \( y = \frac{1}{3 + 2\sqrt{2}} \).
Answer:
\( x = \frac{1}{3 - 2\sqrt{2}} = 3 + 2\sqrt{2} \)
\( y = \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2} \)
\( x + y = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \)
\( xy = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1 \)
(i) \( x^2 + y^2 = (x + y)^2 - 2xy = 6^2 - 2(1) = 36 - 2 = 34 \)
(ii) \( x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 6^3 - 3(1)(6) = 216 - 18 = 198 \)
In simple words: We find that x and y add up to 6 and multiply to 1. This makes the square (34) and the cube (198) easy to find.

📝 Teacher's Note: Rationalizing x and y is the very first step. Once they are simple binomials (\( 3 \pm 2\sqrt{2} \)), the problem is much cleaner.

🎯 Exam Tip: Always show the rationalization of x and y separately before starting the evaluations.