test by ruchi

Exercise 1.3

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

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

\( \implies \frac{\sqrt{15} - \sqrt{21}}{3} \)
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.