OP Malhotra Class 9 Maths Solutions Chapter 1 Rational and Irrational Numbers Exercise 1 (C)

S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational And Irrational Numbers Ex 1(C)

Simplify:

Question 1.
(a) \( \sqrt{\frac{1}{3}} \)
(b) \( \frac{5}{\sqrt{12}} \)
(c) \( \sqrt{1 \frac{46}{75}} \)
Answer:
(a) \( \frac{1}{\sqrt{3}} = \frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
(Multiplying and dividing by \( \sqrt{3} \))
\( \implies \frac{\sqrt{3}}{3} \)
(b) \( \frac{5}{\sqrt{12}} = \frac{5}{\sqrt{2 \times 2 \times 3}} = \frac{5}{2\sqrt{3}} \)
Multiplying and dividing by \( \sqrt{3} \)
\( \implies \frac{5 \times \sqrt{3}}{2\sqrt{3} \times \sqrt{3}} = \frac{5\sqrt{3}}{2 \times 3} = \frac{5\sqrt{3}}{6} \)
(c) \( \sqrt{1 \frac{46}{75}} = \sqrt{\frac{1 \times 75 + 46}{75}} = \sqrt{\frac{121}{75}} = \frac{\sqrt{121}}{\sqrt{75}} = \frac{11}{\sqrt{5 \times 5 \times 3}} = \frac{11}{5\sqrt{3}} \)
Multiplying and dividing by \( \sqrt{3} \)
\( \implies \frac{11 \sqrt{3}}{5\sqrt{3} \times \sqrt{3}} = \frac{11\sqrt{3}}{5 \times 3} = \frac{11\sqrt{3}}{15} \)
In simple words: We rationalized the denominator in each part by multiplying the numerator and denominator by the irrational part of the denominator, simplifying the expression to its most basic form.

🎯 Exam Tip: Always look for perfect square factors inside the radical to simplify the square root first before rationalizing the denominator. This makes calculations easier.

Question 2. \( \sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}} \)
Answer:
\( \sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}} \)
\( = \sqrt{2 \times 2 \times 2 \times 2 \times 7} - \sqrt{3 \times 3 \times 7} + \frac{224}{\sqrt{2 \times 2 \times 7}} \)
\( = 2 \times 2 \sqrt{7} - 3\sqrt{7} + \frac{224}{2\sqrt{7}} \)
\( = 4\sqrt{7}-3\sqrt{7}+\frac{112}{\sqrt{7}} \)
\( = 4\sqrt{7}-3\sqrt{7}+\frac{112\sqrt{7}}{\sqrt{7} \times \sqrt{7}} \)
\( = 4\sqrt{7}-3\sqrt{7}+\frac{112\sqrt{7}}{7} \)
\( = 4\sqrt{7}-3\sqrt{7}+16\sqrt{7} \)
\( = (4-3+16)\sqrt{7} = 17\sqrt{7} \)
In simple words: We simplified each square root by finding perfect square factors, rationalized the third term, and then combined all the similar terms involving \( \sqrt{7} \) to get the final answer.

🎯 Exam Tip: When simplifying expressions with multiple radicals, always factor out perfect squares first. Then, rationalize any denominators containing radicals to make combining like terms easier.

Question 3. \( \frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{\sqrt{3}} \)
Answer:
\( \frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{\sqrt{3}} \)
\( = \frac{4\sqrt{2 \times 3 \times 3}}{\sqrt{2 \times 2 \times 3}} - \frac{8\sqrt{3 \times 5 \times 5}}{\sqrt{2 \times 2 \times 2 \times 2 \times 2}} + \frac{9\sqrt{2}}{\sqrt{3}} \)
\( = \frac{4 \times 3\sqrt{2}}{2\sqrt{3}} - \frac{8 \times 5\sqrt{3}}{4\sqrt{2}} + \frac{9\sqrt{2}}{\sqrt{3}} \)
\( = \frac{6\sqrt{2}}{\sqrt{3}} - \frac{10\sqrt{3}}{\sqrt{2}} + \frac{9\sqrt{2}}{\sqrt{3}} \)
(Rationalising denominator of each)
\( = \frac{6\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} - \frac{10\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} + \frac{9\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( = \frac{6\sqrt{6}}{3} - \frac{10\sqrt{6}}{2} + \frac{9\sqrt{6}}{3} \)
\( = 2\sqrt{6} - 5\sqrt{6} + 3\sqrt{6} \)
\( = (2-5+3)\sqrt{6} \)
\( = 0\sqrt{6} = 0 \)
In simple words: We simplified each term by factoring out perfect squares from the radicals, rationalized the denominators, and then combined all the resulting like terms, which canceled each other out to zero.

🎯 Exam Tip: Always simplify each radical term individually before attempting to combine them. Rationalizing the denominator is a crucial step to correctly combine terms involving square roots.

Question 4. Rationalise the denominators of
(a) \( \frac{1}{4-\sqrt{3}} \)
(b) \( \frac{2}{\sqrt{5}+\sqrt{3}} \)
(c) \( \frac{1}{2 \sqrt{5}-\sqrt{3}} \)
(d) \( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \)
Answer:
(a) \( \frac{1}{4-\sqrt{3}} = \frac{1}{4-\sqrt{3}} \times \frac{4+\sqrt{3}}{4+\sqrt{3}} \)
\( = \frac{4+\sqrt{3}}{(4)^2 - (\sqrt{3})^2} \)
\( = \frac{4+\sqrt{3}}{16-3} = \frac{4+\sqrt{3}}{13} \)
(b) \( \frac{2}{\sqrt{5}+\sqrt{3}} \)
Rationalising factor of \( (\sqrt{5}+\sqrt{3}) \) is \( (\sqrt{5}-\sqrt{3}) \)
\( = \frac{2}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\( = \frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2} = \frac{2(\sqrt{5}-\sqrt{3})}{5-3} \)
\( = \frac{2(\sqrt{5}-\sqrt{3})}{2} = \sqrt{5}-\sqrt{3} \)
(c) \( \frac{1}{2\sqrt{5}-\sqrt{3}} \)
Rationalising factor of \( (2\sqrt{5}-\sqrt{3}) \) is \( (2\sqrt{5}+\sqrt{3}) \)
\( = \frac{1}{2\sqrt{5}-\sqrt{3}} \times \frac{2\sqrt{5}+\sqrt{3}}{2\sqrt{5}+\sqrt{3}} \)
\( = \frac{2\sqrt{5}+\sqrt{3}}{(2\sqrt{5})^2 - (\sqrt{3})^2} = \frac{2\sqrt{5}+\sqrt{3}}{20-3} = \frac{2\sqrt{5}+\sqrt{3}}{17} \)
(d) \( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \)
Rationalising factor of \( (\sqrt{3}-\sqrt{2}) \) is \( (\sqrt{3}+\sqrt{2}) \)
\( = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \)
\( = \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} \)
\( = \frac{3+2+2\sqrt{3 \times 2}}{3-2} = \frac{5+2\sqrt{6}}{1} \)
\( = 5+2\sqrt{6} \)
In simple words: To rationalize the denominator, we multiplied both the numerator and denominator by the conjugate of the denominator, using the difference of squares formula, \( (a-b)(a+b)=a^2-b^2 \), to remove the radical from the bottom.

🎯 Exam Tip: The key to rationalizing binomial denominators is to multiply by the conjugate. Remember that the conjugate of \( a+b\sqrt{c} \) is \( a-b\sqrt{c} \), and for \( \sqrt{a}+\sqrt{b} \) it is \( \sqrt{a}-\sqrt{b} \).

Question 5. Rationalise the denominator and simplify :
(i) \( \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}} \)
(ii) \( \frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}} \)
Answer:
(i) \( \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}} \)
\( = \frac{(4+\sqrt{5})(4+\sqrt{5})}{(4-\sqrt{5})(4+\sqrt{5})} + \frac{(4-\sqrt{5})(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})} \)
(Rationalising the denominator of each)
\( = \frac{(4+\sqrt{5})^2}{(4)^2-(\sqrt{5})^2} + \frac{(4-\sqrt{5})^2}{(4)^2-(\sqrt{5})^2} \)
\( = \frac{16+5+2 \times 4\sqrt{5}}{16-5} + \frac{16+5-2 \times 4\sqrt{5}}{16-5} \)
\( = \frac{21+8\sqrt{5}}{11} + \frac{21-8\sqrt{5}}{11} \)
\( = \frac{21+8\sqrt{5}+21-8\sqrt{5}}{11} = \frac{42}{11} \)
(ii) \( \frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}} \)
\( = \frac{3(5+\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})} + \frac{2(5-\sqrt{3})}{(5+\sqrt{3})(5-\sqrt{3})} \)
(Rationalising the denominator)
\( = \frac{15+3\sqrt{3}}{25-3} + \frac{10-2\sqrt{3}}{25-3} \)
\( = \frac{15+3\sqrt{3}}{22} + \frac{10-2\sqrt{3}}{22} \)
\( = \frac{15+3\sqrt{3}+10-2\sqrt{3}}{22} = \frac{25+\sqrt{3}}{22} \)
In simple words: For each part, we rationalized the denominators of the fractions by multiplying them by their respective conjugates and then combined the resulting fractions, simplifying them to their final form.

🎯 Exam Tip: When simplifying sums or differences of fractions with radical denominators, rationalize each fraction individually first, then find a common denominator (if necessary) to combine them. Keep an eye out for terms that cancel each other out.

Question 6. Find the values of a and b if
(i) \( \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2} \)
(ii) \( \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3} \)
(iii) \( \frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b \sqrt{7} \)
Answer:
(i) \( \frac{3+\sqrt{2}}{3-\sqrt{2}} = a+b\sqrt{2} \)
\( \implies \frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})} = a+b\sqrt{2} \)
\( \implies \frac{(3+\sqrt{2})^2}{9-2} = a+b\sqrt{2} \)
\( \implies \frac{9+2+6\sqrt{2}}{7} = a+b\sqrt{2} \)
\( \implies \frac{11+6\sqrt{2}}{7} = a+b\sqrt{2} \)
\( \implies \frac{11}{7} + \frac{6}{7}\sqrt{2} = a+b\sqrt{2} \)
Comparing, we get
\( a = \frac{11}{7}, b = \frac{6}{7} \)
(ii) \( \frac{5+2\sqrt{3}}{7+4\sqrt{3}} = a+b\sqrt{3} \)
\( \implies \frac{(5+2\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})} = a+b\sqrt{3} \)
(Rationalising the denominator)
\( \implies \frac{35-20\sqrt{3}+14\sqrt{3}-8 \times 3}{(7)^2 - (4\sqrt{3})^2} = a+b\sqrt{3} \)
\( \implies \frac{35-24-6\sqrt{3}}{49-48} = a+b\sqrt{3} \)
\( \implies \frac{11-6\sqrt{3}}{1} = a+b\sqrt{3} \)
\( \implies 11-6\sqrt{3} = a+b\sqrt{3} \)
Comparing, we get
\( a = 11, b = -6 \)
(iii) \( \frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1} = a+b\sqrt{7} \)
\( \implies \frac{(\sqrt{7}-1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)} - \frac{(\sqrt{7}+1)^2}{(\sqrt{7}-1)(\sqrt{7}+1)} = a+b\sqrt{7} \)
\( \implies \frac{(7+1-2\sqrt{7})}{7-1} - \frac{(7+1+2\sqrt{7})}{7-1} = a+b\sqrt{7} \)
\( \implies \frac{8-2\sqrt{7}}{6} - \frac{8+2\sqrt{7}}{6} = a+b\sqrt{7} \)
\( \implies \frac{8-2\sqrt{7}-(8+2\sqrt{7})}{6} = a+b\sqrt{7} \)
\( \implies \frac{8-2\sqrt{7}-8-2\sqrt{7}}{6} = a+b\sqrt{7} \)
\( \implies \frac{-4\sqrt{7}}{6} = a+b\sqrt{7} \)
\( \implies -\frac{2}{3}\sqrt{7} = a+b\sqrt{7} \)
Comparing, we get
\( a = 0, b = -\frac{2}{3} \)
In simple words: For each equation, we rationalized the denominators by multiplying by the conjugate, simplified the expressions, and then equated the rational and irrational parts on both sides to find the values of 'a' and 'b'.

🎯 Exam Tip: When solving for 'a' and 'b', ensure you completely simplify the left-hand side into the form \( A+B\sqrt{C} \) before comparing it with \( a+b\sqrt{C} \). Be careful with signs, especially when subtracting fractions.

Question 7. Rationalise the denominator of \( \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \).
Answer:
\( \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}} = \frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{5}} \)
Rationalising denominator by considering \( (\sqrt{2}+\sqrt{3}) \) as one term.
\( = \frac{(\sqrt{2}+\sqrt{3})-\sqrt{5}}{[(\sqrt{2}+\sqrt{3})+\sqrt{5}][(\sqrt{2}+\sqrt{3})-\sqrt{5}]} \)
\( = \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2} \)
\( = \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2})^2+(\sqrt{3})^2+2\sqrt{2 \times 3}-5} \)
\( = \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{6}-5} \)
\( = \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}} \)
Again rationalising, by multiplying with \( \sqrt{6} \)
\( = \frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})\sqrt{6}}{2\sqrt{6} \times \sqrt{6}} \)
\( = \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{2 \times 6} \)
\( = \frac{\sqrt{4 \times 3}+\sqrt{9 \times 2}-\sqrt{30}}{12} \)
\( = \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12} \)
In simple words: When the denominator has three terms, group two terms to treat it as a binomial, then rationalize using its conjugate. You might need to repeat the rationalization process if a radical remains in the denominator.

🎯 Exam Tip: For denominators with three terms, group two terms together (e.g., \( (\sqrt{2}+\sqrt{3})+\sqrt{5} \)) and use the conjugate. This will eliminate one radical and simplify the expression to a more manageable form, often requiring a second rationalization step.

Question 8. Taking \( \sqrt{2} = 1.414 \) and \( \sqrt{3} = 1.732 \), find without using tables or long division, the value of:
(a) \( \frac{1}{3-\sqrt{2}} \)
(b) \( \frac{2}{\sqrt{3}-\sqrt{2}} \)
Answer:
(a) \( \frac{1}{3-\sqrt{2}} \)
(Rationalising the denominator)
\( = \frac{1}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} \)
\( = \frac{3+\sqrt{2}}{(3)^2-(\sqrt{2})^2} \)
\( = \frac{3+\sqrt{2}}{9-2} = \frac{3+\sqrt{2}}{7} \)
Substitute values: \( = \frac{3+1.414}{7} \)
\( = \frac{4.414}{7} = 0.63057... \approx 0.631 \)
(b) \( \frac{2}{\sqrt{3}-\sqrt{2}} \)
(Rationalising the denominator)
\( = \frac{2}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \)
\( = \frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2} \)
\( = \frac{2(\sqrt{3}+\sqrt{2})}{3-2} \)
\( = \frac{2(\sqrt{3}+\sqrt{2})}{1} = 2(\sqrt{3}+\sqrt{2}) \)
Substitute values: \( = 2(1.732+1.414) \)
\( = 2(3.146) = 6.292 \)
In simple words: We first rationalized the denominator for each expression to remove any square roots from the bottom, then substituted the given approximate values for \( \sqrt{2} \) and \( \sqrt{3} \) and performed the final arithmetic.

🎯 Exam Tip: Always rationalize the denominator first when exact values are given for radicals. This simplifies the expression, making it much easier to substitute and calculate without long division or tables.

Question 9. Express \( \frac{3-5 \sqrt{5}}{3+2 \sqrt{5}} \) in the form \( (a\sqrt{5} - b) \) where a and b are simple fractions.
Answer:
\( \frac{3-5\sqrt{5}}{3+2\sqrt{5}} \)
(Rationalising the denominator)
\( = \frac{3-5\sqrt{5}}{3+2\sqrt{5}} \times \frac{3-2\sqrt{5}}{3-2\sqrt{5}} \)
\( = \frac{(3-5\sqrt{5})(3-2\sqrt{5})}{(3)^2-(2\sqrt{5})^2} \)
\( = \frac{3 \times 3 - 3 \times 2\sqrt{5} - 5\sqrt{5} \times 3 + 5\sqrt{5} \times 2\sqrt{5}}{9-(4 \times 5)} \)
\( = \frac{9-6\sqrt{5}-15\sqrt{5}+10 \times 5}{9-20} \)
\( = \frac{9-21\sqrt{5}+50}{-11} \)
\( = \frac{59-21\sqrt{5}}{-11} \)
\( = -\frac{59}{11} + \frac{21}{11}\sqrt{5} \)
\( = \frac{21}{11}\sqrt{5} - \frac{59}{11} \)
Which is in the form of \( a\sqrt{5}-b \).
Comparing, \( a = \frac{21}{11} \) and \( b = \frac{59}{11} \).
In simple words: We rationalized the denominator by multiplying by its conjugate, expanded and simplified the numerator and denominator, and then rearranged the terms to match the required \( a\sqrt{5}-b \) format to find the fractional values of 'a' and 'b'.

🎯 Exam Tip: Be meticulous with algebraic expansion, especially when dealing with multiple radical terms in the numerator. Ensure all terms are correctly multiplied and combined, and pay close attention to signs when simplifying fractions with negative denominators.

Question 10. Prove that \( \frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1}=\sqrt{2}+\sqrt{3} \).
Answer:
L.H.S. \( = \frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1} \)
(Rationalising the denominators)
\( = \frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} + \frac{2}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \)
\( = \frac{\sqrt{2}+1}{(\sqrt{2})^2-(1)^2} + \frac{2(\sqrt{3}-1)}{(\sqrt{3})^2-(1)^2} \)
\( = \frac{\sqrt{2}+1}{2-1} + \frac{2(\sqrt{3}-1)}{3-1} \)
\( = \frac{\sqrt{2}+1}{1} + \frac{2(\sqrt{3}-1)}{2} \)
\( = \sqrt{2}+1 + \sqrt{3}-1 \)
\( = \sqrt{2}+\sqrt{3} \)
\( = \text{R.H.S.} \)
Hence proved.
In simple words: We simplified the left side by rationalizing each fraction's denominator separately and then combined the resulting terms, which simplified to exactly the right side of the equation.

🎯 Exam Tip: When proving identities, work on one side (usually the more complex one) and simplify it step-by-step until it matches the other side. Rationalization of denominators is often the first key step in such problems.

Question 11. Simplify: \( \frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}} \)
Answer:
\( \frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}} \)
\( = \frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}} \times \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}} - \frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}} \times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} + \frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \)
(Rationalising the denominator of each)
\( = \frac{6\sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3})^2-(\sqrt{6})^2} - \frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^2-(\sqrt{2})^2} + \frac{2\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2} \)
\( = \frac{6\sqrt{6}-6\sqrt{12}}{3-6} - \frac{4\sqrt{18}-4\sqrt{6}}{6-2} + \frac{2\sqrt{12}-2\sqrt{18}}{2-3} \)
\( = \frac{6\sqrt{6}-6 \times 2\sqrt{3}}{-3} - \frac{4 \times 3\sqrt{2}-4\sqrt{6}}{4} + \frac{2 \times 2\sqrt{3}-2 \times 3\sqrt{2}}{-1} \)
\( = \frac{6\sqrt{6}-12\sqrt{3}}{-3} - \frac{12\sqrt{2}-4\sqrt{6}}{4} + \frac{4\sqrt{3}-6\sqrt{2}}{-1} \)
\( = -(2\sqrt{6}-4\sqrt{3}) - (3\sqrt{2}-\sqrt{6}) - (4\sqrt{3}-6\sqrt{2}) \)
\( = -2\sqrt{6}+4\sqrt{3} - 3\sqrt{2}+\sqrt{6} - 4\sqrt{3}+6\sqrt{2} \)
\( = (-2+1)\sqrt{6} + (4-4)\sqrt{3} + (-3+6)\sqrt{2} \)
\( = -\sqrt{6} + 0\sqrt{3} + 3\sqrt{2} \)
\( = 3\sqrt{2}-\sqrt{6} \)
\( = \sqrt{9 \times 2} - \sqrt{6} \)
\( = \sqrt{18} - \sqrt{6} \)
In simple words: We rationalized the denominator of each fraction, simplified the resulting terms, and then combined all like radical terms, collecting coefficients to achieve the final simplified expression.

🎯 Exam Tip: For complex rationalization problems, address each fraction separately first. Pay close attention to negative signs from the denominator and distribute them correctly when combining terms. Simplify radicals at each step to avoid errors.

Question 12. Simplify:
(i) \( \frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}} \)
(ii) \( \frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} \)
Answer:
(i) First term: \( \frac{6}{2\sqrt{3}-\sqrt{6}} \)
\( = \frac{6(2\sqrt{3}+\sqrt{6})}{(2\sqrt{3}-\sqrt{6})(2\sqrt{3}+\sqrt{6})} = \frac{6(2\sqrt{3}+\sqrt{6})}{(2\sqrt{3})^2-(\sqrt{6})^2} \)
\( = \frac{6(2\sqrt{3}+\sqrt{6})}{12-6} = \frac{6(2\sqrt{3}+\sqrt{6})}{6} \)
\( = 2\sqrt{3}+\sqrt{6} \) ....(i)
Second term: \( \frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} \)
\( = \frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{\sqrt{18}-\sqrt{12}}{(\sqrt{3})^2-(\sqrt{2})^2} \)
\( = \frac{3\sqrt{2}-2\sqrt{3}}{3-2} = 3\sqrt{2}-2\sqrt{3} \) ....(ii)
Third term: \( \frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}} \)
\( = \frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{4\sqrt{18}+4\sqrt{6}}{(\sqrt{6})^2-(\sqrt{2})^2} \)
\( = \frac{4 \times 3\sqrt{2}+4\sqrt{6}}{6-2} = \frac{12\sqrt{2}+4\sqrt{6}}{4} \)
\( = 3\sqrt{2}+\sqrt{6} \) ....(iii)
Now, substituting (i), (ii), (iii) into the original expression:
\( (2\sqrt{3}+\sqrt{6}) + (3\sqrt{2}-2\sqrt{3}) - (3\sqrt{2}+\sqrt{6}) \)
\( = 2\sqrt{3}+\sqrt{6}+3\sqrt{2}-2\sqrt{3}-3\sqrt{2}-\sqrt{6} \)
\( = (2\sqrt{3}-2\sqrt{3}) + (\sqrt{6}-\sqrt{6}) + (3\sqrt{2}-3\sqrt{2}) \)
\( = 0+0+0 = 0 \)
(ii) First term: \( \frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} \)
\( = \frac{7\sqrt{3}(\sqrt{10}-\sqrt{3})}{(\sqrt{10}+\sqrt{3})(\sqrt{10}-\sqrt{3})} = \frac{7\sqrt{30}-7 \times 3}{10-3} \)
\( = \frac{7\sqrt{30}-21}{7} = \sqrt{30}-3 \) ....(i)
Second term: \( \frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}} \)
\( = \frac{2\sqrt{5}(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})} = \frac{2\sqrt{30}-2 \times 5}{6-5} \)
\( = \frac{2\sqrt{30}-10}{1} = 2\sqrt{30}-10 \) ....(ii)
Third term: \( \frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} \)
\( = \frac{3\sqrt{2}(\sqrt{15}-3\sqrt{2})}{(\sqrt{15}+3\sqrt{2})(\sqrt{15}-3\sqrt{2})} = \frac{3\sqrt{30}-3 \times 3 \times 2}{(\sqrt{15})^2-(3\sqrt{2})^2} \)
\( = \frac{3\sqrt{30}-18}{15-18} = \frac{3\sqrt{30}-18}{-3} \)
\( = -(\sqrt{30}-6) = -\sqrt{30}+6 \) ....(iii)
Now, substituting (i), (ii), (iii) into the original expression:
\( (\sqrt{30}-3) - (2\sqrt{30}-10) - (-\sqrt{30}+6) \)
\( = \sqrt{30}-3-2\sqrt{30}+10+\sqrt{30}-6 \)
\( = (\sqrt{30}-2\sqrt{30}+\sqrt{30}) + (-3+10-6) \)
\( = 0 + 1 = 1 \)
In simple words: For both parts, we rationalized each individual fraction by multiplying by its conjugate, simplified the terms, and then combined them carefully, paying close attention to signs, to arrive at the final simplified value.

🎯 Exam Tip: When simplifying complex expressions with multiple fractions, always rationalize each fraction separately first. Then combine the simplified terms, ensuring correct distribution of negative signs, and look for opportunities to cancel or combine like terms.

Question 13. If \( x = 2 + \sqrt{3} \), find the value of \( x^2 + \frac{1}{x^2} \).
Answer:
We are given \( x = 2 + \sqrt{3} \). To find \( \frac{1}{x} \), we rationalize the denominator: \( \frac{1}{x} = \frac{1}{2+\sqrt{3}} \) Multiply numerator and denominator by the conjugate \( (2-\sqrt{3}) \): \( \frac{1}{x} = \frac{1 \times (2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \) Using the identity \( (a+b)(a-b) = a^2 - b^2 \): \( \frac{1}{x} = \frac{2-\sqrt{3}}{(2)^2 - (\sqrt{3})^2} \) \( \frac{1}{x} = \frac{2-\sqrt{3}}{4-3} \) \( \frac{1}{x} = \frac{2-\sqrt{3}}{1} \) \( \frac{1}{x} = 2-\sqrt{3} \) Now, find \( x + \frac{1}{x} \): \( x + \frac{1}{x} = (2+\sqrt{3}) + (2-\sqrt{3}) \) \( x + \frac{1}{x} = 2+\sqrt{3}+2-\sqrt{3} \) \( x + \frac{1}{x} = 4 \) To find \( x^2 + \frac{1}{x^2} \), we square both sides of the equation \( x + \frac{1}{x} = 4 \): \( (x + \frac{1}{x})^2 = (4)^2 \) Using the identity \( (a+b)^2 = a^2 + b^2 + 2ab \): \( x^2 + (\frac{1}{x})^2 + 2 \times x \times \frac{1}{x} = 16 \) \( x^2 + \frac{1}{x^2} + 2 = 16 \) Subtract 2 from both sides: \( x^2 + \frac{1}{x^2} = 16 - 2 \) \( x^2 + \frac{1}{x^2} = 14 \)
In simple words: To find this value, first simplify \( \frac{1}{x} \) by rationalizing the denominator. Then, find \( x + \frac{1}{x} \) and square the entire expression to get \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: Remember the algebraic identity \( (a+b)^2 = a^2 + b^2 + 2ab \). Applying this helps simplify calculations when dealing with sums of a number and its reciprocal.

Question 14. If \( x = \sqrt{2} + 1 \), find the value of \( x^2 + \frac{1}{x^2} \).
Answer:
We are given \( x = \sqrt{2} + 1 \). To find \( \frac{1}{x} \), we rationalize the denominator: \( \frac{1}{x} = \frac{1}{\sqrt{2}+1} \) Multiply numerator and denominator by the conjugate \( (\sqrt{2}-1) \): \( \frac{1}{x} = \frac{1 \times (\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} \) Using the identity \( (a+b)(a-b) = a^2 - b^2 \): \( \frac{1}{x} = \frac{\sqrt{2}-1}{(\sqrt{2})^2 - (1)^2} \) \( \frac{1}{x} = \frac{\sqrt{2}-1}{2-1} \) \( \frac{1}{x} = \frac{\sqrt{2}-1}{1} \) \( \frac{1}{x} = \sqrt{2}-1 \) Now, find \( x + \frac{1}{x} \): \( x + \frac{1}{x} = (\sqrt{2}+1) + (\sqrt{2}-1) \) \( x + \frac{1}{x} = \sqrt{2}+1+\sqrt{2}-1 \) \( x + \frac{1}{x} = 2\sqrt{2} \) To find \( x^2 + \frac{1}{x^2} \), we square both sides of the equation \( x + \frac{1}{x} = 2\sqrt{2} \): \( (x + \frac{1}{x})^2 = (2\sqrt{2})^2 \) Using the identity \( (a+b)^2 = a^2 + b^2 + 2ab \): \( x^2 + (\frac{1}{x})^2 + 2 \times x \times \frac{1}{x} = (2^2) \times (\sqrt{2})^2 \) \( x^2 + \frac{1}{x^2} + 2 = 4 \times 2 \) \( x^2 + \frac{1}{x^2} + 2 = 8 \) Subtract 2 from both sides: \( x^2 + \frac{1}{x^2} = 8 - 2 \) \( x^2 + \frac{1}{x^2} = 6 \)
In simple words: First, simplify \( \frac{1}{x} \) by rationalizing its denominator. Then, add \( x \) and \( \frac{1}{x} \), and finally square the result to find \( x^2 + \frac{1}{x^2} \).

🎯 Exam Tip: When dealing with expressions involving square roots, always rationalize the denominator to simplify calculations before performing addition or subtraction.

Question 15. If \( x = \frac{5-\sqrt{21}}{2} \), find the value of (i) \( x + \frac{1}{x} \) and (ii) \( x^2 + \frac{1}{x^2} \).
Answer:
We are given \( x = \frac{5-\sqrt{21}}{2} \). To find \( \frac{1}{x} \): \( \frac{1}{x} = \frac{1}{\frac{5-\sqrt{21}}{2}} \) \( \frac{1}{x} = \frac{2}{5-\sqrt{21}} \) Now, rationalize the denominator: \( \frac{1}{x} = \frac{2 \times (5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})} \) Using the identity \( (a-b)(a+b) = a^2 - b^2 \): \( \frac{1}{x} = \frac{2(5+\sqrt{21})}{(5)^2 - (\sqrt{21})^2} \) \( \frac{1}{x} = \frac{2(5+\sqrt{21})}{25 - 21} \) \( \frac{1}{x} = \frac{2(5+\sqrt{21})}{4} \) \( \frac{1}{x} = \frac{5+\sqrt{21}}{2} \) (i) Now, find \( x + \frac{1}{x} \): \( x + \frac{1}{x} = \frac{5-\sqrt{21}}{2} + \frac{5+\sqrt{21}}{2} \) Since the denominators are the same, combine the numerators: \( x + \frac{1}{x} = \frac{(5-\sqrt{21}) + (5+\sqrt{21})}{2} \) \( x + \frac{1}{x} = \frac{5-\sqrt{21}+5+\sqrt{21}}{2} \) \( x + \frac{1}{x} = \frac{10}{2} \) \( x + \frac{1}{x} = 5 \) (ii) To find \( x^2 + \frac{1}{x^2} \), we square both sides of the equation \( x + \frac{1}{x} = 5 \): \( (x + \frac{1}{x})^2 = (5)^2 \) Using the identity \( (a+b)^2 = a^2 + b^2 + 2ab \): \( x^2 + (\frac{1}{x})^2 + 2 \times x \times \frac{1}{x} = 25 \) \( x^2 + \frac{1}{x^2} + 2 = 25 \) Subtract 2 from both sides: \( x^2 + \frac{1}{x^2} = 25 - 2 \) \( x^2 + \frac{1}{x^2} = 23 \) Therefore, \( x + \frac{1}{x} = 5 \) and \( x^2 + \frac{1}{x^2} = 23 \).
In simple words: First, simplify \( \frac{1}{x} \) by removing the square root from the denominator. Then, add \( x \) and \( \frac{1}{x} \) to find the first part. For the second part, square the sum \( (x + \frac{1}{x}) \) and adjust the result.

🎯 Exam Tip: When a question asks for multiple related values, calculate the simplest intermediate value first (like \( x + \frac{1}{x} \)), as it often helps quickly find the more complex ones (like \( x^2 + \frac{1}{x^2} \)).