Rational and Irrational Numbers

Q1: Multiple Choice Type

a. If \( x = \sqrt{5} – 2 \), \( x + \frac{1}{x} \) is equal to:

Step 1: Find \( \frac{1}{x} \) \[ \frac{1}{x} = \frac{1}{\sqrt{5} – 2} \]Rationalize: \[ = \frac{\sqrt{5} + 2}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{\sqrt{5} + 2}{5 – 4} = \sqrt{5} + 2 \]Step 2: Add \[ x + \frac{1}{x} = (\sqrt{5} – 2) + (\sqrt{5} + 2) = 2\sqrt{5} \]Answer: i. \( 2\sqrt{5} \)

b. If \( x = 1 + \sqrt{2} \), then \( \left(x + \frac{1}{x}\right)^2 \) is:

Step 1: Find \( \frac{1}{x} \) \[ \frac{1}{x} = \frac{1}{1+\sqrt{2}} = \frac{1-\sqrt{2}}{1-2} = \sqrt{2} – 1 \]Step 2: Add \[ x + \frac{1}{x} = (1+\sqrt{2}) + (\sqrt{2}-1) = 2\sqrt{2} \]Step 3: Square \[ (2\sqrt{2})^2 = 8 \]Answer: ii. 8

c. \( \frac{2\sqrt{27} + 3\sqrt{12}} {4\sqrt{3}} \) is equal to:

Step 1: Simplify roots \[ \sqrt{27} = 3\sqrt{3},\quad \sqrt{12} = 2\sqrt{3} \]Step 2: Substitute \[ \frac{2(3\sqrt{3}) + 3(2\sqrt{3})}{4\sqrt{3}} = \frac{6\sqrt{3} + 6\sqrt{3}}{4\sqrt{3}} = \frac{12\sqrt{3}}{4\sqrt{3}} = 3 \]Answer: iii. 3

d. \( (\sqrt{5} – \sqrt{3})^2 \) is:

Step 1: Use identity \[ (a-b)^2 = a^2 – 2ab + b^2 \\ = 5 – 2\sqrt{15} + 3 = 8 – 2\sqrt{15} \]Answer: iv. \( 8 – 2\sqrt{15} \)

e. \( \frac{3}{4+\sqrt{7}} \) is equal to:

Step 1: Rationalize \[ \frac{3}{4+\sqrt{7}} \times \frac{4-\sqrt{7}}{4-\sqrt{7}} = \frac{3(4-\sqrt{7})}{16 – 7} = \frac{3(4-\sqrt{7})}{9} \\ = \frac{1}{3}(4-\sqrt{7}) \]Answer: i. \( \frac{1}{3}(4-\sqrt{7}) \)

f. \( \frac{1}{7-\sqrt{5}} \) is equal to:

Step 1: Rationalize \[ \frac{1}{7-\sqrt{5}} \times \frac{7+\sqrt{5}}{7+\sqrt{5}} = \frac{7+\sqrt{5}}{49 – 5} = \frac{7+\sqrt{5}}{44} \\ = \frac{1}{44}(7+\sqrt{5}) \]Answer: ii. \( \frac{1}{44}(7+\sqrt{5}) \)

g. If \( x = \sqrt{2} – 1 \), then \( \left(x – \frac{1}{x}\right)^2 \) is:

Step 1: Find \( \frac{1}{x} \) \[ \frac{1}{x} = \frac{1}{\sqrt{2}-1} = \sqrt{2} + 1 \]Step 2: Subtract \[ x – \frac{1}{x} = (\sqrt{2}-1) – (\sqrt{2}+1) = -2 \]Step 3: Square \[ (-2)^2 = 4 \]Answer: iii. 4

h. \( \frac{5-\sqrt{7}}{5+\sqrt{7}} – \frac{5+\sqrt{7}}{5-\sqrt{7}} \) is equal to:

Step 1: Take LCM \[ = \frac{(5-\sqrt{7})^2 – (5+\sqrt{7})^2}{(5+\sqrt{7})(5-\sqrt{7})} \]Step 2: Expand \[ (5-\sqrt{7})^2 = 25 – 10\sqrt{7} + 7 = 32 – 10\sqrt{7} \\ (5+\sqrt{7})^2 = 32 + 10\sqrt{7} \\ \text{Numerator} = (32 – 10\sqrt{7}) – (32 + 10\sqrt{7}) = -20\sqrt{7} \\ \text{Denominator} = 25 – 7 = 18 \\ = \frac{-20\sqrt{7}}{18} = \frac{-10\sqrt{7}}{9} \]Answer: iv. \( \frac{-10\sqrt{7}}{9} \)

Q2: State, with reason, which of the following are surds and which are not:

i. \( \sqrt{180} \)

Step 1: Simplify \[ \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \]Step 2: Conclusion
Contains irrational part ⇒ surd
Answer: Surd

ii. \( \sqrt[4]{27} \)

Step 1: Check perfect power \[ 27 = 3^3 \]Not a perfect 4th power
Step 2: Conclusion
Irrational ⇒ surd
Answer: Surd

iii. \( \sqrt[5]{128} \)

Step 1: Express as power \[ 128 = 2^7 \\ \sqrt[5]{128} = 2^{7/5} \]Not a rational number
Step 2: Conclusion
Irrational ⇒ surd
Answer: Surd

iv. \( \sqrt[3]{64} \)

Step 1: Evaluate \[ \sqrt[3]{64} = 4 \]Step 2: Conclusion
Rational number ⇒ not a surd
Answer: Not a surd

v. \( \sqrt[3]{25} \cdot \sqrt[3]{40} \)

Step 1: Combine \[ \sqrt[3]{25} \cdot \sqrt[3]{40} = \sqrt[3]{1000} \]Step 2: Evaluate \[ \sqrt[3]{1000} = 10 \]Step 3: Conclusion
Rational ⇒ not a surd
Answer: Not a surd

vi. \( \sqrt[3]{-125} \)

Step 1: Evaluate \[ \sqrt[3]{-125} = -5 \]Step 2: Conclusion
Rational ⇒ not a surd
Answer: Not a surd

vii. \( \sqrt{\pi} \)

Step 1: Note
\( \pi \) is irrational
Step 2: Conclusion \[ \sqrt{\pi} \text{ is irrational} \]Irrational ⇒ not a surd
Answer: Not a surd

viii. \( \sqrt{3+\sqrt{2}} \)

Step 1: Observe
\( 3 + \sqrt{2} \) is irrational
Step 2: Conclusion \[ \sqrt{3+\sqrt{2}} \text{ is irrational} \]Irrational ⇒ not a surd
Answer: Not a surd

Q3: Write the lowest rationalizing factor of:

i. \( 5\sqrt{2} \)

Step 1: Multiply by \( \sqrt{2} \) \[ 5\sqrt{2} \times \sqrt{2} = 10 \]Step 2: Result is rational
Answer: \( \sqrt{2} \)

ii. \( \sqrt{24} \)

Step 1: Simplify \[ \sqrt{24} = 2\sqrt{6} \]Step 2: Multiply by \( \sqrt{6} \) \[ 2\sqrt{6} \times \sqrt{6} = 12 \]Answer: \( \sqrt{6} \)

iii. \( \sqrt{5} – 3 \)

Step 1: Take conjugate \[ \sqrt{5} + 3 \]Step 2: Multiply \[ (\sqrt{5}-3)(\sqrt{5}+3) = 5 – 9 = -4 \]Answer: \( \sqrt{5} + 3 \)

iv. \( 7 – \sqrt{7} \)

Step 1: Take conjugate \[ 7 + \sqrt{7} \]Step 2: Multiply \[ (7-\sqrt{7})(7+\sqrt{7}) = 49 – 7 = 42 \]Answer: \( 7 + \sqrt{7} \)

v. \( \sqrt{18} – \sqrt{50} \)

Step 1: Simplify \[ \sqrt{18} = 3\sqrt{2}, \quad \sqrt{50} = 5\sqrt{2} \\ = 3\sqrt{2} – 5\sqrt{2} = -2\sqrt{2} \]Step 2: Multiply by \( \sqrt{2} \) \[ -2\sqrt{2} \times \sqrt{2} = -4 \]Answer: \( \sqrt{2} \)

vi. \( \sqrt{5} – \sqrt{2} \)

Step 1: Take conjugate \[ \sqrt{5} + \sqrt{2} \]Step 2: Multiply \[ (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}) = 5 – 2 = 3 \]Answer: \( \sqrt{5} + \sqrt{2} \)

vii. \( \sqrt{13} + 3 \)

Step 1: Take conjugate \[ \sqrt{13} – 3 \]Step 2: Multiply \[ (\sqrt{13}+3)(\sqrt{13}-3) = 13 – 9 = 4 \]Answer: \( \sqrt{13} – 3 \)

Q4: Rationalize the denominators of:

i. \( \frac{2\sqrt{3}}{\sqrt{5}} \)

Step 1: Multiply numerator and denominator by \( \sqrt{5} \) \[ \frac{2\sqrt{3}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \]Step 2: Simplify \[ = \frac{2\sqrt{3}\sqrt{5}}{5} \\ = \frac{2\sqrt{15}}{5} \]Answer: \( \frac{2\sqrt{15}}{5} \)

ii. \( \frac{\sqrt{6} – \sqrt{5}}{\sqrt{6} + \sqrt{5}} \)

Step 1: Multiply by conjugate \[ \frac{\sqrt{6} – \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} – \sqrt{5}}{\sqrt{6} – \sqrt{5}} \]Step 2: Apply identity \[ (a+b)(a-b) = a^2 – b^2 \]Step 3: Simplify denominator \[ (\sqrt{6})^2 – (\sqrt{5})^2 = 6 – 5 = 1 \]Step 4: Simplify numerator \[ (\sqrt{6} – \sqrt{5})^2 = 6 – 2\sqrt{30} + 5 = 11 – 2\sqrt{30} \]Answer: \( 11 – 2\sqrt{30} \)

Q5: Find the values of \(‘a’\) and \(‘b’\) in each of the following:

i. \( \frac{2+\sqrt{3}}{2-\sqrt{3}} = a + b\sqrt{3} \)

Step 1: Rationalize denominator \[ \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \]Step 2: Simplify denominator \[ (2-\sqrt{3})(2+\sqrt{3}) = 4 – 3 = 1 \]Step 3: Expand numerator \[ (2+\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]Step 4: Compare \[ a + b\sqrt{3} = 7 + 4\sqrt{3} \\ a = 7,\quad b = 4 \]Answer: \( a = 7,\ b = 4 \)

ii. \( \frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b \)

Step 1: Rationalize denominator \[ \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} \]Step 2: Simplify denominator \[ (\sqrt{7})^2 – 2^2 = 7 – 4 = 3 \]Step 3: Expand numerator \[ (\sqrt{7}-2)^2 = 7 – 4\sqrt{7} + 4 = 11 – 4\sqrt{7} \\ = \frac{11 – 4\sqrt{7}}{3} \]Step 4: Rearrange \[ = \frac{-4}{3}\sqrt{7} + \frac{11}{3} \]Step 5: Compare \[ a = -\frac{4}{3},\quad b = \frac{11}{3} \]Answer: \( a = -\frac{4}{3},\ b = \frac{11}{3} \)

iii. \( \frac{3}{\sqrt{3} – \sqrt{2}} = a\sqrt{3} – b\sqrt{2} \)

Step 1: Rationalize denominator \[ \frac{3}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \]Step 2: Simplify denominator \[ 3 – 2 = 1 \]Step 3: Multiply numerator \[ = 3(\sqrt{3}+\sqrt{2}) = 3\sqrt{3} + 3\sqrt{2} \]Step 4: Compare \[ a\sqrt{3} – b\sqrt{2} = 3\sqrt{3} + 3\sqrt{2} \\ a = 3,\quad -b = 3 \Rightarrow b = -3 \]Answer: \( a = 3,\ b = -3 \)

Q6: Simplify:

i. \( \frac{22}{2\sqrt{3}+1} + \frac{17}{2\sqrt{3}-1} \)

Step 1: Take LCM \[ = \frac{22(2\sqrt{3}-1) + 17(2\sqrt{3}+1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)} \]Step 2: Simplify denominator \[ (2\sqrt{3})^2 – 1^2 = 12 – 1 = 11 \]Step 3: Expand numerator \[ 22(2\sqrt{3}-1) = 44\sqrt{3} – 22 \\ 17(2\sqrt{3}+1) = 34\sqrt{3} + 17 \\ \text{Numerator} = (44\sqrt{3} + 34\sqrt{3}) + (-22 + 17) \\ = 78\sqrt{3} – 5 \]Step 4: Final result \[ = \frac{78\sqrt{3} – 5}{11} \]Answer: \( \frac{78\sqrt{3} – 5}{11} \)

ii. \( \frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}} – \frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}} \)

Step 1: Take LCM \[ = \frac{\sqrt{2}(\sqrt{6}+\sqrt{2}) – \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} \]Step 2: Simplify denominator \[ 6 – 2 = 4 \]Step 3: Expand numerator \[ \sqrt{2}(\sqrt{6}+\sqrt{2}) = \sqrt{12} + 2 = 2\sqrt{3} + 2 \\ \sqrt{3}(\sqrt{6}-\sqrt{2}) = \sqrt{18} – \sqrt{6} = 3\sqrt{2} – \sqrt{6} \\ \text{Numerator} = (2\sqrt{3} + 2) – (3\sqrt{2} – \sqrt{6}) \\ = 2\sqrt{3} + 2 – 3\sqrt{2} + \sqrt{6} \]Step 4: Final result \[ = \frac{2\sqrt{3} + 2 – 3\sqrt{2} + \sqrt{6}}{4} \]Answer: \( \frac{2\sqrt{3} + 2 – 3\sqrt{2} + \sqrt{6}}{4} \)

Q7: If \( x=\frac{\sqrt{5}-2}{\sqrt{5}+2} \) and \( y=\frac{\sqrt{5}+2}{\sqrt{5}-2} \); find:

i. \( x^2 \)

Step 1: Rationalize \( x \) \[ x = \frac{\sqrt{5}-2}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2} \\ = \frac{(\sqrt{5}-2)^2}{5 – 4} = (\sqrt{5}-2)^2 \\ = 5 – 4\sqrt{5} + 4 = 9 – 4\sqrt{5} \]Step 2: Square \[ x^2 = (9 – 4\sqrt{5})^2 \\ = 81 – 72\sqrt{5} + 80 = 161 – 72\sqrt{5} \]Answer: \( x^2 = 161 – 72\sqrt{5} \)

ii. \( y^2 \)

Step 1: Rationalize \( y \) \[ y = \frac{\sqrt{5}+2}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2} \\ = (\sqrt{5}+2)^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5} \]Step 2: Square \[ y^2 = (9 + 4\sqrt{5})^2 \\ = 81 + 72\sqrt{5} + 80 = 161 + 72\sqrt{5} \]Answer: \( y^2 = 161 + 72\sqrt{5} \)

iii. \( xy \)

Step 1: Multiply \[ xy = \frac{\sqrt{5}-2}{\sqrt{5}+2} \times \frac{\sqrt{5}+2}{\sqrt{5}-2} \\ = 1 \]Answer: \( xy = 1 \)

iv. \( x^2 + y^2 + xy \)

Step 1: Substitute values \[ x^2 = 161 – 72\sqrt{5},\quad y^2 = 161 + 72\sqrt{5},\quad xy = 1 \]Step 2: Add \[ x^2 + y^2 = (161 – 72\sqrt{5}) + (161 + 72\sqrt{5}) = 322 \\ x^2 + y^2 + xy = 322 + 1 = 323 \]Answer: \( 323 \)

Q8: If \( m=\frac{1}{3-2\sqrt{2}} \) and \( n=\frac{1}{3+2\sqrt{2}} \), find:

i. \( m^2 \)

Step 1: Rationalize \( m \) \[ m = \frac{1}{3-2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} \\ = \frac{3+2\sqrt{2}}{9 – 8} = 3 + 2\sqrt{2} \]Step 2: Square \[ m^2 = (3 + 2\sqrt{2})^2 \\ = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \]Answer: \( m^2 = 17 + 12\sqrt{2} \)

ii. \( n^2 \)

Step 1: Rationalize \( n \) \[ n = \frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ = \frac{3-2\sqrt{2}}{9 – 8} = 3 – 2\sqrt{2} \]Step 2: Square \[ n^2 = (3 – 2\sqrt{2})^2 \\ = 9 – 12\sqrt{2} + 8 = 17 – 12\sqrt{2} \]Answer: \( n^2 = 17 – 12\sqrt{2} \)

iii. \( mn \)

Step 1: Multiply \[ mn = \frac{1}{3-2\sqrt{2}} \times \frac{1}{3+2\sqrt{2}} \\ = \frac{1}{(3-2\sqrt{2})(3+2\sqrt{2})} \\ = \frac{1}{9 – 8} = 1 \]Answer: \( mn = 1 \)

Q9: If \( x = 2\sqrt{3} + 2\sqrt{2} \), find:

i. \( \frac{1}{x} \)

Step 1: Factor out 2 \[ x = 2(\sqrt{3} + \sqrt{2}) \]Step 2: Take reciprocal \[ \frac{1}{x} = \frac{1}{2(\sqrt{3}+\sqrt{2})} \]Step 3: Rationalize \[ = \frac{1}{2(\sqrt{3}+\sqrt{2})} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ = \frac{\sqrt{3}-\sqrt{2}}{2(3-2)} = \frac{\sqrt{3}-\sqrt{2}}{2} \]Answer: \( \frac{1}{x} = \frac{\sqrt{3}-\sqrt{2}}{2} \)

ii. \( x + \frac{1}{x} \)

Step 1: Substitute values \[ x = 2\sqrt{3} + 2\sqrt{2}, \quad \frac{1}{x} = \frac{\sqrt{3}-\sqrt{2}}{2} \]Step 2: Add \[ x + \frac{1}{x} = (2\sqrt{3} + 2\sqrt{2}) + \frac{\sqrt{3}-\sqrt{2}}{2} \]Take LCM = 2 \[ = \frac{4\sqrt{3} + 4\sqrt{2} + \sqrt{3} – \sqrt{2}}{2} \\ = \frac{5\sqrt{3} + 3\sqrt{2}}{2} \]Answer: \( \frac{5\sqrt{3} + 3\sqrt{2}}{2} \)

iii. \( \left(x + \frac{1}{x}\right)^2 \)

Step 1: Square the result \[ \left(\frac{5\sqrt{3} + 3\sqrt{2}}{2}\right)^2 \\ = \frac{(5\sqrt{3} + 3\sqrt{2})^2}{4} \]Step 2: Expand \[ (5\sqrt{3})^2 = 75,\quad (3\sqrt{2})^2 = 18,\quad 2 \cdot 5\sqrt{3} \cdot 3\sqrt{2} = 30\sqrt{6} \\ = \frac{75 + 18 + 30\sqrt{6}}{4} \\ = \frac{93 + 30\sqrt{6}}{4} \]Answer: \( \frac{93 + 30\sqrt{6}}{4} \)

Q10: If \( x = 1 – \sqrt{2} \), find the value of \( \left(x – \frac{1}{x}\right)^3 \).

Step 1: Find \( \frac{1}{x} \) \[ \frac{1}{x} = \frac{1}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} \\ = \frac{1+\sqrt{2}}{1 – 2} = -(1+\sqrt{2}) = -1 – \sqrt{2} \]Step 2: Find \( x – \frac{1}{x} \) \[ x – \frac{1}{x} = (1 – \sqrt{2}) – (-1 – \sqrt{2}) \\ = 1 – \sqrt{2} + 1 + \sqrt{2} = 2 \]Step 3: Cube \[ \left(x – \frac{1}{x}\right)^3 = 2^3 = 8 \]Answer: \( 8 \)

Q11: If \( x = 5 – 2\sqrt{6} \), find \( x^2 + \frac{1}{x^2} \)

Step 1: Find \( \frac{1}{x} \) \[ \frac{1}{x} = \frac{1}{5 – 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\ = \frac{5 + 2\sqrt{6}}{25 – 24} = 5 + 2\sqrt{6} \]Step 2: Find \( x + \frac{1}{x} \) \[ x + \frac{1}{x} = (5 – 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10 \]Step 3: Use identity \[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2 \\ = 10^2 – 2 = 100 – 2 = 98 \]Answer: \( 98 \)

Q12: If \( \sqrt{2} = 1.4 \) and \( \sqrt{3} = 1.7 \), find the value of each of the following, correct to one decimal place:

i. \( \frac{1}{\sqrt{3} – \sqrt{2}} \)

Step 1: Rationalize \[ \frac{1}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{3}+\sqrt{2}}{3 – 2} \\ = \sqrt{3} + \sqrt{2} \]Step 2: Substitute values \[ = 1.7 + 1.4 = 3.1 \]Answer: \( 3.1 \)

ii. \( \frac{1}{3 + 2\sqrt{2}} \)

Step 1: Rationalize \[ \frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} = \frac{3-2\sqrt{2}}{9 – 8} \\ = 3 – 2\sqrt{2} \]Step 2: Substitute values \[ = 3 – 2(1.4) = 3 – 2.8 = 0.2 \]Answer: \( 0.2 \)

iii. \( \frac{2 – \sqrt{3}}{\sqrt{3}} \)

Step 1: Split fraction \[ = \frac{2}{\sqrt{3}} – 1 \]Step 2: Rationalize \( \frac{2}{\sqrt{3}} \) \[ \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]Step 3: Substitute values \[ = \frac{2(1.7)}{3} – 1 = \frac{3.4}{3} – 1 \\ = 1.13 – 1 = 0.13 \]Rounded to one decimal place: \[ = 0.1 \]Answer: \( 0.1 \)

Q13: Evaluate: \( \frac{4-\sqrt{5}}{4+\sqrt{5}} + \frac{4+\sqrt{5}}{4-\sqrt{5}} \)

Step 1: Take LCM \[ = \frac{(4-\sqrt{5})^2 + (4+\sqrt{5})^2}{(4+\sqrt{5})(4-\sqrt{5})} \]Step 2: Simplify denominator \[ (4+\sqrt{5})(4-\sqrt{5}) = 16 – 5 = 11 \]Step 3: Expand numerator \[ (4-\sqrt{5})^2 = 16 – 8\sqrt{5} + 5 = 21 – 8\sqrt{5} \\ (4+\sqrt{5})^2 = 21 + 8\sqrt{5} \\ \text{Numerator} = (21 – 8\sqrt{5}) + (21 + 8\sqrt{5}) = 42 \]Step 4: Final result \[ = \frac{42}{11} = 3 \frac{9}{11} \]Answer: \( \frac{42}{11} = 3\frac{9}{11} \)

Q14: If \( \frac{2+\sqrt{5}}{2-\sqrt{5}} = x \) and \( \frac{2-\sqrt{5}}{2+\sqrt{5}} = y \); find the value of \( x^2 – y^2 \)

Step 1: Rationalize \( x \) and \( y \) \[ x = \frac{2+\sqrt{5}}{2-\sqrt{5}} \times \frac{2+\sqrt{5}}{2+\sqrt{5}} = \frac{(2+\sqrt{5})^2}{4 – 5} \\ = \frac{4 + 4\sqrt{5} + 5}{-1} = -9 – 4\sqrt{5} \\ y = \frac{2-\sqrt{5}}{2+\sqrt{5}} \times \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{(2-\sqrt{5})^2}{4 – 5} \\ = \frac{4 – 4\sqrt{5} + 5}{-1} = -9 + 4\sqrt{5} \]Step 2: Use identity \[ x^2 – y^2 = (x – y)(x + y) \]Step 3: Find \( x – y \) \[ x – y = (-9 – 4\sqrt{5}) – (-9 + 4\sqrt{5}) = -8\sqrt{5} \]Step 4: Find \( x + y \) \[ x + y = (-9 – 4\sqrt{5}) + (-9 + 4\sqrt{5}) = -18 \]Step 5: Multiply \[ x^2 – y^2 = (-8\sqrt{5})(-18) = 144\sqrt{5} \]Answer: \( 144\sqrt{5} \)