Exercise: 1-D
Q1: Rationalise the denominator of: \(\frac{2}{\sqrt{6}}\)
Step 1: Multiply the numerator and the denominator by \(\sqrt{6}\).
\(\frac{2}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}\)
Step 2: Simplify the products.
\(\frac{2\sqrt{6}}{(\sqrt{6})^2} = \frac{2\sqrt{6}}{6}\)
Step 3: Reduce the fraction to its lowest terms by dividing by 2.
\(\frac{1\sqrt{6}}{3} = \frac{\sqrt{6}}{3}\)
Answer: \(\frac{\sqrt{6}}{3}\)
Q2: Rationalise the denominator of: \(\frac{\sqrt{2}}{2\sqrt{3}}\)
Step 1: Multiply the numerator and the denominator by \(\sqrt{3}\) to remove the square root from the bottom.
\(\frac{\sqrt{2}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
Step 2: Multiply the terms in the numerator and the denominator.
\(\frac{\sqrt{2 \times 3}}{2 \times (\sqrt{3})^2}\)
Step 3: Simplify the expression.
\(\frac{\sqrt{6}}{2 \times 3} = \frac{\sqrt{6}}{6}\)
Answer: \(\frac{\sqrt{6}}{6}\)
Q3: Rationalise the denominator of: \(\frac{1}{3 + \sqrt{5}}\)
Step 1: Identify the conjugate of the denominator \(3 + \sqrt{5}\), which is \(3 – \sqrt{5}\).
Step 2: Multiply both the numerator and the denominator by \(3 – \sqrt{5}\).
\(\frac{1}{3 + \sqrt{5}} \times \frac{3 – \sqrt{5}}{3 – \sqrt{5}}\)
Step 3: Apply the identity \((a + b)(a – b) = a^2 – b^2\) to the denominator.
\(\frac{3 – \sqrt{5}}{(3)^2 – (\sqrt{5})^2}\)
Step 4: Simplify the squares in the denominator.
\(\frac{3 – \sqrt{5}}{9 – 5} = \frac{3 – \sqrt{5}}{4}\)
Answer: \(\frac{3 – \sqrt{5}}{4}\)
Q4: Rationalise the denominator of: \(\frac{1}{\sqrt{3} – 1}\)
Step 1: Identify the conjugate of the denominator \(\sqrt{3} – 1\), which is \(\sqrt{3} + 1\).
Step 2: Multiply both the numerator and the denominator by \(\sqrt{3} + 1\).
\(\frac{1}{\sqrt{3} – 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\)
Step 3: Use the identity \((a – b)(a + b) = a^2 – b^2\) for the denominator.
\(\frac{\sqrt{3} + 1}{(\sqrt{3})^2 – (1)^2}\)
Step 4: Simplify the squares in the denominator.
\(\frac{\sqrt{3} + 1}{3 – 1} = \frac{\sqrt{3} + 1}{2}\)
Answer: \(\frac{\sqrt{3} + 1}{2}\)
Q5: Rationalise the denominator of: \(\frac{1}{4 + 2\sqrt{3}}\)
Step 1: Identify the conjugate of the denominator \(4 + 2\sqrt{3}\), which is \(4 – 2\sqrt{3}\).
Step 2: Multiply both the numerator and the denominator by \(4 – 2\sqrt{3}\).
\(\frac{1}{4 + 2\sqrt{3}} \times \frac{4 – 2\sqrt{3}}{4 – 2\sqrt{3}}\)
Step 3: Apply the identity \((a + b)(a – b) = a^2 – b^2\) to the denominator.
\(\frac{4 – 2\sqrt{3}}{(4)^2 – (2\sqrt{3})^2}\)
Step 4: Simplify the squares in the denominator.
\((4)^2 = 16\) and \((2\sqrt{3})^2 = 4 \times 3 = 12\).
\(\frac{4 – 2\sqrt{3}}{16 – 12} = \frac{4 – 2\sqrt{3}}{4}\)
Step 5: Factor out 2 from the numerator and simplify.
\(\frac{2(2 – \sqrt{3})}{4} = \frac{2 – \sqrt{3}}{2}\)
Answer: \(\frac{2 – \sqrt{3}}{2}\)
Q6: Rationalise the denominator of: \(\frac{1}{\sqrt{6} – \sqrt{3}}\)
Step 1: Identify the conjugate of the denominator \(\sqrt{6} – \sqrt{3}\), which is \(\sqrt{6} + \sqrt{3}\).
Step 2: Multiply both the numerator and the denominator by \(\sqrt{6} + \sqrt{3}\).
\(\frac{1}{\sqrt{6} – \sqrt{3}} \times \frac{\sqrt{6} + \sqrt{3}}{\sqrt{6} + \sqrt{3}}\)
Step 3: Apply the identity \((a – b)(a + b) = a^2 – b^2\) to the denominator.
\(\frac{\sqrt{6} + \sqrt{3}}{(\sqrt{6})^2 – (\sqrt{3})^2}\)
Step 4: Simplify the squares in the denominator.
\(\frac{\sqrt{6} + \sqrt{3}}{6 – 3} = \frac{\sqrt{6} + \sqrt{3}}{3}\)
Step 5: Simplify \(\sqrt{6}\) as \(\sqrt{2 \times 3}\). No further simplification is possible in the denominator.
Answer: \(\frac{\sqrt{6} + \sqrt{3}}{3}\)
Q7: Rationalise the denominator of: \(\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)
Step 1: Identify the conjugate of the denominator \(\sqrt{3} + 1\), which is \(\sqrt{3} – 1\).
Step 2: Multiply both the numerator and the denominator by \(\sqrt{3} – 1\).
\(\frac{\sqrt{3} – 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} – 1}{\sqrt{3} – 1}\)
Step 3: Use the identity \((a – b)^2 = a^2 – 2ab + b^2\) for the numerator and \((a + b)(a – b) = a^2 – b^2\) for the denominator.
Numerator: \((\sqrt{3})^2 – 2(\sqrt{3})(1) + (1)^2 = 3 – 2\sqrt{3} + 1 = 4 – 2\sqrt{3}\)
Denominator: \((\sqrt{3})^2 – (1)^2 = 3 – 1 = 2\)
Step 4: Write the simplified fraction.
\(\frac{4 – 2\sqrt{3}}{2}\)
Step 5: Factor out 2 from the numerator and divide.
\(\frac{2(2 – \sqrt{3})}{2} = 2 – \sqrt{3}\)
Answer: 2 – \(\sqrt{3}\)
Q8: Rationalise the denominator of: \(\frac{3 – 2\sqrt{2}}{3 + 2\sqrt{2}}\)
Step 1: Identify the conjugate of the denominator \(3 + 2\sqrt{2}\), which is \(3 – 2\sqrt{2}\).
Step 2: Multiply both the numerator and the denominator by \(3 – 2\sqrt{2}\).
\(\frac{3 – 2\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 – 2\sqrt{2}}{3 – 2\sqrt{2}}\)
Step 3: Use the identity \((a – b)^2 = a^2 – 2ab + b^2\) for the numerator.
\((3 – 2\sqrt{2})^2 = (3)^2 – 2(3)(2\sqrt{2}) + (2\sqrt{2})^2\)
\(= 9 – 12\sqrt{2} + (4 \times 2)\)
\(= 9 – 12\sqrt{2} + 8 = 17 – 12\sqrt{2}\)
Step 4: Use the identity \((a + b)(a – b) = a^2 – b^2\) for the denominator.
\((3)^2 – (2\sqrt{2})^2 = 9 – (4 \times 2)\)
\(= 9 – 8 = 1\)
Step 5: Write the final simplified expression.
\(\frac{17 – 12\sqrt{2}}{1} = 17 – 12\sqrt{2}\)
Answer: 17 – 12\(\sqrt{2}\)
Q9: Rationalise the denominator of: \(\frac{1}{2\sqrt{5} – \sqrt{3}}\)
Step 1: Identify the conjugate of the denominator \(2\sqrt{5} – \sqrt{3}\), which is \(2\sqrt{5} + \sqrt{3}\).
Step 2: Multiply both the numerator and the denominator by \(2\sqrt{5} + \sqrt{3}\).
\(\frac{1}{2\sqrt{5} – \sqrt{3}} \times \frac{2\sqrt{5} + \sqrt{3}}{2\sqrt{5} + \sqrt{3}}\)
Step 3: Apply the identity \((a – b)(a + b) = a^2 – b^2\) to the denominator.
\(\frac{2\sqrt{5} + \sqrt{3}}{(2\sqrt{5})^2 – (\sqrt{3})^2}\)
Step 4: Simplify the squares in the denominator.
\((2\sqrt{5})^2 = 4 \times 5 = 20\)
\(( \sqrt{3})^2 = 3\)
\(\frac{2\sqrt{5} + \sqrt{3}}{20 – 3} = \frac{2\sqrt{5} + \sqrt{3}}{17}\)
Answer: \(\frac{2\sqrt{5} + \sqrt{3}}{17}\)
Q10: Rationalise the denominator of: \(\frac{1}{1 + \sqrt{5} + \sqrt{3}}\)
Step 1: Group the terms as \((1 + \sqrt{5}) + \sqrt{3}\). Multiply the numerator and denominator by the conjugate \((1 + \sqrt{5}) – \sqrt{3}\).
\(\frac{1}{(1 + \sqrt{5}) + \sqrt{3}} \times \frac{(1 + \sqrt{5}) – \sqrt{3}}{(1 + \sqrt{5}) – \sqrt{3}}\)
Step 2: Use the identity \((a + b)(a – b) = a^2 – b^2\) for the denominator.
Denominator: \((1 + \sqrt{5})^2 – (\sqrt{3})^2 = (1 + 5 + 2\sqrt{5}) – 3 = 3 + 2\sqrt{5}\).
The expression becomes: \(\frac{1 + \sqrt{5} – \sqrt{3}}{3 + 2\sqrt{5}}\).
Step 3: Multiply the new expression by the conjugate of the new denominator: \(3 – 2\sqrt{5}\).
\(\frac{1 + \sqrt{5} – \sqrt{3}}{3 + 2\sqrt{5}} \times \frac{3 – 2\sqrt{5}}{3 – 2\sqrt{5}}\)
Step 4: Simplify the denominator: \(3^2 – (2\sqrt{5})^2 = 9 – 20 = -11\).
Step 5: Multiply the numerator terms:
\(1(3 – 2\sqrt{5}) + \sqrt{5}(3 – 2\sqrt{5}) – \sqrt{3}(3 – 2\sqrt{5})\)
\(= 3 – 2\sqrt{5} + 3\sqrt{5} – 10 – 3\sqrt{3} + 2\sqrt{15}\)
\(= -7 + \sqrt{5} – 3\sqrt{3} + 2\sqrt{15}\).
Step 6: Write the final fraction with the positive denominator by adjusting the signs in the numerator.
\(\frac{-7 + \sqrt{5} – 3\sqrt{3} + 2\sqrt{15}}{-11} = \frac{7 – \sqrt{5} + 3\sqrt{3} – 2\sqrt{15}}{11}\)
Answer: \(\frac{7 – \sqrt{5} + 3\sqrt{3} – 2\sqrt{15}}{11}\)
Q11: Rationalise the denominator of: \(\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} – \sqrt{5}}\)
Step 1: Group the first two terms of the denominator: \((\sqrt{2} + \sqrt{3}) – \sqrt{5}\). Multiply the numerator and denominator by the conjugate \((\sqrt{2} + \sqrt{3}) + \sqrt{5}\).
\(\frac{\sqrt{2}[(\sqrt{2} + \sqrt{3}) + \sqrt{5}]}{[(\sqrt{2} + \sqrt{3}) – \sqrt{5}][(\sqrt{2} + \sqrt{3}) + \sqrt{5}]}\)
Step 2: Expand the numerator: \(\sqrt{2}(\sqrt{2}) + \sqrt{2}(\sqrt{3}) + \sqrt{2}(\sqrt{5}) = 2 + \sqrt{6} + \sqrt{10}\).
Step 3: Simplify the denominator using \((a-b)(a+b) = a^2 – b^2\):
\((\sqrt{2} + \sqrt{3})^2 – (\sqrt{5})^2\)
\(= (2 + 3 + 2\sqrt{6}) – 5\)
\(= 5 + 2\sqrt{6} – 5 = 2\sqrt{6}\).
Step 4: The expression now is: \(\frac{2 + \sqrt{6} + \sqrt{10}}{2\sqrt{6}}\).
Step 5: Multiply the numerator and denominator by \(\sqrt{6}\) to remove the remaining radical.
\(\frac{(2 + \sqrt{6} + \sqrt{10}) \times \sqrt{6}}{2\sqrt{6} \times \sqrt{6}}\)
Step 6: Distribute \(\sqrt{6}\) in the numerator: \(2\sqrt{6} + 6 + \sqrt{60}\).
Simplify \(\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}\).
Numerator = \(6 + 2\sqrt{6} + 2\sqrt{15}\).
Step 7: Simplify the denominator: \(2 \times 6 = 12\).
Step 8: Divide the entire expression by 2 to reach the simplest form:
\(\frac{2(3 + \sqrt{6} + \sqrt{15})}{12} = \frac{3 + \sqrt{6} + \sqrt{15}}{6}\).
Answer: \(\frac{3 + \sqrt{6} + \sqrt{15}}{6}\)
Q12: If \(\frac{\sqrt{3}+1}{\sqrt{3}-1} = a + b\sqrt{3}\), find the values of a and b.
Step 1: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{3} + 1\).
\(\frac{\sqrt{3} + 1}{\sqrt{3} – 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\)
Step 2: Apply the identity \((a + b)^2 = a^2 + 2ab + b^2\) for the numerator.
Numerator: \((\sqrt{3})^2 + 2(\sqrt{3})(1) + (1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}\).
Step 3: Apply the identity \((a – b)(a + b) = a^2 – b^2\) for the denominator.
Denominator: \((\sqrt{3})^2 – (1)^2 = 3 – 1 = 2\).
Step 4: Simplify the fraction.
\(\frac{4 + 2\sqrt{3}}{2} = \frac{4}{2} + \frac{2\sqrt{3}}{2} = 2 + \sqrt{3}\).
Step 5: Equate the simplified L.H.S. to the R.H.S.
\(2 + \sqrt{3} = a + b\sqrt{3}\).
Step 6: By comparing the rational and irrational parts on both sides:
\(a = 2\)
\(b\sqrt{3} = 1\sqrt{3} \Rightarrow b = 1\).
Answer: a = 2, b = 1
Q13: If \(\frac{3+\sqrt{2}}{3-\sqrt{2}} = a + b\sqrt{2}\), find the values of a and b.
Step 1: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(3 + \sqrt{2}\).
\(\frac{3 + \sqrt{2}}{3 – \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}\)
Step 2: Apply the identity \((a + b)^2 = a^2 + 2ab + b^2\) for the numerator.
Numerator: \((3)^2 + 2(3)(\sqrt{2}) + (\sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}\).
Step 3: Apply the identity \((a – b)(a + b) = a^2 – b^2\) for the denominator.
Denominator: \((3)^2 – (\sqrt{2})^2 = 9 – 2 = 7\).
Step 4: Write the expression as separate terms.
\(\frac{11 + 6\sqrt{2}}{7} = \frac{11}{7} + \frac{6}{7}\sqrt{2}\).
Step 5: Equate the simplified L.H.S. to the R.H.S.
\(\frac{11}{7} + \frac{6}{7}\sqrt{2} = a + b\sqrt{2}\).
Step 6: By comparing the rational and irrational parts on both sides:
\(a = \frac{11}{7}\)
\(b = \frac{6}{7}\).
Answer: a = \(\frac{11}{7}\), b = \(\frac{6}{7}\)
Q14: If \(\frac{5-\sqrt{6}}{5+\sqrt{6}} = a – b\sqrt{6}\), find the values of a and b.
Step 1: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(5 – \sqrt{6}\).
\(\frac{5 – \sqrt{6}}{5 + \sqrt{6}} \times \frac{5 – \sqrt{6}}{5 – \sqrt{6}}\)
Step 2: Apply the identity \((a – b)^2 = a^2 – 2ab + b^2\) for the numerator.
Numerator: \((5)^2 – 2(5)(\sqrt{6}) + (\sqrt{6})^2 = 25 – 10\sqrt{6} + 6 = 31 – 10\sqrt{6}\).
Step 3: Apply the identity \((a + b)(a – b) = a^2 – b^2\) for the denominator.
Denominator: \((5)^2 – (\sqrt{6})^2 = 25 – 6 = 19\).
Step 4: Write the simplified L.H.S. as separate terms.
\(\frac{31 – 10\sqrt{6}}{19} = \frac{31}{19} – \frac{10}{19}\sqrt{6}\).
Step 5: Equate the simplified L.H.S. to the R.H.S.
\(\frac{31}{19} + (-\frac{10}{19})\sqrt{6} = a – b\sqrt{6}\).
Step 6: By comparing the rational and irrational parts on both sides:
\(a = \frac{31}{19}\)
\(b = \frac{10}{19}\).
Answer: a = \(\frac{31}{19}\), b = \(\frac{10}{19}\)
Q15: If \(\frac{5+2\sqrt{3}}{7+4\sqrt{3}} = a – b\sqrt{3}\), find the values of a and b.
Step 1: Multiply the numerator and the denominator by the conjugate of the denominator, which is \(7 – 4\sqrt{3}\).
\(\frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \frac{7 – 4\sqrt{3}}{7 – 4\sqrt{3}}\)
Step 2: Multiply the terms in the numerator.
\(5(7 – 4\sqrt{3}) + 2\sqrt{3}(7 – 4\sqrt{3})\)
\(= 35 – 20\sqrt{3} + 14\sqrt{3} – 8(3)\)
\(= 35 – 6\sqrt{3} – 24 = 11 – 6\sqrt{3}\).
Step 3: Apply the identity \((a + b)(a – b) = a^2 – b^2\) for the denominator.
Denominator: \((7)^2 – (4\sqrt{3})^2 = 49 – (16 \times 3) = 49 – 48 = 1\).
Step 4: Simplify the fraction.
\(\frac{11 – 6\sqrt{3}}{1} = 11 – 6\sqrt{3}\).
Step 5: Equate the simplified L.H.S. to the R.H.S.
\(11 – 6\sqrt{3} = a – b\sqrt{3}\).
Step 6: By comparing the rational and irrational parts on both sides:
\(a = 11\)
\(b = 6\).
Answer: a = 11, b = 6
Q16: Simplify: \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} + \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
Step 1: The L.C.M. of the denominators \((\sqrt{5}-\sqrt{3})\) and \((\sqrt{5}+\sqrt{3})\) is their product.
Denominator: \((\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}) = (\sqrt{5})^2 – (\sqrt{3})^2 = 5 – 3 = 2\).
Step 2: Cross-multiply to find the new numerators.
Numerator: \((\sqrt{5}+\sqrt{3})^2 + (\sqrt{5}-\sqrt{3})^2\)
Step 3: Use the identity \((a+b)^2 + (a-b)^2 = 2(a^2 + b^2)\).
Numerator: \(2((\sqrt{5})^2 + (\sqrt{3})^2) = 2(5 + 3) = 2(8) = 16\).
Step 4: Place the simplified numerator over the common denominator.
\(\frac{16}{2} = 8\).
Answer: 8
Q17: Simplify: \( \frac{7+3\sqrt{5}}{3+\sqrt{5}} – \frac{7-3\sqrt{5}}{3-\sqrt{5}} \)
Step 1: Take LCM of denominators:
LCM = \( (3+\sqrt{5})(3-\sqrt{5}) \)
Step 2: Write expression with common denominator:
\[
= \frac{(7+3\sqrt{5})(3-\sqrt{5}) – (7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}
\]Step 3: Expand first term:
\[
(7+3\sqrt{5})(3-\sqrt{5}) = 21 – 7\sqrt{5} + 9\sqrt{5} – 15 = 6 + 2\sqrt{5}
\]Step 4: Expand second term:
\[
(7-3\sqrt{5})(3+\sqrt{5}) = 21 + 7\sqrt{5} – 9\sqrt{5} – 15 = 6 – 2\sqrt{5}
\]Step 5: Substitute in numerator:
\[
= (6 + 2\sqrt{5}) – (6 – 2\sqrt{5}) = 4\sqrt{5}
\]Step 6: Simplify denominator:
\[
(3+\sqrt{5})(3-\sqrt{5}) = 9 – 5 = 4
\]Step 7: Final result:
\[
= \frac{4\sqrt{5}}{4} = \sqrt{5}
\]Answer: \( \sqrt{5} \)
Q18: Show that: \(\frac{1}{3-\sqrt{8}} + \frac{1}{\sqrt{7}-\sqrt{6}} + \frac{1}{\sqrt{5}-2} – \frac{1}{\sqrt{8}-\sqrt{7}} – \frac{1}{\sqrt{6}-\sqrt{5}} = 5\)
Step 1: Rationalise the first term: \(\frac{1}{3-\sqrt{8}} \times \frac{3+\sqrt{8}}{3+\sqrt{8}} = \frac{3+\sqrt{8}}{9-8} = 3+\sqrt{8}\).
Step 2: Rationalise the second term: \(\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} = \frac{\sqrt{7}+\sqrt{6}}{7-6} = \sqrt{7}+\sqrt{6}\).
Step 3: Rationalise the third term: \(\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{\sqrt{5}+2}{5-4} = \sqrt{5}+2\).
Step 4: Rationalise the fourth term: \(\frac{1}{\sqrt{8}-\sqrt{7}} \times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}} = \frac{\sqrt{8}+\sqrt{7}}{8-7} = \sqrt{8}+\sqrt{7}\).
Step 5: Rationalise the fifth term: \(\frac{1}{\sqrt{6}-\sqrt{5}} \times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} = \frac{\sqrt{6}+\sqrt{5}}{6-5} = \sqrt{6}+\sqrt{5}\).
Step 6: Substitute the rationalised values back into the equation:
\((3+\sqrt{8}) + (\sqrt{7}+\sqrt{6}) + (\sqrt{5}+2) – (\sqrt{8}+\sqrt{7}) – (\sqrt{6}+\sqrt{5})\)
Step 7: Remove the brackets and cancel out opposite terms:
\(3 + \sqrt{8} + \sqrt{7} + \sqrt{6} + \sqrt{5} + 2 – \sqrt{8} – \sqrt{7} – \sqrt{6} – \sqrt{5}\)
Step 8: All radical terms cancel each other out, leaving only the integers:
\(3 + 2 = 5\).
Answer: L.H.S. = R.H.S. = 5. Hence Proved.
Q19: If \(x = 3 + \sqrt{8}\), find the value of \(x^2 + \frac{1}{x^2}\).
Step 1: Write \(\frac{1}{x}\) and rationalise the denominator by multiplying with the conjugate \(3 – \sqrt{8}\).
\(\frac{1}{3 + \sqrt{8}} \times \frac{3 – \sqrt{8}}{3 – \sqrt{8}}\)
Step 2: Simplify using the identity \((a + b)(a – b) = a^2 – b^2\).
Denominator: \(3^2 – (\sqrt{8})^2 = 9 – 8 = 1\).
So, \(\frac{1}{x} = 3 – \sqrt{8}\).
Step 3: Add the values of \(x\) and \(\frac{1}{x}\).
\(x + \frac{1}{x} = (3 + \sqrt{8}) + (3 – \sqrt{8})\)
\(x + \frac{1}{x} = 6\).
Step 4: Use the identity: \(x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2\).
Substitute the value of \(x + \frac{1}{x}\) into the identity:
\(x^2 + \frac{1}{x^2} = (6)^2 – 2\)
Step 5: Perform the final calculation.
\(36 – 2 = 34\).
Answer: 34
Q20: If \(x = 4 – \sqrt{15}\), find the value of \(x^2 + \frac{1}{x^2}\).
Step 1: Write \(\frac{1}{x}\) and rationalise the denominator by multiplying with the conjugate \(4 + \sqrt{15}\).
\(\frac{1}{4 – \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}}\)
Step 2: Use the identity \((a – b)(a + b) = a^2 – b^2\) for the denominator.
Denominator: \(4^2 – (\sqrt{15})^2 = 16 – 15 = 1\).
So, \(\frac{1}{x} = 4 + \sqrt{15}\).
Step 3: Add the values of \(x\) and \(\frac{1}{x}\).
\(x + \frac{1}{x} = (4 – \sqrt{15}) + (4 + \sqrt{15})\)
\(x + \frac{1}{x} = 8\).
Step 4: Use the algebraic identity: \(x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2\).
Substitute the value \(x + \frac{1}{x} = 8\) into the identity:
\(x^2 + \frac{1}{x^2} = (8)^2 – 2\)
Step 5: Perform the final calculation.
\(64 – 2 = 62\).
Answer: 62
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