Here the complete solution of Exercise 1.2 from the new Math 9th syllabus, designed by the Punjab Curriculum and Textbook Board (PCTB), Lahore.
1. Rationalize the denominator
(i) \( \frac{13}{4 + \sqrt{5}} \)
\[\frac{13}{4 + \sqrt{5}} \times \frac{4 – \sqrt{5}}{4 – \sqrt{5}} = \frac{13(4 – \sqrt{5})}{16 – 5} = \frac{13(4 – \sqrt{5})}{11} = \frac{52}{11} – \frac{13\sqrt{5}}{11}\]
(ii) \( \frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}} \)
\[\frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6} + \sqrt{15}}{3}\]
(iii) \( \frac{\sqrt{2} – 1}{\sqrt{5}} \)
\[\frac{\sqrt{2} – 1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{10} – \sqrt{5}}{5}\]
(iv) \( \frac{6 – 4\sqrt{2}}{6 + 4\sqrt{2}} \)
\[\frac{6 – 4\sqrt{2}}{6 + 4\sqrt{2}} \times \frac{6 – 4\sqrt{2}}{6 – 4\sqrt{2}} = \frac{(6 – 4\sqrt{2})^2}{36 – (4\sqrt{2})^2} = \frac{36 – 48\sqrt{2} + 32}{36 – 32} = \frac{68 – 48\sqrt{2}}{4} = 17 – 12\sqrt{2}\]
(v) \( \frac{\sqrt{3} – \sqrt{2}}{\sqrt{3} + \sqrt{2}} \)
\[\frac{\sqrt{3} – \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} – \sqrt{2}}{\sqrt{3} – \sqrt{2}} = \frac{(\sqrt{3} – \sqrt{2})^2}{3 – 2} = \frac{3 – 2\sqrt{6} + 2}{1} = 5 – 2\sqrt{6}\]
(vi) \( \frac{4\sqrt{3}}{\sqrt{7} + \sqrt{5}} \)
\[\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{5}} \times \frac{\sqrt{7} – \sqrt{5}}{\sqrt{7} – \sqrt{5}} = \frac{4\sqrt{3}(\sqrt{7} – \sqrt{5})}{7 – 5} = 2\sqrt{21} – 2\sqrt{15}\]
2. Simplify the following
(i) \( \left( \frac{81}{16} \right)^{-\frac{3}{4}} \)
\[\left( \frac{81}{16} \right)^{-\frac{3}{4}} = \left( \frac{3^4}{2^4} \right)^{-\frac{3}{4}} = \left( \frac{3}{2} \right)^{-3} = \frac{2^3}{3^3} = \frac{8}{27}\]
(ii) \( \left( \frac{3}{4} \right)^{-2} \div \left( \frac{4}{9} \right)^3 \times \frac{16}{27} \)
\[\left( \frac{3}{4} \right)^{-2} \div \left( \frac{4}{9} \right)^3 \times \frac{16}{27}\]
\[= \left( \frac{4}{3} \right)^2 \times \left( \frac{9}{4} \right)^3 \times \frac{16}{27}\]
\[= \left( \frac{2^2}{3} \right)^2 \times \left( \frac{3^2}{2^2} \right)^3 \times \frac{2^4}{3^3}\]
\[= \frac{2^4}{3^2} \times \frac{3^6}{2^6} \times \frac{2^4}{3^3}\]
\[2^{4-6+4}\times 3^{6-2-3}\]
\[= 2^2 \times 3\]
\[= 4 \times 3\]
\[= 12\]
(iii) \( (0.027)^{-\frac{1}{3}} \)
\[(0.027)^{-\frac{1}{3}} = \left( \frac{27}{1000} \right)^{-\frac{1}{3}} = \left( \frac{3^3}{10^3} \right)^{-\frac{1}{3}} = \frac{10}{3}\]
(iv) \( \sqrt[7]{\frac{x^{14} \times y^{21} \times z^{35}}{y^{14} \times z^7}} \)
\[\frac{x^{14} \times y^{21} \times z^{35}}{y^{14} \times z^7} = x^{14} \times y^7 \times z^{28}\]
\[\sqrt[7]{x^{14} \times y^7 \times z^{28}} = x^2 \times y \times z^4\]
(v) \( \frac{5 \cdot (25)^{n+1} – 25 \cdot (5)^{2n}}{5 \cdot (5)^{2n+3} – (25)^{n+1}} \)
\[\frac{5 \cdot (25)^{n+1} – 25 \cdot (5)^n}{5 \cdot (5)^{2n+3} – (25)^{n+1}}\]
\[= \frac{5 \cdot (5^2)^{n+1} – 5^2 \cdot (5)^n}{5 \cdot (5)^{2n+3} – (5^2)^{n+1}}\]
\[= \frac{5 \cdot 5^{2n+2} – 5^2 \cdot 5^n}{5 \cdot 5^{2n+3} – 5^{2n+2}}\]
\[= \frac{5^{2n+3} – 5^{2n+2}}{5^{2n+4} – 5^{2n+2}}\]
\[= \frac{5^{2n+2} (5 – 1)}{5^{2n+2} (5^2 – 1)}\]
\[= \frac{5 – 1}{25 – 1}\]
\[= \frac{4}{24}\]
\[= \frac{1}{6}\]
(vi) \( \frac{(16)^{x+1} \cdot 20(4^{2x})}{2^{x-3} \cdot 8^{x+2}} \)
\[\frac{(16)^{x+1} + 20(4^{2x})}{2^{x-3} \times 8^{x+2}}\]
\[= \frac{(2^4)^{x+1} + (2^2 \times 5) (2^2)^{2x}}{2^{x-3} \times (2^3)^{x+2}}\]
\[= \frac{2^{4x+4} + 2^2 \cdot 5 \cdot 2^{4x}}{2^{x-3} \cdot 2^{3x+6}}\]
\[= \frac{2^{4x+4} + 5 \cdot 2^{4x+2}}{2^{x-3+3x+6}}\]
\[= \frac{2^{4x+4} + 5 \cdot 2^{4x+2}}{2^{4x+3}}\]
Split the numerator it follows that:
\[= \frac{2^{4x+4}}{2^{4x+3}} + \frac{5 \cdot 2^{4x+2}}{2^{4x+3}}\]
\[= 2^{4x+4-(4x+3)} + 5 \cdot 2^{4x+2-(4x+3)}\]
\[= 2^1 + 5 \cdot 2^{-1}\]
\[= 2 + \frac{5}{2}\]
\[= \frac{4}{2} + \frac{5}{2}\]
\[= \frac{9}{2}\]
(vii) \( \frac{(64)^{-\frac{2}{3}}}{(9)^{-\frac{3}{2}}} \)
\[(64)^{-\frac{2}{3}} = \left( 4^3 \right)^{-\frac{2}{3}} = 4^{-2} = \frac{1}{16}, \quad (9)^{-\frac{3}{2}} = (3^2)^{-\frac{3}{2}} = 3^{-3} = \frac{1}{27}\]
\[\frac{\frac{1}{16}}{\frac{1}{27}} = \frac{27}{16}\]
(viii) \( \frac{3^n \cdot 9^{n+1}}{3^n \cdot 9^{n-1}} \)
\[\frac{3^n \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}\]
\[= \frac{3^n \times (3^2)^{n+1}}{3^{n-1} \times (3^2)^{n-1}}\]
\[= \frac{3^n \times 3^{2n+2}}{3^{n-1} \times 3^{2n-2}}\]
\[= \frac{3^{n + 2n + 2}}{3^{n-1 + 2n-2}}\]
\[= \frac{3^{3n+2}}{3^{3n-3}}\]
\[= 3^{3n+2 – (3n-3)}\]
\[= 3^{3n+2 – 3n+3}\]
\[= 3^5\]
\[= 243\]
(ix) \( \frac{5^{n+3} – 6 \cdot 5^{n+1}}{9 \cdot 5^n – 2^n \cdot 5^n} \)
\[\frac{5^{n+3} – 6 \cdot 5^{n+1}}{9 \cdot 5^n – 2^n \cdot 5^n}\]
\[= \frac{5^n \cdot (5^3 – 6 \cdot 5^1)}{5^n \cdot (9 – 2^n)}\]
\[= \frac{5^3 – 6 \cdot 5}{9 – 2^n}\]
\[= \frac{125 – 30}{9 – 2^n}\]
\[= \frac{95}{9 – 2^n}\]
Question 3
Given \( x = 3 + \sqrt{8} \), we also know that:
\[\frac{1}{x} = \frac{1}{3 + \sqrt{8}} \cdot \frac{3 – \sqrt{8}}{3 – \sqrt{8}} = \frac{3 – \sqrt{8}}{9 – 8} = 3 – \sqrt{8}\]
Now let’s calculate each part.
(i) \( x + \frac{1}{x} \)
\[x + \frac{1}{x} = (3 + \sqrt{8}) + (3 – \sqrt{8}) = 6.\]
(ii) \( x – \frac{1}{x} \)
\[x – \frac{1}{x} = (3 + \sqrt{8}) – (3 – \sqrt{8}) = 2\sqrt{8} = 4\sqrt{2}\]
(iii) \( x^2 + \frac{1}{x^2} \)
\[x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 – 2 = 6^2 – 2 = 36 – 2 = 34\]
(iv) \( x^2 – \frac{1}{x^2} \)
\[x^2 – \frac{1}{x^2} = (x+\frac{1}{x})(x-\frac{1}{x}) = 6\times 4\sqrt{2}=24\sqrt{2}\]
(v) \( x^4 + \frac{1}{x^4} \)
\[x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 – 2 = 34^2 – 2 = 1156 – 2 = 1154\]
(vi) \( \left(x – \frac{1}{x}\right)^2 \)
\[\left(x – \frac{1}{x}\right)^2 = (4\sqrt{2})^2 = 32\]
Find the rational numbers \( p \) and \( q \) such that \(\frac{8 – 3\sqrt{2}}{4 + 3\sqrt{2}} = p + q\sqrt{2}\)
\[\frac{8 – 3\sqrt{2}}{4 + 3\sqrt{2}}.\]
\[\frac{8 – 3\sqrt{2}}{4 + 3\sqrt{2}} \cdot \frac{4 – 3\sqrt{2}}{4 – 3\sqrt{2}} = \frac{(8 – 3\sqrt{2})(4 – 3\sqrt{2})}{(4 + 3\sqrt{2})(4 – 3\sqrt{2})}\]
Simplify the denominator
\[(4 + 3\sqrt{2})(4 – 3\sqrt{2}) = 4^2 – (3\sqrt{2})^2 = 16 – 18 = -2\]
Simplify the numerator:
\[(8 – 3\sqrt{2})(4 – 3\sqrt{2}) = 8 \cdot 4 – 8 \cdot 3\sqrt{2} – 3\sqrt{2} \cdot 4 + (3\sqrt{2})^2 = 32 – 24\sqrt{2} – 12\sqrt{2} + 18\]
Combine terms:
\[32 + 18 – 36\sqrt{2} = 50 – 36\sqrt{2}.\]
Thus:
\[\frac{8 – 3\sqrt{2}}{4 + 3\sqrt{2}} = \frac{50 – 36\sqrt{2}}{-2} = -\frac{50}{2} + \frac{36\sqrt{2}}{2} = -25 + 18\sqrt{2}\]
\[-25 + 18 \sqrt{2}= p + q \sqrt{2}\]
Compare the above for \(p) and \(q):
\[p = -25, \quad q = 18\]
Question 5
(i) \[\frac{\left(25\right)^{\frac{3}{2}} \times \left(243\right)^{\frac{3}{5}}}{\left(16\right)^{\frac{5}{4}} \times \left(8\right)^{\frac{4}{3}}}\]
\[25^{\frac{3}{2}} = 125, \quad 243^{\frac{3}{5}} = 27, \quad 16^{\frac{5}{4}} = 32, \quad 8^{\frac{4}{3}} = 16\]
\[\frac{125 \times 27}{32 \times 16} = \frac{3375}{512}\]
Answer:
\[\frac{3375}{512}\]
(ii) \[\frac{54 \times \sqrt[3]{(27)^{2x}}}{9^{x+1} + 216 \times 3^{2x-1}}\]
\[\frac{54 \cdot 3^x \cdot \sqrt[3]{(27)^{2x}}}{9^{x+1} + 216 \cdot (3^{2x-1})}\]
\[= \frac{2 \cdot 3^3 \cdot \left( (3^3)^{2x} \right)^{1/3}}{(3^2)^{x+1} + 2^3 \cdot 3^3 \cdot (3^{2x-1})}\]
\[= \frac{2 \cdot 3^3 \cdot (3^{2x})}{3^{2x+2} + 2^3 \cdot 3^3 \cdot 3^{2x-1}}\]
\[= \frac{2 \cdot 3^{2x+3}}{3^{2x+2} + 2^3 \cdot 3^{2x+2}}\]
\[= \frac{2 \cdot 3^{2x+3}}{3^{2x+2} (1 + 2^3)}\]
\[= \frac{2 \cdot 3^{2x+3}}{3^{2x+2} \cdot 9}\]
\[= \frac{2 \cdot 3}{9}\]
\[= \frac{6}{9}\]
\[= \frac{2}{3}\]
(iii) \[\sqrt{\frac{(216)^{\frac{2}{3}} \times (25)^{\frac{1}{2}}}{(0.04)^{-\frac{3}{2}}}}\]
\[\sqrt{\frac{(216)^{2/3} \cdot (25)^{1/2}}{(0.04)^{-3/2}}}\]
\[= \left[ \frac{(6^3)^{2/3} \cdot (5^2)^{1/2}}{\left(\frac{4}{100}\right)^{-3/2}} \right]^{1/2}\]
\[= \left[ \frac{6^2 \cdot 5}{(1/25)^{-3/2}} \right]^{1/2}\]
\[= \left[ \frac{6^2 \cdot 5}{(5^{-2})^{-3/2}} \right]^{1/2}\]
\[= \left[ \frac{6^2 \cdot 5}{5^3} \right]^{1/2}\]
\[= \left( \frac{6^2}{5^2} \right)^{1/2}\]
\[= \frac{6}{5}\]
(iv) \[(a^{\frac{1}{3}} + b^{\frac{2}{3}}) \times (a^{\frac{2}{3}} – a^{\frac{1}{3}}b^{\frac{2}{3}} + b^{\frac{4}{3}})\]
\[= a^{\frac{1}{3}} \times a^{\frac{2}{3}} + a^{\frac{1}{3}} \times (-a^{\frac{1}{3}}b^{\frac{2}{3}}) + a^{\frac{1}{3}} \times b^{\frac{4}{3}}\]
\[+ b^{\frac{2}{3}} \times a^{\frac{2}{3}} + b^{\frac{2}{3}} \times (-a^{\frac{1}{3}}b^{\frac{2}{3}}) + b^{\frac{2}{3}} \times b^{\frac{4}{3}}\]
Combine powers:
\[= a^{1} – a^{\frac{2}{3}}b^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{4}{3}} + a^{\frac{2}{3}}b^{\frac{2}{3}} – a^{\frac{1}{3}}b^{\frac{4}{3}} + b^{2}\]
Cancel terms:
\[= a^{1} + b^{2}\]
Answer:
\[a + b^2\]
