Exercise 1.2 — Class 9 Mathematics Solutions

Solution

1. Identify the conjugate of the denominator.

   The denominator is $4 + \sqrt{3}$. Its conjugate is $4 - \sqrt{3}$.

2. Multiply numerator and denominator by the conjugate.

   $$\frac{13}{4+\sqrt{3}} \times \frac{4-\sqrt{3}}{4-\sqrt{3}} = \frac{13(4-\sqrt{3})}{(4+\sqrt{3})(4-\sqrt{3})}$$

3. Apply the difference of squares identity $(a+b)(a-b) = a^2 - b^2$.

   $$\frac{13(4-\sqrt{3})}{4^2 - (\sqrt{3})^2} = \frac{13(4-\sqrt{3})}{16 - 3} = \frac{13(4-\sqrt{3})}{13}$$

4. Simplify the fraction.

   Canceling the common factor $13$ from the numerator and denominator:

   $$= 4 - \sqrt{3}$$

Solution

1. Identify the rationalizing factor.

   The monomial denominator is $\sqrt{3}$. We multiply both numerator and denominator by $\sqrt{3}$.

2. Multiply numerator and denominator by $\sqrt{3}$.

   $$\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{(\sqrt{2}+\sqrt{5})\sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$

3. Distribute $\sqrt{3}$ across the numerator.

   $$\frac{\sqrt{2 \times 3} + \sqrt{5 \times 3}}{3} = \frac{\sqrt{6} + \sqrt{15}}{3}$$

Solution

1. Multiply numerator and denominator by $\sqrt{5}$.

   $$\frac{\sqrt{2}-1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{(\sqrt{2}-1)\sqrt{5}}{(\sqrt{5})^2}$$

2. Distribute and simplify.

   $$\frac{\sqrt{2} \times \sqrt{5} - 1 \times \sqrt{5}}{5} = \frac{\sqrt{10} - \sqrt{5}}{5}$$

Solution

1. Multiply numerator and denominator by conjugate $6-4\sqrt{2}$.

   $$\frac{6-4\sqrt{2}}{6+4\sqrt{2}} \times \frac{6-4\sqrt{2}}{6-4\sqrt{2}} = \frac{(6-4\sqrt{2})^2}{(6+4\sqrt{2})(6-4\sqrt{2})}$$

2. Expand the numerator using $(a-b)^2 = a^2 - 2ab + b^2$ and denominator using $a^2 - b^2$.

   $$\text{Numerator} = 6^2 - 2(6)(4\sqrt{2}) + (4\sqrt{2})^2 = 36 - 48\sqrt{2} + 16(2) = 36 - 48\sqrt{2} + 32 = 68 - 48\sqrt{2}$$ $$\text{Denominator} = 6^2 - (4\sqrt{2})^2 = 36 - 16(2) = 36 - 32 = 4$$

3. Divide each term by the denominator $4$.

   $$\frac{68 - 48\sqrt{2}}{4} = \frac{68}{4} - \frac{48\sqrt{2}}{4} = 17 - 12\sqrt{2}$$

Solution

1. Multiply numerator and denominator by conjugate $\sqrt{3}-\sqrt{2}$.

   $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$$

2. Expand numerator and denominator.

   $$\text{Numerator} = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}$$ $$\text{Denominator} = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$$

3. Write the final simplified form.

   $$\frac{5 - 2\sqrt{6}}{1} = 5 - 2\sqrt{6}$$

Solution

1. Multiply numerator and denominator by conjugate $\sqrt{7}-\sqrt{5}$.

   $$\frac{4\sqrt{3}}{\sqrt{7}+\sqrt{5}} \times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} = \frac{4\sqrt{3}(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^2 - (\sqrt{5})^2}$$

2. Simplify the denominator.

   $$(\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2$$

3. Cancel the common factor $2$.

   $$\frac{4\sqrt{3}(\sqrt{7}-\sqrt{5})}{2} = 2\sqrt{3}(\sqrt{7}-\sqrt{5})$$

Solution

1. Convert negative exponent to positive by taking the reciprocal.

   $$\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{16}{81}\right)^{\frac{3}{4}}$$

2. Express $16$ and $81$ as powers of $2$ and $3$.

   Since $16 = 2^4$ and $81 = 3^4$:

   $$\frac{16}{81} = \left(\frac{2}{3}\right)^4$$

3. Apply the power of a power rule $(a^m)^n = a^{mn}$.

   $$\left[\left(\frac{2}{3}\right)^4\right]^{\frac{3}{4}} = \left(\frac{2}{3}\right)^{4 \times \frac{3}{4}} = \left(\frac{2}{3}\right)^3$$

4. Calculate the final value.

   $$\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$$

Solution

1. Simplify each component into prime factors of $2$ and $3$.

   • $\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^2 = \frac{4^2}{3^2} = \frac{(2^2)^2}{3^2} = \frac{2^4}{3^2}$
   • $\left(\frac{4}{9}\right)^3 = \frac{4^3}{9^3} = \frac{(2^2)^3}{(3^2)^3} = \frac{2^6}{3^6}$
   • $\frac{16}{27} = \frac{2^4}{3^3}$

2. Change division into multiplication by inverting $\left(\frac{4}{9}\right)^3$.

   $$\frac{2^4}{3^2} \times \frac{3^6}{2^6} \times \frac{2^4}{3^3}$$

3. Combine powers of like bases using exponent laws.

   $$\frac{2^{4+4} \times 3^6}{2^6 \times 3^{2+3}} = \frac{2^8 \times 3^6}{2^6 \times 3^5} = 2^{8-6} \times 3^{6-5} = 2^2 \times 3^1$$

4. Compute the final product.

   $$2^2 \times 3 = 4 \times 3 = 12$$

Solution

1. Express decimal as a fraction.

   $$0.027 = \frac{27}{1000}$$

2. Express $27$ and $1000$ as cubes.

   $$\frac{27}{1000} = \left(\frac{3}{10}\right)^3$$

3. Apply the power $-\frac{1}{3}$.

   $$(0.027)^{-\frac{1}{3}} = \left[\left(\frac{3}{10}\right)^3\right]^{-\frac{1}{3}} = \left(\frac{3}{10}\right)^{3 \times \left(-\frac{1}{3}\right)} = \left(\frac{3}{10}\right)^{-1}$$

4. Invert the fraction.

   $$\left(\frac{3}{10}\right)^{-1} = \frac{10}{3}$$

Solution

1. Simplify the terms inside the radical using $\frac{a^m}{a^n} = a^{m-n}$.

   $$\frac{x^{14} \cdot y^{21} \cdot z^{35}}{y^{14} \cdot z^7} = x^{14} \cdot y^{21-14} \cdot z^{35-7} = x^{14} y^7 z^{28}$$

2. Convert radical to rational exponent $\frac{1}{7}$.

   $$\sqrt[7]{x^{14} y^7 z^{28}} = \left(x^{14} y^7 z^{28}\right)^{\frac{1}{7}}$$

3. Distribute the exponent to each variable.

   $$x^{14 \times \frac{1}{7}} \cdot y^{7 \times \frac{1}{7}} \cdot z^{28 \times \frac{1}{7}} = x^2 y^1 z^4 = x^2 y z^4$$

Solution

1. Express all bases in terms of prime base $5$.

   Since $25 = 5^2$:

   • $(25)^{n+1} = (5^2)^{n+1} = 5^{2n+2}$
   • $25 \cdot (5)^{2n} = 5^2 \cdot 5^{2n} = 5^{2n+2}$
   • $5 \cdot (5)^{2n+3} = 5^1 \cdot 5^{2n+3} = 5^{2n+4}$

2. Substitute into the numerator and denominator.

   $$\text{Numerator} = 5^1 \cdot 5^{2n+2} - 5^{2n+2} = 5^{2n+3} - 5^{2n+2} = 5^{2n+2}(5^1 - 1) = 5^{2n+2}(4)$$ $$\text{Denominator} = 5^{2n+4} - 5^{2n+2} = 5^{2n+2}(5^2 - 1) = 5^{2n+2}(25 - 1) = 5^{2n+2}(24)$$

3. Cancel common factors and simplify.

   $$\frac{5^{2n+2} \times 4}{5^{2n+2} \times 24} = \frac{4}{24} = \frac{1}{6}$$

Solution

1. Express all terms with base $2$.

   • $(16)^{x+1} = (2^4)^{x+1} = 2^{4x+4} = 2^{4x} \cdot 2^4 = 16 \cdot 2^{4x}$
   • $20(4^{2x}) = 20((2^2)^{2x}) = 20(2^{4x})$
   • $2^{x-3} \times 8^{x+2} = 2^{x-3} \times (2^3)^{x+2} = 2^{x-3} \times 2^{3x+6} = 2^{(x-3)+(3x+6)} = 2^{4x+3} = 8 \cdot 2^{4x}$

2. Combine terms in numerator.

   $$\text{Numerator} = 16 \cdot 2^{4x} + 20 \cdot 2^{4x} = (16+20) \cdot 2^{4x} = 36 \cdot 2^{4x}$$

3. Simplify the quotient.

   $$\frac{36 \cdot 2^{4x}}{8 \cdot 2^{4x}} = \frac{36}{8} = \frac{9}{2}$$

Solution

1. Evaluate each exponential term separately.

   • $(64)^{-\frac{2}{3}} = (4^3)^{-\frac{2}{3}} = 4^{3 \times \left(-\frac{2}{3}\right)} = 4^{-2} = \frac{1}{4^2} = \frac{1}{16}$
   • $(9)^{-\frac{3}{2}} = (3^2)^{-\frac{3}{2}} = 3^{2 \times \left(-\frac{3}{2}\right)} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}$

2. Perform division.

   $$(64)^{-\frac{2}{3}} \div (9)^{-\frac{3}{2}} = \frac{1}{16} \div \frac{1}{27} = \frac{1}{16} \times \frac{27}{1} = \frac{27}{16}$$

Solution

1. Express $9$ as $3^2$.

   • $9^{n+1} = (3^2)^{n+1} = 3^{2(n+1)} = 3^{2n+2}$
   • $9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2}$

2. Combine exponents in numerator and denominator.

   $$\text{Numerator} = 3^n \times 3^{2n+2} = 3^{n + 2n + 2} = 3^{3n+2}$$ $$\text{Denominator} = 3^{n-1} \times 3^{2n-2} = 3^{(n-1) + (2n-2)} = 3^{3n-3}$$

3. Apply quotient law $\frac{a^m}{a^n} = a^{m-n}$.

   $$\frac{3^{3n+2}}{3^{3n-3}} = 3^{(3n+2) - (3n-3)} = 3^{3n+2 - 3n + 3} = 3^5$$

4. Calculate the final value.

   $$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$$

Solution

1. Split the exponents in the numerator to isolate $5^n$.

   • $5^{n+3} = 5^n \cdot 5^3 = 125 \cdot 5^n$
   • $6 \cdot 5^{n+1} = 6 \cdot (5^n \cdot 5^1) = 30 \cdot 5^n$

2. Factor out $5^n$ from numerator and denominator.

   $$\text{Numerator} = 5^n(125 - 30) = 5^n(95)$$ $$\text{Denominator} = 5^n(9 - 4) = 5^n(5)$$

3. Cancel $5^n$ and simplify the fraction.

   $$\frac{5^n \times 95}{5^n \times 5} = \frac{95}{5} = 19$$

Solution

1. Calculate $\frac{1}{x}$ by rationalizing.

   Given $x = 3 + \sqrt{8}$:

   $$\frac{1}{x} = \frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}} = \frac{3-\sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3-\sqrt{8}}{9 - 8} = 3 - \sqrt{8}$$

2. Compute $x + \frac{1}{x}$.

   $$x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 3 + \sqrt{8} + 3 - \sqrt{8} = 6$$

Solution

1. Calculate $\frac{1}{x}$ by rationalizing the denominator.

   Given $x = 3 + \sqrt{8}$:

   $$\frac{1}{x} = \frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}} = \frac{3-\sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3-\sqrt{8}}{9 - 8} = 3 - \sqrt{8}$$

2. Find the sum $x + \frac{1}{x}$.

   $$x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 3 + 3 = 6$$

3. Take the square on both sides using $(a + b)^2 = a^2 + 2ab + b^2$.

   Squaring both sides of $x + \frac{1}{x} = 6$:

   $$\left(x + \frac{1}{x}\right)^2 = (6)^2$$

   Expanding the left hand side:

   $$x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 36$$

   Since $x \times \frac{1}{x} = 1$:

   $$x^2 + 2(1) + \frac{1}{x^2} = 36$$ $$x^2 + 2 + \frac{1}{x^2} = 36$$

4. Subtract $2$ from both sides to isolate $x^2 + \frac{1}{x^2}$.

   $$x^2 + \frac{1}{x^2} = 36 - 2 = 34$$

Solution

1. Simplify each factor in prime exponential form.

   • $(25)^{\frac{3}{2}} = (5^2)^{\frac{3}{2}} = 5^{2 \times \frac{3}{2}} = 5^3 = 125$
   • $(243)^{\frac{3}{5}} = (3^5)^{\frac{3}{5}} = 3^{5 \times \frac{3}{5}} = 3^3 = 27$
   • $(16)^{\frac{5}{4}} = (2^4)^{\frac{5}{4}} = 2^{4 \times \frac{5}{4}} = 2^5 = 32$
   • $(8)^{\frac{4}{3}} = (2^3)^{\frac{4}{3}} = 2^{3 \times \frac{4}{3}} = 2^4 = 16$

2. Substitute simplified values into the expression.

   $$\frac{125 \times 27}{32 \times 16}$$

3. Multiply out numerator and denominator.

   $$\text{Numerator} = 125 \times 27 = 3375$$ $$\text{Denominator} = 32 \times 16 = 512$$

4. Final irreducible fraction.

   $$= \frac{3375}{512}$$

Solution

1. Express all terms with base $3$.

   • $\sqrt[3]{(27)^{2x}} = \sqrt[3]{(3^3)^{2x}} = \sqrt[3]{3^{6x}} = (3^{6x})^{\frac{1}{3}} = 3^{2x}$
   • $9^{x+1} = (3^2)^{x+1} = 3^{2x+2} = 3^{2x} \cdot 3^2 = 9 \cdot 3^{2x}$
   • $216(3^{2x-1}) = 216 \left(\frac{3^{2x}}{3^1}\right) = \frac{216}{3} \cdot 3^{2x} = 72 \cdot 3^{2x}$

2. Combine denominator terms.

   $$\text{Denominator} = 9 \cdot 3^{2x} + 72 \cdot 3^{2x} = (9 + 72) \cdot 3^{2x} = 81 \cdot 3^{2x}$$

3. Form the quotient and cancel $3^{2x}$.

   $$\frac{54 \times 3^{2x}}{81 \times 3^{2x}} = \frac{54}{81}$$

4. Reduce fraction by dividing numerator and denominator by $27$.

   $$\frac{54 \div 27}{81 \div 27} = \frac{2}{3}$$

Solution

1. Simplify each factor inside the square root.

   • $(216)^{\frac{2}{3}} = (6^3)^{\frac{2}{3}} = 6^{3 \times \frac{2}{3}} = 6^2 = 36$
   • $(25)^{\frac{1}{2}} = \sqrt{25} = 5$
   • $(0.04)^{-\frac{3}{2}} = \left(\frac{4}{100}\right)^{-\frac{3}{2}} = \left(\frac{1}{25}\right)^{-\frac{3}{2}} = (25)^{\frac{3}{2}} = (5^2)^{\frac{3}{2}} = 5^3 = 125$

2. Substitute values inside the square root.

   $$\sqrt{\frac{36 \times 5}{125}}$$

3. Simplify fraction inside the root.

   $$\frac{36 \times 5}{125} = \frac{36}{25}$$

4. Take the square root.

   $$\sqrt{\frac{36}{25}} = \frac{\sqrt{36}}{\sqrt{25}} = \frac{6}{5}$$

Solution

1. Identify the algebraic pattern.

   Recall the algebraic identity:

   $$(x + y)(x^2 - xy + y^2) = x^3 + y^3$$

2. Set substitutions.

   Let:

   • $x = a^{\frac{1}{3}} \implies x^2 = \left(a^{\frac{1}{3}}\right)^2 = a^{\frac{2}{3}}$
   • $y = b^{\frac{2}{3}} \implies y^2 = \left(b^{\frac{2}{3}}\right)^2 = b^{\frac{4}{3}}$
   • $xy = a^{\frac{1}{3}} b^{\frac{2}{3}}$

3. Apply the identity.

   $$\left(a^{\frac{1}{3}} + b^{\frac{2}{3}}\right) \left(a^{\frac{2}{3}} - a^{\frac{1}{3}}b^{\frac{2}{3}} + b^{\frac{4}{3}}\right) = \left(a^{\frac{1}{3}}\right)^3 + \left(b^{\frac{2}{3}}\right)^3$$

4. Simplify powers.

   $$= a^{3 \times \frac{1}{3}} + b^{3 \times \frac{2}{3}} = a^1 + b^2 = a + b^2$$