Maharashtra Board Class 9 Maths Chapter 2 Set 2.4 Algebra Standard Part 1 Real Numbers Solutions

Question 1. Multiply.
(i) \( \sqrt{3}(\sqrt{7} - \sqrt{3}) \)
(ii) \( (\sqrt{5} - \sqrt{7})\sqrt{2} \)
(iii) \( (3\sqrt{2} - \sqrt{3})(4\sqrt{3} - \sqrt{2}) \)
Answer:
(i) \( \sqrt{3}(\sqrt{7} - \sqrt{3}) = \sqrt{3} \times \sqrt{7} - \sqrt{3} \times \sqrt{3} = \sqrt{21} - 3 \)
(ii) \( (\sqrt{5} - \sqrt{7})\sqrt{2} = \sqrt{5} \times \sqrt{2} - \sqrt{7} \times \sqrt{2} = \sqrt{10} - \sqrt{14} \)
(iii) \( (3\sqrt{2} - \sqrt{3})(4\sqrt{3} - \sqrt{2}) = 3\sqrt{2}(4\sqrt{3} - \sqrt{2}) - \sqrt{3}(4\sqrt{3} - \sqrt{2}) = 12\sqrt{6} - 6 - 12 + \sqrt{6} = 13\sqrt{6} - 18 \). Multiplying surds involves distributing the terms and simplifying the square roots where possible.
In simple words: To multiply these expressions, we distribute the term outside the bracket to each term inside, just like normal algebra, and then simplify the square roots.

🎯 Exam Tip: Remember that \( \sqrt{a} \times \sqrt{a} = a \). Always simplify the terms inside the square roots completely to avoid calculation errors.

Question 1. Multiply:
(i) \( \sqrt{3}(\sqrt{7} - \sqrt{3}) \)
(ii) \( (\sqrt{5} - \sqrt{7})\sqrt{2} \)
(iii) \( (3\sqrt{2} - \sqrt{3})(4\sqrt{3} - \sqrt{2}) \)
Answer:
(i) \( \sqrt{3}(\sqrt{7} - \sqrt{3}) = \sqrt{3} \times \sqrt{7} - \sqrt{3} \times \sqrt{3} \)
\( = \sqrt{3 \times 7} - \sqrt{3 \times 3} \)
\( \therefore \sqrt{3}(\sqrt{7} - \sqrt{3}) = \sqrt{21} - 3 \)

(ii) \( (\sqrt{5} - \sqrt{7})\sqrt{2} = \sqrt{5} \times \sqrt{2} - \sqrt{7} \times \sqrt{2} \)
\( = \sqrt{5 \times 2} - \sqrt{7 \times 2} \)
\( \therefore (\sqrt{5} - \sqrt{7})\sqrt{2} = \sqrt{10} - \sqrt{14} \)

(iii) \( (3\sqrt{2} - \sqrt{3})(4\sqrt{3} - \sqrt{2}) \)
\( = 3\sqrt{2}(4\sqrt{3} - \sqrt{2}) - \sqrt{3}(4\sqrt{3} - \sqrt{2}) \)
\( = 3\sqrt{2} \times 4\sqrt{3} - 3\sqrt{2} \times \sqrt{2} - \sqrt{3} \times 4\sqrt{3} + \sqrt{3} \times \sqrt{2} \)
\( = 12\sqrt{2 \times 3} - 3\sqrt{2 \times 2} - 4\sqrt{3 \times 3} + \sqrt{3 \times 2} \)
\( = 12\sqrt{6} - (3 \times 2) - (4 \times 3) + \sqrt{6} \)
\( = 12\sqrt{6} - 6 - 12 + \sqrt{6} \)
\( = (12 + 1)\sqrt{6} - 6 - 12 \)
\( \therefore (3\sqrt{2} - \sqrt{3})(4\sqrt{3} - \sqrt{2}) = 13\sqrt{6} - 18 \)
This step-by-step expansion helps simplify complex surds systematically.
In simple words: To multiply surds, distribute the terms outside the bracket to each term inside, and multiply the numbers under the square root signs together.

🎯 Exam Tip: Remember that \( \sqrt{a} \times \sqrt{a} = a \). Keeping this simple rule in mind prevents common calculation errors during expansion.

Question 2. Rationalize the denominator.
(i) \( \frac{1}{\sqrt{7} + \sqrt{2}} \)
(ii) \( \frac{3}{2\sqrt{5} - 3\sqrt{2}} \)
(iii) \( \frac{4}{7 + 4\sqrt{3}} \)
(iv) \( \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \)
Answer:
(i) \( \frac{1}{\sqrt{7} + \sqrt{2}} = \frac{1}{\sqrt{7} + \sqrt{2}} \times \frac{\sqrt{7} - \sqrt{2}}{\sqrt{7} - \sqrt{2}} \) [Multiplying the numerator and denominator by the conjugate \( (\sqrt{7} - \sqrt{2}) \)]
\( = \frac{\sqrt{7} - \sqrt{2}}{(\sqrt{7})^2 - (\sqrt{2})^2} \)
\( = \frac{\sqrt{7} - \sqrt{2}}{7 - 2} \)
\( \therefore \frac{1}{\sqrt{7} + \sqrt{2}} = \frac{\sqrt{7} - \sqrt{2}}{5} \)

(ii) \( \frac{3}{2\sqrt{5} - 3\sqrt{2}} = \frac{3}{2\sqrt{5} - 3\sqrt{2}} \times \frac{2\sqrt{5} + 3\sqrt{2}}{2\sqrt{5} + 3\sqrt{2}} \) [Multiplying the numerator and denominator by the conjugate \( (2\sqrt{5} + 3\sqrt{2}) \)]
\( = \frac{3(2\sqrt{5} + 3\sqrt{2})}{(2\sqrt{5})^2 - (3\sqrt{2})^2} \)
\( = \frac{3(2\sqrt{5} + 3\sqrt{2})}{4(5) - 9(2)} \)
\( = \frac{3(2\sqrt{5} + 3\sqrt{2})}{20 - 18} \)
\( \therefore \frac{3}{2\sqrt{5} - 3\sqrt{2}} = \frac{3(2\sqrt{5} + 3\sqrt{2})}{2} \)

(iii) \( \frac{4}{7 + 4\sqrt{3}} = \frac{4}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} \) [Multiplying the numerator and denominator by the conjugate \( (7 - 4\sqrt{3}) \)]
\( = \frac{4(7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \)
\( = \frac{4(7 - 4\sqrt{3})}{49 - 16(3)} \)
\( = \frac{4(7 - 4\sqrt{3})}{49 - 48} \)
\( \therefore \frac{4}{7 + 4\sqrt{3}} = 4(7 - 4\sqrt{3}) = 28 - 16\sqrt{3} \)

(iv) \( \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} \) [Multiplying the numerator and denominator by the conjugate \( (\sqrt{5} - \sqrt{3}) \)]
\( = \frac{(\sqrt{5} - \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} \)
\( = \frac{(\sqrt{5})^2 - 2\sqrt{5}\sqrt{3} + (\sqrt{3})^2}{5 - 3} \)
\( = \frac{5 - 2\sqrt{15} + 3}{2} \)
\( = \frac{8 - 2\sqrt{15}}{2} \)
\( \therefore \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = 4 - \sqrt{15} \)
Rationalizing the denominator is a fundamental technique used to simplify expressions containing surds.
In simple words: To rationalize a denominator, multiply both the top and bottom of the fraction by the conjugate of the denominator. This removes the square roots from the bottom of the fraction.

🎯 Exam Tip: Always use the identity \( (a+b)(a-b) = a^2 - b^2 \) to simplify the denominator. Double-check your signs when multiplying conjugates to avoid simple arithmetic errors.

Question 1. Rationalize the denominator:
(i) \( \frac{1}{\sqrt{7}+\sqrt{2}} \)
(ii) \( \frac{3}{2\sqrt{5}-3\sqrt{2}} \)
Answer:
(i) \( \frac{1}{\sqrt{7}+\sqrt{2}} \)
Multiplying the numerator and denominator by \( (\sqrt{7}-\sqrt{2}) \):
\( = \frac{1}{\sqrt{7}+\sqrt{2}} \times \frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}} \)
\( = \frac{\sqrt{7}-\sqrt{2}}{(\sqrt{7})^2-(\sqrt{2})^2} \)
\( \text{... } [\because (a-b)(a+b) = a^2 - b^2] \)
\( = \frac{\sqrt{7}-\sqrt{2}}{7-2} \)
\( \therefore \frac{1}{\sqrt{7}+\sqrt{2}} = \frac{\sqrt{7}-\sqrt{2}}{5} \)

(ii) \( \frac{3}{2\sqrt{5}-3\sqrt{2}} \)
Multiplying the numerator and denominator by \( (2\sqrt{5}+3\sqrt{2}) \):
\( = \frac{3}{2\sqrt{5}-3\sqrt{2}} \times \frac{2\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+3\sqrt{2}} \)
\( \text{... [Multiplying the numerator and denominator by } (2\sqrt{5}+3\sqrt{2})] \)
\( = \frac{3(2\sqrt{5}+3\sqrt{2})}{(2\sqrt{5})^2-(3\sqrt{2})^2} \)
\( \text{... } [\times \text{ because } (a-b)(a+b) = a^2 - b^2] \)
\( = \frac{3(2\sqrt{5}+3\sqrt{2})}{(4 \times 5)-(9 \times 2)} \)
\( = \frac{3(2\sqrt{5}+3\sqrt{2})}{20-18} \)
\( \therefore \frac{3}{2\sqrt{5}-3\sqrt{2}} = \frac{3(2\sqrt{5}+3\sqrt{2})}{2} \)
In simple words: To rationalize a denominator, we multiply both the top and bottom of the fraction by the conjugate of the denominator (changing the sign in the middle). This helps eliminate the square roots from the bottom using the formula \( (a-b)(a+b) = a^2 - b^2 \).

🎯 Exam Tip: Always remember to change the sign between the terms when finding the conjugate to multiply. Double-check your calculations when squaring terms like \( (2\sqrt{5})^2 \) to avoid simple arithmetic errors.

Question 1. Rationalize the denominator:
(iii) \( \frac{4}{7 + 4\sqrt{3}} \)
(iv) \( \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \)
Answer:
(iii)
\( \frac{4}{7 + 4\sqrt{3}} = \frac{4}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} \) [Multiplying the numerator and denominator by \( (7 - 4\sqrt{3}) \)]
\( = \frac{4(7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \) [\( \because (a - b)(a + b) = a^2 - b^2 \)]
\( = \frac{4(7 - 4\sqrt{3})}{49 - (16 \times 3)} \)
\( = \frac{4(7 - 4\sqrt{3})}{49 - 48} \)
\( = \frac{4(7 - 4\sqrt{3})}{1} \)
\( = 28 - 16\sqrt{3} \)
\( \therefore \frac{4}{7 + 4\sqrt{3}} = 28 - 16\sqrt{3} \)

(iv)
\( \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3})} \times \frac{(\sqrt{5} - \sqrt{3})}{(\sqrt{5} - \sqrt{3})} \) [Multiplying the numerator and denominator by \( (\sqrt{5} - \sqrt{3}) \)]
\( = \frac{(\sqrt{5} - \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} \) [\( \because (a - b)(a + b) = a^2 - b^2 \)]
\( = \frac{(\sqrt{5})^2 - 2 \times \sqrt{5} \times \sqrt{3} + (\sqrt{3})^2}{5 - 3} \) [\( \because (a - b)^2 = a^2 - 2ab + b^2 \)]
\( = \frac{5 - 2\sqrt{15} + 3}{2} \)
\( = \frac{8 - 2\sqrt{15}}{2} \)
\( = \frac{2(4 - \sqrt{15})}{2} \)
\( = 4 - \sqrt{15} \)
\( \therefore \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = 4 - \sqrt{15} \)
In simple words: To rationalize a denominator, we multiply both the top and bottom of the fraction by the conjugate of the denominator. This eliminates the square roots from the bottom, making the expression simpler to work with.

🎯 Exam Tip: Always remember to change the sign of the middle term when finding the conjugate of the denominator (e.g., the conjugate of \( a + b \) is \( a - b \)). This ensures that you can use the identity \( (a-b)(a+b) = a^2 - b^2 \) to eliminate the square roots.