UP Board Solutions Class 9 Maths Chapter 3 Rationalisation Ex 3.1

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.1 परिमेयीकरण

Exercise 3.1 Rationalisation अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. [latex]\sqrt{10} \times \sqrt{15}[/latex] का मान ज्ञात कीजिए। हलः
Answer:
In simple words: This question asks for the value of the product of two square roots, \(\sqrt{10}\) and \(\sqrt{15}\). To solve this, we would multiply the numbers inside the square roots.

🎯 Exam Tip: Remember that \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\) is a fundamental property for simplifying radical expressions.

Question 2. [latex]\sqrt{12} \times \sqrt{18}[/latex] का मान ज्ञात कीजिए। हलः
Answer:
In simple words: The question requires finding the product of \(\sqrt{12}\) and \(\sqrt{18}\). This can be done by multiplying the numbers under the square root and then simplifying the resulting square root.

🎯 Exam Tip: Always simplify the radicals first before multiplying for easier calculations, or multiply directly and then simplify the combined radical.

Question 3. [latex]2+\sqrt{3}[/latex] का परिमेयीकरण गुणनखण्ड ज्ञात कीजिए । हल: [latex]2+\sqrt{3}[/latex] का परिमेयीकरण गुणनखण्ड या संयुग्मी = [latex]2-\sqrt{3}[/latex]
Answer: [latex]2+\sqrt{3}[/latex] का परिमेयीकरण गुणनखण्ड या संयुग्मी = [latex]2-\sqrt{3}[/latex]
In simple words: The rationalizing factor or conjugate of an expression like \(a+\sqrt{b}\) is \(a-\sqrt{b}\). Multiplying an expression by its conjugate eliminates the radical from the denominator, making the expression rational.

🎯 Exam Tip: The rationalizing factor (conjugate) for a binomial of the form \(a+\sqrt{b}\) is \(a-\sqrt{b}\) and vice-versa, utilizing the difference of squares identity \((a+b)(a-b) = a^2-b^2\).

Question 4. [latex]\sqrt{5}[/latex] का परिमेयीकरण गुणनखण्ड ज्ञात कीजिए । हलः [latex]\sqrt{5}[/latex] का परिमेयीकरण गुणनखण्ड = [latex]\frac{1}{\sqrt{5}}[/latex]
Answer: [latex]\sqrt{5}[/latex] का परिमेयीकरण गुणनखण्ड = [latex]\frac{1}{\sqrt{5}}[/latex]
In simple words: To rationalize a single square root term like \(\sqrt{5}\), you multiply it by itself. The rationalizing factor is \(\sqrt{5}\) itself (to make the denominator rational) or, if looking for a factor to eliminate it from the numerator, \(\frac{1}{\sqrt{5}}\).

🎯 Exam Tip: When asked for a rationalizing factor, the simplest answer for \(\sqrt{a}\) is \(\sqrt{a}\), because \(\sqrt{a} \times \sqrt{a} = a\). However, sometimes the context implies a factor for a denominator, which is \(\sqrt{a}\) itself to remove the radical.

Question 5. [latex]\sqrt[5]{6} \times \sqrt[5]{6}[/latex] का मान ज्ञात कीजिए। हलः
Answer:
In simple words: This question asks to find the value of the product of two fifth roots of 6. When multiplying two identical radicals, the result is the base raised to the power of (1/n + 1/n).

🎯 Exam Tip: For any \(n^{th}\) root, \(\sqrt[n]{a} \times \sqrt[n]{a} = (\sqrt[n]{a})^2\). If it's `\(\sqrt[n]{a} \times \sqrt[n]{a}\)` as `\((a^{1/n}) \times (a^{1/n})\)`, the result is `\(a^{2/n}\)`. If \(n=2\), it is simply `\(a\)`. For general \(n\), it is `\(a^{2/n}\)`. Pay attention to the index of the root.

Exercise 3.1 Rationalisation लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

निम्न के मान ज्ञात कीजिए- (प्रश्न 6 – 13)

Question 6. [latex](3+2 \sqrt{2})(3-2 \sqrt{2})[/latex] हलः
Answer:
In simple words: This question asks to simplify the product of two binomials that are conjugates of each other. Using the identity \((a+b)(a-b) = a^2-b^2\) simplifies the expression, removing the radical.

🎯 Exam Tip: Recognize the form \((a+b)(a-b)\). Applying the difference of squares formula \(a^2-b^2\) is crucial for quickly solving such problems and rationalizing denominators.

Question 7. [latex](\sqrt{5}-2)(\sqrt{3}-\sqrt{5})[/latex] हलः
Answer:
In simple words: This problem involves multiplying two binomials containing radicals. You need to use the distributive property (FOIL method) to multiply each term in the first binomial by each term in the second.

🎯 Exam Tip: When multiplying binomials with radicals, apply the distributive property carefully. Remember to simplify any resulting radicals (e.g., \(\sqrt{a} \times \sqrt{a} = a\)).

Question 8. [latex](5+\sqrt{7})(5-\sqrt{7})[/latex] हलः
Answer:
In simple words: This expression is in the form of \((a+b)(a-b)\), which simplifies to \(a^2-b^2\). Applying this formula will quickly solve the problem.

🎯 Exam Tip: The difference of squares identity \((a+b)(a-b) = a^2-b^2\) is very common in rationalization problems. Master its application to save time and avoid errors.

Question 9. [latex](\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})[/latex] हलः
Answer:
In simple words: This question involves multiplying two conjugate binomials, similar to the form \((a-b)(a+b)\). Use the difference of squares identity, where \(a = \sqrt{8}\) and \(b = \sqrt{2}\).

🎯 Exam Tip: Before applying the difference of squares, consider simplifying the radicals if possible (e.g., \(\sqrt{8} = 2\sqrt{2}\)) to potentially simplify calculations further. However, direct application of the formula also works effectively here.

Question 10. [latex](3+\sqrt{3})(2+\sqrt{2})[/latex] हलः
Answer:
In simple words: This problem asks to multiply two binomials each containing a radical. Use the FOIL method (First, Outer, Inner, Last) to multiply each term in the first binomial by each term in the second.

🎯 Exam Tip: Be meticulous with the distributive property. Ensure all four products are correctly calculated and then combine any like terms if they exist (though not in this specific case, as the radicals are different).

Question 11. [latex](3+\sqrt{3})(3-\sqrt{3})[/latex] हलः
Answer:
In simple words: This expression is a classic example of the difference of squares formula, \((a+b)(a-b) = a^2-b^2\). Here, \(a=3\) and \(b=\sqrt{3}\).

🎯 Exam Tip: Identifying expressions in the form of \((a+b)(a-b)\) allows for rapid simplification to \(a^2-b^2\), which is often used in rationalizing denominators.

Question 12. [latex](\sqrt{5}+\sqrt{2})^{2}[/latex] हलः
Answer:
In simple words: This question asks to expand the square of a binomial, which follows the identity \((a+b)^2 = a^2+2ab+b^2\). Here, \(a=\sqrt{5}\) and \(b=\sqrt{2}\).

🎯 Exam Tip: Remember the square of a binomial formula, \((a+b)^2 = a^2+2ab+b^2\), for expanding expressions like this. Don't forget the \(2ab\) term!

Question 13. [latex](\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})[/latex] हलः
Answer:
In simple words: This problem presents a product of two conjugate binomials. Apply the difference of squares identity, \((a-b)(a+b) = a^2-b^2\), for direct simplification.

🎯 Exam Tip: Recognizing the difference of squares pattern \((a-b)(a+b)\) is a shortcut to avoid lengthy multiplication and directly get \(a^2-b^2\).

Exercise 3.1 Rationalisation लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Question 14. निम्न व्यंजकों को सरल कीजिए।
Answer:हलः
(i) \(\sqrt{16} \times \sqrt{2} = \sqrt{2 \times 2 \times 2 \times 2 \times 2} = \sqrt{2^5} = (2^5)^{\frac{1}{2}} = 2\)
(ii) \(\sqrt[4]{4} \times \sqrt[4]{16} = \sqrt[4]{2 \times 2 \times 2 \times 2 \times 2} = \sqrt[4]{2^5}\)
\(\implies \sqrt[4]{(2 \times 2)^3} = (4^3)^{\frac{1}{4}} = 4\)
(iii) \(\frac{\sqrt[3]{243}}{\sqrt[3]{3}} = \sqrt[3]{\frac{243}{3}} = \sqrt[3]{81} = \sqrt[3]{3 \times 3 \times 3 \times 3} = \sqrt[3]{3^4} = (3^4)^{\frac{1}{4}} = 3\)
(iv) \(\frac{\sqrt[3]{343}}{\sqrt[3]{7}} = \sqrt[3]{\frac{343}{7}} = \sqrt[3]{49} = (49)^{\frac{1}{3}}\)
In simple words: This question requires simplifying various radical expressions using properties of roots and exponents. Each sub-part involves either multiplying, dividing, or simplifying powers of roots.

🎯 Exam Tip: For problems involving simplification of radicals, remember the properties \(\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}\) and \(\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}\). Also, convert radicals to exponential form (\(\sqrt[n]{a} = a^{1/n}\)) for easier calculations, especially when dealing with different indices.

Question 15. सरल कीजिए। हलः
Answer:
In simple words: This question asks for simplification. Without specific expressions provided, simplification generally means performing indicated operations and reducing the expression to its simplest form.

🎯 Exam Tip: Always look for common factors, combine like terms, and rationalize denominators or numerators as needed to simplify expressions.

Question 16. सरल कीजिए। हलः
Answer:
In simple words: This task requires simplifying a given mathematical expression. This typically involves applying rules of arithmetic, algebra, or radicals to reach the most concise form.

🎯 Exam Tip: Ensure all steps are logical and follow mathematical rules. Double-check for any potential errors in calculation or application of properties.

Question 17. सरल कीजिए। हलः
Answer:
(i) \(( \sqrt{3} + \sqrt{7} ) ( \sqrt{3} + \sqrt{7} ) = ( \sqrt{3} + \sqrt{7} )^2\)
\(= 3 + 7 + 2\sqrt{3 \times 7} = 10 + 2\sqrt{21}\)
(ii) \(( 2\sqrt{5} + 3\sqrt{2} )^2 = ( 2\sqrt{5} + 3\sqrt{2} ) ( 2\sqrt{5} + 3\sqrt{2} )\)
\(= 20 + 6\sqrt{10} + 6\sqrt{10} + 9 \times 2\)
\(= 20 + 12\sqrt{10} + 18 = 38 + 12\sqrt{10}\)
In simple words: This question involves simplifying expressions that are squares of binomials containing radicals. For part (i), \((a+b)^2\) form is used directly. For part (ii), the same square of a binomial rule applies, requiring careful distribution and combination of terms.

🎯 Exam Tip: When squaring binomials with radicals, remember the formula \((a+b)^2 = a^2+2ab+b^2\). Pay close attention to simplifying \(( \sqrt{x} )^2 = x\) and combining like radical terms (e.g., \(6\sqrt{10} + 6\sqrt{10} = 12\sqrt{10}\)).