UP Board Solutions Class 9 Maths Chapter 4 Algebraic Identities Ex 4.1

Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.1 बीजगणितीय सर्वसमिकाएँ

Ex 4.1 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. \(\left(\sqrt{5} a+\sqrt{6} b\right)\) का वर्ग ज्ञात कीजिए।
Answer:
वर्ग करने पर:
\(\left(\sqrt{5} a+\sqrt{6} b\right)^2 = (\sqrt{5}a)^2 + 2(\sqrt{5}a)(\sqrt{6}b) + (\sqrt{6}b)^2\)
\( = 5a^2 + 2\sqrt{30}ab + 6b^2\)
In simple words: This question asks to find the square of the given binomial expression using the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\).

🎯 Exam Tip: Remember to apply the \((a+b)^2 = a^2 + 2ab + b^2\) identity carefully when squaring binomials, especially with square root terms. Ensure that coefficients and variables are correctly multiplied.

Question 2. (x + 4)(x + 10) का मान ज्ञात कीजिए।
Answer: \((x + 4)(x + 10) = (x)^2 + (4 + 10)x + 4 \times 10 = x^2 + 14x + 40\)
In simple words: To find the value of this product, use the identity \((x+a)(x+b) = x^2 + (a+b)x + ab\).

🎯 Exam Tip: Ensure correct signs and multiplication when expanding binomials. Practice mental math for simple additions and multiplications.

Question 3. (x + 8)(x - 10) का मान ज्ञात कीजिए।
Answer: \((x + 8)(x - 10) = x^2 + (8-10)x - 8 \times 10 = x^2 - 2x - 80\)
In simple words: This problem uses the identity \((x+a)(x+b) = x^2 + (a+b)x + ab\), where \(b\) is negative, to expand the given expression.

🎯 Exam Tip: Pay close attention to the signs of the constants when applying the \((x+a)(x+b)\) identity, especially when one constant is negative.

Question 4. (3x + 4) (3x - 5) का मान ज्ञात कीजिए।
Answer: \((3x + 4) (3x - 5) = (3x)^2 + (4 - 5)3x - 4 \times 5 = 9x^2 - 3x - 20\)
In simple words: This problem utilizes the identity \((x+a)(x+b)\) by considering \(x\) as \(3x\), to expand the product.

🎯 Exam Tip: When the variable term itself has a coefficient (like \(3x\)), remember to square the entire term \((3x)^2\) and apply the distributive property correctly for the middle term.

Ex 4.1 Algebraic Identities लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

Question 5. निम्न के मान ज्ञात कीजिए।
(i) \(\left(4x - \frac{1}{2x}\right)^2\)
(ii) \(\left(5x^2 - 7y^2\right)^2\)
(iii) \((x - 0.1)(x + 0.1)\)
Answer:
(i) \(\left(4x - \frac{1}{2x}\right)^2 = (4x)^2 - 2(4x)\left(\frac{1}{2x}\right) + \left(\frac{1}{2x}\right)^2 = 16x^2 - 4 + \frac{1}{4x^2}\)
(ii) \(\left(5x^2 - 7y^2\right)^2 = (5x^2)^2 - 2(5x^2)(7y^2) + (7y^2)^2 = 25x^4 - 70x^2y^2 + 49y^4\)
(iii) \((x-0.1)(x+0.1) = x^2 - (0.1)^2 = x^2 - 0.01\)
In simple words: This question requires applying binomial expansion identities like \((a-b)^2 = a^2 - 2ab + b^2\) and \((a-b)(a+b) = a^2 - b^2\) to simplify the given expressions.

🎯 Exam Tip: Be careful with the signs and coefficients when expanding, especially when terms involve fractions or decimals. Remember that \((0.1)^2 = 0.01\).

Question 6. सर्वसमिकाओं का प्रयोग कर गुणनफल ज्ञात कीजिए।
Answer:
(i) \(111 \times 102 = (100 + 11) \times (100 + 2) = (100)^2 + (11 + 2) \times 100 + 11 \times 2 = 10000 + 1300 + 22 = 11322\)
(ii) \(105 \times 97 = (100 + 5) \times (100 - 3) = (100)^2 + (5 - 3) \times 100 - 5 \times 3 = 10000 + 200 - 15 = 10185\)
(iii) \(85 \times 96 = (100 - 15) \times (100 - 4) = (100)^2 + (-15 - 4) \times 100 + 15 \times 4 = 10000 - 1900 + 60 = 8160\)
(iv) \(10008 \times 995 = (1000 + 8) \times (1000 - 5) = (1000)^2 + (8 - 5) \times 1000 - 8 \times 5 = 1000000 + 3000 - 40 = 1002960\)
(v) \(991 \times 982 = (1000 - 9) \times (1000 - 18) = (1000)^2 - (9 + 18) \times 1000 + 9 \times 18 = 1000000 - 27000 + 162 = 973162\)
(vi) \((0.98)^2 = (1 - 0.02)^2 = (1)^2 - 2 \times 1 \times 0.02 + (0.02)^2 = 1 - 0.04 + 0.0004 = 0.9604\)
In simple words: This question demonstrates how to use various algebraic identities like \((x+a)(x+b)\) and \((a-b)^2\) to simplify multiplication of numbers by expressing them in terms of a base number (like 100 or 1000).

🎯 Exam Tip: Choose the base number (e.g., 100, 1000) strategically to make calculations easier. Be careful with signs, especially when both numbers are less than the base (e.g., \(100-a\) and \(100-b\)).

Question 7. सर्वसमिकाओं का प्रयोग कर सरल कीजिए।
(i) \(\frac{7.83 \times 7.83 - 1.17 \times 1.17}{6.66}\)
(ii) \(185 \times 185 - 115 \times 115\)
(iii) \(322 \times 322 - 2 \times 322 \times 22 + 22 \times 22\)
Answer:
(i) माना \(7.83 = a\) तथा \(1.17 = b\)
\(\frac{a \times a - b \times b}{a - b} = \frac{a^2 - b^2}{a - b}\)

\( \implies \frac{(a+b)(a-b)}{(a-b)} = a+b\)
\( = 7.83 + 1.17 = 9\)
(ii) माना \(185 = a\) तथा \(115 = b\)
\(a \times a - b \times b = a^2 - b^2 = (a + b)(a - b) = (185 + 115)(185 - 115) = (300)(70) = 21000\)
(iii) माना \(322 = a\) तथा \(22 = b\)
\(a \times a - 2 \times a \times b + b \times b = a^2 - 2ab + b^2 = (a - b)^2 = (322 - 22)^2 = (300)^2 = 90000\)
In simple words: This question demonstrates the application of algebraic identities such as \((a^2 - b^2) = (a+b)(a-b)\) and \((a-b)^2 = a^2 - 2ab + b^2\) to simplify complex numerical expressions.

🎯 Exam Tip: Identify the correct algebraic identity to apply based on the structure of the given expression. Substitute values carefully and perform arithmetic operations accurately.

Question 8. यदि \(x-\frac{1}{x}=6\), तब \(x^{4}+\frac{1}{x^{4}}\) का मान ज्ञात कीजिए।
Answer:
वर्ग करने पर,
\(\left(x - \frac{1}{x}\right)^2 = (6)^2\)
\(x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} = 36\)

\( \implies x^2 + \frac{1}{x^2} = 36 + 2 = 38\) ...(1)
पुनः वर्ग करने पर
\(\left(x^2 + \frac{1}{x^2}\right)^2 = (38)^2\)
\(x^4 + \frac{1}{x^4} + 2 \times x^2 \times \frac{1}{x^2} = 1444\)

\( \implies x^4 + \frac{1}{x^4} = 1444 - 2 = 1442\)
In simple words: To find \(x^4 + \frac{1}{x^4}\) from \(x-\frac{1}{x}\), we square the given expression twice, applying the identity \((a-b)^2\) and then \((a+b)^2\).

🎯 Exam Tip: This problem involves sequential squaring. Remember that \((x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2\). Be careful with the intermediate calculation results.

Ex 4.1 Algebraic Identities लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Question 9. यदि \(x+\frac{1}{x}=\sqrt{8}\), तब निम्न के मान ज्ञात कीजिए।
Answer:
(i) वर्ग करने पर
\(\left(x+\frac{1}{x}\right)^2 = (\sqrt{8})^2\)
\(x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} = 8\)

\( \implies x^2 + \frac{1}{x^2} = 8 - 2 = 6\) ...(2)
(ii) समी० (2) का वर्ग करने पर
\(\left(x^2 + \frac{1}{x^2}\right)^2 = (6)^2\)
\(x^4 + \frac{1}{x^4} + 2 \times x^2 \times \frac{1}{x^2} = 36\)

\( \implies x^4 + \frac{1}{x^4} = 36 - 2 = 34\)
In simple words: Given \(x+\frac{1}{x}\), we find \(x^2+\frac{1}{x^2}\) by squaring the expression once, and then \(x^4+\frac{1}{x^4}\) by squaring the result again, using the identity \((a+b)^2\).

🎯 Exam Tip: For problems involving powers of \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\), remember that squaring eliminates the reciprocal term in the middle (\(2 \cdot x \cdot \frac{1}{x} = 2\)).

Question 10. यदि \(x-\frac{1}{x}=3\), तब निम्न के मान ज्ञात कीजिए ।
Answer:
(i) वर्ग करने पर
\(\left(x - \frac{1}{x}\right)^2 = (3)^2\)
\(x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} = 9\)

\( \implies x^2 + \frac{1}{x^2} = 9 + 2 = 11\) ...(1)
(ii) दोनों पक्षों का वर्ग करने पर
\(\left(x^2 + \frac{1}{x^2}\right)^2 = (11)^2\)
\(x^4 + \frac{1}{x^4} + 2 \times x^2 \times \frac{1}{x^2} = 121\)

\( \implies x^4 + \frac{1}{x^4} = 121 - 2 = 119\)
In simple words: Similar to previous problems, we find higher powers of \(x\) and \(\frac{1}{x}\) by repeatedly squaring the given expression \(x-\frac{1}{x}=3\), first to find \(x^2+\frac{1}{x^2}\) and then \(x^4+\frac{1}{x^4}\).

🎯 Exam Tip: Understand the difference in formulas when squaring \((x-\frac{1}{x})\) versus \((x+\frac{1}{x})\). The sign of the middle term (\(-2\) or \(+2\)) changes accordingly.

Question 11. यदि \(x^2 + \frac{1}{x^2} = 27\), तब निम्न के मान ज्ञात कीजिए।
Answer:
(i) \(\left(x+\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2x \times \frac{1}{x}\)

\( \implies \left(x+\frac{1}{x}\right)^2 = 27 + 2 = 29\)

\( \implies x+\frac{1}{x} = \sqrt{29}\)
(ii) \(\left(x-\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2x \times \frac{1}{x}\)

\( \implies \left(x-\frac{1}{x}\right)^2 = 27 - 2 = 25\)

\( \implies x-\frac{1}{x} = \sqrt{25} = 5\)
In simple words: Given \(x^2+\frac{1}{x^2}\), we can find \(x+\frac{1}{x}\) and \(x-\frac{1}{x}\) by recognizing that \((a+b)^2 = a^2+b^2+2ab\) and \((a-b)^2 = a^2+b^2-2ab\).

🎯 Exam Tip: When given a sum of squares and asked for the sum or difference of variables, remember to use the \((a \pm b)^2\) identities. The value of \(2ab\) becomes \(2\) for \(x\) and \(\frac{1}{x}\) terms.

Question 12. यदि x = 5 और y = 7 तब \(49x^2 - 70xy + 25y^2\) का मान ज्ञात कीजिए।
Answer:
\(x = 5\) तथा \(y = 7\)
\(49x^2 - 70xy + 25y^2 = (7x)^2 - 2 \times 7x \times 5y + (5y)^2\)
\( = (7x - 5y)^2 = (7 \times 5 - 5 \times 7)^2\)
\( = (35 - 35)^2 = 0\)
In simple words: This problem involves recognizing the given expression as the expansion of \((a-b)^2\), substituting the values of \(x\) and \(y\), and then simplifying.

🎯 Exam Tip: Always try to factorize or identify an algebraic identity before substituting numerical values. This often simplifies calculations significantly.

Ex 4.1 Algebraic Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

Question 13. यदि x = 15 व y = 27 तब \(81x^2 - 90xy + 25y^2\) का मान ज्ञात कीजिए।
Answer:
\(81x^2 - 90xy + 25y^2 = (9x)^2 - 2 \times 9x \times 5y + (5y)^2\)
\( = (9x - 5y)^2 = (9 \times 15 - 5 \times 27)^2\)
\( = (135 - 135)^2 = 0\)
In simple words: This problem is solved by recognizing the expression as \((a-b)^2\), where \(a=9x\) and \(b=5y\), then substituting the given values of \(x\) and \(y\).

🎯 Exam Tip: Look for perfect square trinomials in expressions like \(Ax^2 + Bxy + Cy^2\). Often they simplify to \((ax \pm by)^2\), making substitution and calculation straightforward.

Question 14. यदि x + y = 12 व xy = 32, तब \(x^2 + y^2\) का मान ज्ञात कीजिए।
Answer:
\(x + y = 12\) ...(1)
\(xy = 32\) ...(2)
समी० (1) का वर्ग करने पर
\((x + y)^2 = (12)^2\)
\(x^2 + y^2 + 2xy = 144\)
\(x^2 + y^2 + 2 \times 32 = 144\)
\(x^2 + y^2 = 144 - 64 = 80\)
In simple words: Given the sum and product of two variables, we can find the sum of their squares by squaring the sum and using the identity \((x+y)^2 = x^2+y^2+2xy\).

🎯 Exam Tip: This is a standard problem type where \((x+y)^2 = x^2+y^2+2xy\) is crucial. Be careful with arithmetic when substituting the values.

Question 15. निम्न को सरल कीजिए।
Answer:
(i) \(\left(\frac{x}{2} - \frac{2}{5}\right)^2 - x^2 + 2x\)
\( = \left(\frac{x}{2}\right)^2 - 2\left(\frac{x}{2}\right)\left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^2 - x^2 + 2x\)
\( = \frac{x^2}{4} - \frac{2x}{5} + \frac{4}{25} - x^2 + 2x\)
\( = \left(\frac{x^2}{4} - x^2\right) + \left(2x - \frac{2x}{5}\right) + \frac{4}{25}\)
\( = \left(\frac{x^2 - 4x^2}{4}\right) + \left(\frac{10x - 2x}{5}\right) + \frac{4}{25}\)
\( = -\frac{3x^2}{4} + \frac{8x}{5} + \frac{4}{25}\)
(ii) \(\left(m+\frac{n}{7}\right)\left(m-\frac{n}{7}\right) = m^2 - \left(\frac{n}{7}\right)^2 = m^2 - \frac{n^2}{49}\)
In simple words: This problem involves simplifying algebraic expressions by applying identities like \((a-b)^2 = a^2 - 2ab + b^2\) and \((a+b)(a-b) = a^2 - b^2\), then combining like terms.

🎯 Exam Tip: For expressions with multiple terms and operations, expand using identities first, then meticulously combine like terms by finding common denominators where necessary. Pay attention to signs.