UP Board Solutions Class 9 Maths Chapter 4 Algebraic Identities Ex 4.2

Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.2 बीजगणितीय सर्वसमिकाएँ

Exercise 4.2 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. (x + 2y + 4z) का विस्तार ज्ञात कीजिए ।
Answer: हलः \((x + 2y + 4z)^2 = x^2 + (2y)^2 + (4z)^2 + 2x \cdot 2y + 2 \cdot 2y \cdot 4z + 2 \cdot x \cdot 4z = x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz\)
In simple words: This question asks to expand a trinomial squared using the identity \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\). Substitute the given terms into the formula and simplify.

🎯 Exam Tip: Remember the expansion formula for \((a+b+c)^2\) and pay close attention to signs when multiplying terms, especially with negative numbers.

Question 2. (-2x + 3y + 2z)2 का विस्तार ज्ञात कीजिए।
Answer: हलः \((-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(2z) = 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz\)
In simple words: Expand the given expression by applying the identity \((a+b+c)^2\), where `a`, `b`, and `c` are \(-2x\), \(3y\), and \(2z\) respectively, then perform the multiplications and additions.

🎯 Exam Tip: Be careful with the signs when squaring negative terms and when multiplying terms. A common mistake is to forget the negative sign during multiplication.

Exercise 4.2 Algebraic Identities लघु उत्तरीय प्रश्न (Short Answer Type Questions)

Question 3. निम्न के विस्तार ज्ञात कीजिए।
Answer: हलः
(i) \((-3x + y + z)^2 = (-3x)^2 + (y)^2 +(z)^2 + 2(-3x)(y) + 2(y)(z) + 2(-3x)(z) = 9x^2 + y^2 + z^2 - 6xy + 2yz - 6xz\)
(ii) \((-x + 2y + z)^2 = (-x)^2 + (2y)^2 +(z)^2 + 2(-x)(2y) + 2(2y)(z) + 2(-x)(z) = x^2 + 4y^2 + z^2 - 4xy + 4yz - 2xz\)
(iii) \((3x + 2y - z)^2 = (3x)^2 + (2y)^2 +(-z)^2 + 2(3x)(2y) + 2(2y)(-z) + 2(3x)(-z) = 9x^2 + 4y^2 + z^2 + 12xy - 4yz - 6xz\)
(iv) \((2 + x - 2y)^2 = (2)^2 + (x)^2 + (-2y)^2 + 2 \cdot 2 \cdot x + 2 \cdot x(-2y) + 2 \cdot 2(-2y) = 4 + x^2 + 4y^2 + 4x - 4xy - 8y\)
(v) \((m + 2n - 5p)^2 = (m)^2 + (2n)^2 + (-5p)^2 + 2 \cdot m \cdot 2n + 2 \cdot 2n(-5p) + 2 \cdot m(-5p) = m^2 + 4n^2 + 25p^2 + 4mn - 20np - 10mp\)
(vi) \((ab + bc + ca)^2 = (ab)^2 + (bc)^2 + (ca)^2 + 2(ab)(bc) + 2(bc)(ca) + 2(ca)(ab) = a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2abc^2 + 2a^2bc\)
In simple words: This question requires expanding several trinomials squared using the standard algebraic identity \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\). Each part involves identifying the `a`, `b`, and `c` terms, substituting them into the formula, and simplifying the resulting expression, paying close attention to signs and variable products.

🎯 Exam Tip: For expressions with multiple variables or negative terms, it's crucial to substitute carefully and simplify each term systematically to avoid errors. Double-check the signs in the final answer.

Question 4. सरल कीजिए।
Answer: हलः
(i) \((2x + p - c)^2 - (2x - p + c)^2\)
Use the identity \(A^2 - B^2 = (A + B)(A - B)\) where \(A = (2x + p - c)\) and \(B = (2x - p + c)\)
\(= ((2x + p - c) + (2x - p + c))((2x + p - c) - (2x - p + c))\)
\(= (2x + p - c + 2x - p + c)(2x + p - c - 2x + p - c)\)
\(= (4x)(2p - 2c)\)
\(= (4x)2(p - c)\)
\(= 8x(p - c)\)
(ii) \((x^2 + y^2 - z^2)^2 - (x^2 - y^2 + z^2)^2\)
Use the identity \(A^2 - B^2 = (A + B)(A - B)\) where \(A = (x^2 + y^2 - z^2)\) and \(B = (x^2 - y^2 + z^2)\)
\(= ((x^2 + y^2 - z^2) + (x^2 - y^2 + z^2))((x^2 + y^2 - z^2) - (x^2 - y^2 + z^2))\)
\(= (x^2 + y^2 - z^2 + x^2 - y^2 + z^2)(x^2 + y^2 - z^2 - x^2 + y^2 - z^2)\)
\(= (2x^2)(2y^2 - 2z^2)\)
\(= (2x^2)2(y^2 - z^2)\)
\(= 4x^2(y^2 - z^2)\)
(iii) \((a + b + c)^2 + (a - b + c)^2 + (a + b - c)^2\)
\(= (a^2 + b^2 + c^2 + 2ab + 2bc + 2ac) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ac) + (a^2 + b^2 + c^2 + 2ab - 2bc - 2ac)\)
\(= 3(a^2 + b^2 + c^2) + 2ab - 2bc + 2ac\)
In simple words: This question involves simplifying algebraic expressions. Part (i) and (ii) use the difference of squares identity \((A^2 - B^2) = (A-B)(A+B)\) to simplify. Part (iii) requires expanding three trinomials squared and then combining like terms.

🎯 Exam Tip: For simplification problems, always look for common algebraic identities like difference of squares or perfect squares. Careful handling of negative signs during expansion and combination of terms is crucial.

Exercise 4.2 Algebraic Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

Question 5. यदि \(a^2 + b^2 + c^2 = 20\) व \(a + b + c = 0\), तब \(ab + bc + ca\) का मान ज्ञात कीजिए।
Answer: हलः
दिया है: \(a + b + c = 0\)
वर्ग करने पर
\((a + b + c)^2 = 0^2\)
\(a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0\)
ज्ञात है कि \(a^2 + b^2 + c^2 = 20\)
\(20 + 2(ab + bc + ca) = 0\)
\(2(ab + bc + ca) = 0 - 20\)
\(2(ab + bc + ca) = -20\)
\(ab + bc + ca = \frac{-20}{2}\)
\(ab + bc + ca = -10\)
In simple words: Given the sum of squares and that the sum of the variables is zero, we need to find the sum of pairwise products. We use the identity \((a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\). Since \((a+b+c)=0\), its square is also 0. Substituting the given value of \(a^2+b^2+c^2\), we can solve for \(ab+bc+ca\).

🎯 Exam Tip: This type of problem often tests your knowledge of the \((a+b+c)^2\) identity. Ensure you substitute values correctly and handle the arithmetic carefully, especially with negative numbers.

Question 6. यदि \(a + b + c = 9\) व \(ab + bc + ca = 40\), तब \(a^2 + b^2 + c^2\) का मान ज्ञात कीजिए।
Answer: हलः
यदि \(a + b + c = 9\)
वर्ग करने पर
\((a + b + c)^2 = 9^2\)
\(a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81\)
ज्ञात है कि \(ab + bc + ca = 40\)
\(a^2 + b^2 + c^2 + 2 \times 40 = 81\)
\(a^2 + b^2 + c^2 + 80 = 81\)
\(a^2 + b^2 + c^2 = 81 - 80\)
\(a^2 + b^2 + c^2 = 1\)
In simple words: We are given the sum of three variables and the sum of their pairwise products. To find the sum of their squares, we use the identity \((a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\). By substituting the known values into this identity, we can solve for \(a^2+b^2+c^2\).

🎯 Exam Tip: Directly apply the trinomial square identity. Remember to square the sum correctly and perform the arithmetic operations accurately to find the required value.

Question 7. यदि \(a^2 + b^2 + c^2 = 16\) व \(ab + bc + ca = 10\), तब \(a + b + c\) का मान ज्ञात कीजिए ।
Answer: हलः
ज्ञात है कि \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
दिए गए मानों को प्रतिस्थापित करने पर:
\((a + b + c)^2 = 16 + 2 \times 10\)
\((a + b + c)^2 = 16 + 20\)
\((a + b + c)^2 = 36\)
दोनों पक्षों का वर्गमूल लेने पर:
\(a + b + c = \sqrt{36}\)
\(a + b + c = 6\)
In simple words: This problem asks us to find the sum of three variables given the sum of their squares and the sum of their pairwise products. We utilize the algebraic identity \((a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\). By substituting the provided values, we can calculate \((a+b+c)^2\) and then take the square root to find \(a+b+c\).

🎯 Exam Tip: This question is a direct application of the \((a+b+c)^2\) identity. Be sure to perform the arithmetic correctly, especially the multiplication before addition, and remember to take the square root for the final answer.

Balaji Publications Mathematics Class 9 Solutions