GSEB Class 8 Maths Solutions Chapter 9 બૈજિક પદાવલિઓ અને નિત્યસમ Exercise 9.5

Question 1. યોગ્ય નિત્યસમનો ઉપયોગ કરીને નીચેના ગુણાકાર મેળવો:
(i) (x + 3) (x + 3)
Answer:
\( = (x + 3)^2 \)
\( = (x)^2 + 2 (x)(3) + (3)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = x^2 + 6x + 9 \)
In simple words: This problem uses the identity for squaring a sum. We expanded the expression by applying the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), which gave us the final simplified form.

Exam Tip: Remember to identify the correct algebraic identity to apply based on the given expression. Pay attention to the signs and terms to prevent calculation errors.

Question 1.
(ii) (2y + 5) (2y + 5)
Answer:
\( = (2y + 5)^2 \)
\( = (2y)^2 + 2(2y)(5) + (5)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = 4y^2 + 20y + 25 \)
In simple words: We used the sum-square identity for this. By squaring the binomial and applying the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we obtained the final expanded form of the expression.

Exam Tip: When squaring a term like \( (2y)^2 \), make sure to square both the coefficient (2) and the variable (y).

Question 1.
(iii) (2a - 7) (2a – 7)
Answer:
\( = (2a - 7)^2 \)
\( = (2a)^2 – 2(2a)(7) + (7)^2 \)
\( [:: (a - b)^2 = a^2 - 2ab + b^2] \)
\( = 4a^2 - 28a + 49 \)
In simple words: This uses the identity for squaring a difference. We expanded the expression by using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), leading to the simplified form.

Exam Tip: For expressions involving subtraction, the middle term in the expansion will have a negative sign.

Question 1.
(iv) \( (3a - \frac {1}{2}) (3a - \frac {1}{2}) \)
Answer:
\( = (3a - \frac {1}{2})^2 \)
\( = (3a)^2 - 2(3a)(\frac {1}{2}) + (\frac {1}{2})^2 \)
\( [:: (a - b)^2 = a^2 - 2ab + b^2] \)
\( = 9a^2 - 3a + \frac {1}{4} \)
In simple words: We applied the identity for squaring a difference to this expression. By using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), we got the final simplified answer.

Exam Tip: Be careful with fractions when squaring and multiplying terms to avoid mistakes.

Question 1.
(v) (1.1m - 0.4) (1.1m + 0.4)
Answer:
\( = (1.1m)^2 – (0.4)^2 \)
\( [:: (a + b) (a – b) = a^2 – b^2] \)
\( = 1.21m^2 - 0.16 \)
In simple words: This problem uses the difference of squares identity. We multiplied the terms by applying the formula \( (a + b)(a - b) = a^2 - b^2 \), giving us the expanded form.

Exam Tip: This identity is useful for quickly multiplying binomials where one is a sum and the other is a difference of the same two terms.

Question 1.
(vi) (a² + b²) (-a² + b²)
Answer:
\( = (b^2 + a^2)(b^2 – a^2) \)
\( = (b^2)^2 – (a^2)^2 \)
\( [:: (a + b) (a – b) = a^2 – b^2] \)
\( = b^4 - a^4 \)
In simple words: We rearranged the terms and then used the difference of squares identity. By applying \( (a + b)(a - b) = a^2 - b^2 \), we got the final simplified result.

Exam Tip: Always rearrange terms if needed to match a known identity. Here, \( (a^2 + b^2)(-a^2 + b^2) \) is the same as \( (b^2 + a^2)(b^2 - a^2) \).

Question 1.
(vii) (6x - 7) (6x + 7)
Answer:
\( = (6x)^2 – (7)^2 \)
\( [:: (a + b) (a – b) = a^2 – b^2] \)
\( = 36x^2 - 49 \)
In simple words: We used the difference of squares identity here. Multiplying the given terms with the formula \( (a + b)(a - b) = a^2 - b^2 \) gave us the expanded expression.

Exam Tip: Recognize the pattern \( (X - Y)(X + Y) \) immediately as \( X^2 - Y^2 \) to save time in calculations.

Question 1.
(viii) (- a + c) (- a + c)
Answer:
\( = (-a + c)^2 \)
\( = (-a)^2 + 2(-a)(c) + (c)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = a^2 - 2ac + c^2 \)
બીજી રીતઃ
\( = (c - a)^2 \)
\( = (c)^2 – 2(c)(a) + (a)^2 \)
\( = c^2 - 2ca + a^2 \)
In simple words: We used the identity for squaring a sum. By expanding the expression with the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we got the final simplified form. The alternative method shows that rearranging terms to \( (c-a)^2 \) gives the same result.

Exam Tip: Note that \( (-a + c)^2 \) is equal to \( (c - a)^2 \). Both forms will lead to the same expanded result. This shows flexibility in how you define 'a' and 'b' in the identity.

Question 1.
(ix) \( (\frac{x}{2}+\frac{3 y}{4}) (\frac{x}{2}+\frac{3 y}{4}) \)
Answer:
\( = (\frac{x}{2}+\frac{3 y}{4})^2 \)
\( = (\frac{x}{2})^2 + 2(\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = \frac{x^{2}}{4}+\frac{3 x y}{4}+\frac{9 y^{2}}{16} \)
In simple words: This problem uses the identity for squaring a sum. We expanded the expression by applying the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), which gave us the final simplified form with fractional terms.

Exam Tip: Be extra careful when squaring fractions and multiplying them to avoid calculation errors. Remember to simplify the middle term.

Question 1.
(x) (7a - 9b) (7a – 9b)
Answer:
\( = (7a - 9b)^2 \)
\( = (7a)^2 - 2(7a)(9b) + (9b)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 49a^2 - 126ab + 81b^2 \)
In simple words: We applied the identity for squaring a difference. By using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), we expanded the given expression to get the final simplified result.

Exam Tip: Ensure that the square of both the coefficient and the variable is taken correctly for each term (e.g., \( (7a)^2 = 49a^2 \)).

Question 2. (x + a) (x + b) = x² + (a + b)x + ab નિત્યસમનો ઉપયોગ કરીને નીચેના ગુણાકાર શોધો:
(i) (x + 3) (x + 7)
Answer:
(x + a) (x + b) = x² + (a + b) x + ab નિત્યસમનો ઉપયોગ કરીને.
\( = (x)^2 + (3 + 7) x + (3)(7) \)
\( = x^2 + (10)x + 21 \)
\( = x^2 + 10x + 21 \)
In simple words: We used the given identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) to multiply these two binomials. We substituted the values of 'a' and 'b' and then simplified the expression.

Exam Tip: Clearly identify 'a' and 'b' in the given expression before applying the identity to prevent errors in addition and multiplication.

Question 2.
(ii) (4x + 5) (4x + 1)
Answer:
\( = (4x)^2 + (5 + 1)4x + (5)(1) \)
\( = 16x^2 + (6)4x + 5 \)
\( = 16x^2 + 24x + 5 \)
In simple words: We applied the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x \) is \( 4x \), and 'a' and 'b' are 5 and 1 respectively. We then calculated and simplified the terms.

Exam Tip: When the 'x' term in the identity is a complex expression like \( 4x \), be sure to substitute it correctly in all parts of the formula, especially when squaring.

Question 2.
(iii) (4x - 5) (4x – 1)
Answer:
\( = (4x)^2 + (- 5 - 1) 4x + (-5) (-1) \)
\( = 16x^2 + (-6)4x + 5 \)
\( = 16x^2 - 24x + 5 \)
In simple words: We used the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x \) is \( 4x \), and 'a' and 'b' are -5 and -1. We carried out the calculations carefully, minding the negative signs.

Exam Tip: Pay close attention to negative signs when substituting 'a' and 'b' values into the identity. A common mistake is to miss a negative sign, leading to an incorrect result.

Question 2.
(iv) (4x + 5) (4x - 1)
Answer:
\( = (4x)^2 + (5 – 1) 4x + (5)(- 1) \)
\( = 16x^2 + (4)4x – 5 \)
\( = 16x^2 + 16x – 5 \)
In simple words: We applied the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). In this case, \( x \) is \( 4x \), 'a' is 5, and 'b' is -1. We then calculated the terms and simplified the expression.

Exam Tip: Remember to apply the correct sign to the product 'ab' and the sum '(a+b)x' when 'a' or 'b' is negative.

Question 2.
(v) (2x + 5y) (2x + 3y)
Answer:
\( = (2x)^2 + (5y + 3y)2x + (5y)(3y) \)
\( = 4x^2 + (8y)2x + 15y^2 \)
\( = 4x^2 + 16xy + 15y^2 \)
In simple words: This problem uses the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x \) is \( 2x \), 'a' is \( 5y \), and 'b' is \( 3y \). We substituted these values and simplified the expression.

Exam Tip: When 'a' and 'b' themselves contain variables (like \( 5y \) and \( 3y \)), remember to multiply the variables as well, for example, \( (5y)(3y) = 15y^2 \).

Question 2.
(vi) (2a² + 9) (2a² + 5)
Answer:
\( = (2a^2)^2 + (9 + 5) 2a^2 + (9)(5) \)
\( = 4a^4 + (14)2a^2 + 45 \)
\( = 4a^4 + 28a^2 + 45 \)
In simple words: We used the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x \) is \( 2a^2 \), 'a' is 9, and 'b' is 5. We then calculated and simplified all the terms.

Exam Tip: When the 'x' term in the identity involves a power (like \( 2a^2 \)), ensure you correctly square it to get \( (2a^2)^2 = 4a^4 \).

Question 2.
(vii) (xyz – 4) (xyz – 2)
Answer:
\( = (xyz)^2 + (- 4 – 2) xyz + (-4)(-2) \)
\( = x^2y^2z^2 + (-6)xyz + 8 \)
\( = x^2y^2z^2 - 6xyz + 8 \)
In simple words: We applied the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). In this problem, \( x \) is \( xyz \), 'a' is -4, and 'b' is -2. We then performed the necessary calculations, remembering the rules for signs.

Exam Tip: Even with multiple variables like \( xyz \), treat them as a single term 'x' for the identity. Be careful with the multiplication of negative constants.

Question 3. નિત્યસમનો ઉપયોગ કરીને નીચેના વર્ગ શોધોઃ
(i) (b - 7)²
Answer:
\( = (b)^2 – 2(b)(7) + (7)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = b^2 - 14b + 49 \)
In simple words: We used the identity for squaring a difference. By applying the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), we found the square of the given expression.

Exam Tip: When applying \( (a-b)^2 \), always ensure the middle term \( 2ab \) has a negative sign and the last term \( b^2 \) is positive.

Question 3.
(ii) (xy + 3z)²
Answer:
\( = (xy)^2 + 2(xy)(3z) + (3z)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = x^2y^2 + 6xyz + 9z^2 \)
In simple words: This problem uses the identity for squaring a sum. We found the square of the expression by applying the formula \( (a + b)^2 = a^2 + 2ab + b^2 \) and then simplified.

Exam Tip: Remember to square both the coefficient and the variable(s) for each term when expanding, e.g., \( (3z)^2 = 9z^2 \).

Question 3.
(iii) (6x² – 5y)²
Answer:
\( = (6x^2)^2 - 2(6x^2)(5y) + (5y)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 36x^4 – 60x^2y + 25y^2 \)
In simple words: We used the identity for squaring a difference. We expanded the expression by applying the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), leading to the final simplified form.

Exam Tip: When a term involves both a coefficient and a variable with a power, such as \( (6x^2) \), ensure that both are correctly squared, so \( (6x^2)^2 = 36x^4 \).

Question 3.
(iv) \( (\frac {2}{3}m + \frac {3}{2}n)^2 \)
Answer:
\( = (\frac {2}{3}m)^2 + 2(\frac {2}{3}m)(\frac {3}{2}n) + (\frac {3}{2}n)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = \frac {4}{9}m^2 + 2mn + \frac {9}{4}n^2 \)
In simple words: We applied the identity for squaring a sum. By expanding the expression with the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), we got the final simplified form with fractional terms.

Exam Tip: When dealing with fractional coefficients, remember to square both the numerator and the denominator. Simplify the middle term carefully by cancelling common factors.

Question 3.
(v) (0.4p – 0.5q)²
Answer:
\( = (0.4p)^2 – 2(0.4p)(0.5q) + (0.5q)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 0.16p^2 – 0.4pq + 0.25q^2 \)
In simple words: We used the identity for squaring a difference. By applying the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), we found the square of the given expression involving decimals.

Exam Tip: Be careful with decimal multiplication and squaring. For example, \( (0.4)^2 = 0.16 \), not 1.6 or 0.016.

Question 3.
(vi) (2xy + 5y)²
Answer:
\( = (2xy)^2 + 2(2xy)(5y) + (5y)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = 4x^2y^2 + 20xy^2 + 25y^2 \)
In simple words: This problem uses the identity for squaring a sum. We expanded the expression by applying the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), which gave us the final simplified form with multiple variables.

Exam Tip: Ensure that all variables in a term are squared when applying the identity, e.g., \( (2xy)^2 = 4x^2y^2 \). Also, combine like variables correctly in the middle term, such as \( y \times y = y^2 \).

Question 4. સાદું રૂપ આપોઃ
(i) (a²-b²)²
Answer:
\( = (a^2)^2 – 2(a^2)(b^2) + (b^2)^2 \)
\( = a^4 - 2a^2b^2 + b^4 \)
In simple words: We used the identity for squaring a difference. By applying the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), where 'a' is \( a^2 \) and 'b' is \( b^2 \), we expanded and simplified the expression.

Exam Tip: When terms already have exponents, remember the rule \( (x^m)^n = x^{m \times n} \). So, \( (a^2)^2 = a^4 \).

Question 4.
(ii) \( (2x + 5)^2 – (2x – 5)^2 \)
Answer:
\( = [(2x)^2 + 2(2x)(5) + (5)^2] – [(2x)^2 – 2(2x)(5) + (5)^2] \)
\( = [4x^2 + 20x + 25] – [4x^2 – 20x + 25] \)
\( = 4x^2 + 20x + 25 – 4x^2 + 20x – 25 \)
\( = 4x^2 - 4x^2 + 20x + 20x + 25 - 25 \)
\( = 0 + 40x + 0 \)
\( = 40x \)
In simple words: This problem involves subtracting the square of a difference from the square of a sum. We expanded both parts using the identities \( (a + b)^2 \) and \( (a - b)^2 \), then subtracted the second expanded expression from the first. Many terms cancelled out, leaving a simple result.

Exam Tip: A useful shortcut for this type of problem is \( (a+b)^2 - (a-b)^2 = 4ab \). Applying this directly here, \( 4(2x)(5) = 40x \), which confirms the detailed steps.

Question 4.
(iii) \( (7m – 8n)^2 - (7m + 8n)^2 \)
Answer:
\( = [(7m)^2 – 2(7m)(8n) + (8n)^2] - [(7m)^2 + 2(7m)(8n) + (8n)^2] \)
\( = [49m^2 – 112mn + 64n^2] - [49m^2 + 112mn + 64n^2] \)
\( = 49m^2 – 112mn + 64n^2 – 49m^2 - 112mn - 64n^2 \)
\( = 49m^2 – 49m^2 – 112mn – 112mn + 64n^2 – 64n^2 \)
\( = 0 – 224mn + 0 \)
\( = -224mn \)
In simple words: This problem requires us to subtract the square of a sum from the square of a difference. We expanded both expressions using the respective identities, then carefully subtracted the second expanded form from the first. Many terms cancelled, giving a simplified negative result.

Exam Tip: Be very careful with the signs when removing parentheses after subtraction. All terms inside the second bracket will have their signs flipped when the bracket is removed.

Question 4.
(iv) (4m + 5n)² + (5m + 4n)²
Answer:
\( = [(4m)^2 + 2(4m)(5n) + (5n)^2] + [(5m)^2 + 2(5m)(4n) + (4n)^2] \)
\( = [16m^2 + 40mn + 25n^2] + [25m^2 + 40mn + 16n^2] \)
\( = 16m^2 + 40mn + 25n^2 + 25m^2 + 40mn + 16n^2 \)
\( = 16m^2 + 25m^2 + 40mn + 40mn + 25n^2 + 16n^2 \)
\( = 41m^2 + 80mn + 41n^2 \)
In simple words: We need to add two squared binomials. We expanded both terms using the \( (a + b)^2 \) identity, then collected and combined the like terms ( \( m^2 \), \( mn \), and \( n^2 \) ) to reach the final simplified expression.

Exam Tip: Group and combine like terms carefully after expanding. Make sure to add the coefficients of the same variable parts correctly.

Question 4.
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
Answer:
\( = [(2.5p)^2 – 2(2.5p)(1.5q) + (1.5q)^2] – [(1.5p)^2 – 2(1.5p)(2.5q) + (2.5q)^2] \)
\( = [6.25p^2 – 7.5pq + 2.25q^2] – [2.25p^2 – 7.5pq + 6.25q^2] \)
\( = 6.25p^2 – 7.5pq + 2.25q^2 – 2.25p^2 + 7.5pq – 6.25q^2 \)
\( = 6.25p^2 – 2.25p^2 – 7.5pq + 7.5pq + 2.25q^2 – 6.25q^2 \)
\( = 4p^2 + 0 - 4q^2 \)
\( = 4p^2 - 4q^2 \)
In simple words: This problem involves subtracting two squared binomials with decimal coefficients. We expanded both expressions using the identity for squaring a difference, then carefully subtracted the second expanded form from the first, making sure to handle the decimal arithmetic correctly. Many terms cancelled out.

Exam Tip: Remember that \( (a-b)^2 - (c-d)^2 \) requires expanding each square separately before subtracting. Watch out for sign changes after the subtraction operation.

Question 4.
(vi) (ab + bc)² – 2ab²c
Answer:
\( = [(ab)^2 + 2(ab)(bc) + (bc)^2] – 2ab^2c \)
\( = a^2b^2 + 2ab^2c + b^2c^2 – 2ab^2c \)
\( = a^2b^2 + 2ab^2c – 2ab^2c + b^2c^2 \)
\( = a^2b^2 + 0 + b^2c^2 \)
\( = a^2b^2 + b^2c^2 \)
In simple words: We expanded the first term using the identity for squaring a sum. Then, we combined the like terms. The middle term \( 2ab^2c \) and \( -2ab^2c \) cancelled each other out, leaving the simplified expression.

Exam Tip: Be mindful of all variables and their powers when multiplying terms (e.g., \( (ab)(bc) = ab^2c \)). Carefully identify and combine only the terms that are exactly alike.

Question 4.
(vii) \( (m^2 - n^2m)^2 + 2m^3n^2 \)
Answer:
\( = [(m^2)^2 - 2(m^2)(n^2m) + (n^2m)^2] + 2m^3n^2 \)
\( = m^4 – 2m^3n^2 + n^4m^2 + 2m^3n^2 \)
\( = m^4 – 2m^3n^2 + 2m^3n^2 + n^4m^2 \)
\( = m^4 + 0 + n^4m^2 \)
\( = m^4 + m^2n^4 \)
In simple words: We first expanded the squared term using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \). After expanding, we combined the like terms. The terms \( -2m^3n^2 \) and \( +2m^3n^2 \) cancelled each other out, simplifying the expression to its final form.

Exam Tip: When simplifying, remember to group terms with identical variables and powers. Pay attention to how exponents combine in multiplication (e.g., \( m^2 \times m = m^3 \)).

Question 5. સાબિત કરોઃ
(i) \( (3x + 7)^2 – 84x = (3x – 7)^2 \)
Answer:
ડા.બા. \( = (3x + 7)^2 – 84x \)
\( = (3x)^2 + 2(3x)(7) + (7)^2 – 84x \)
\( = 9x^2 + 42x + 49 – 84x \)
\( = 9x^2 + 42x – 84x + 49 \)
\( = 9x^2 - 42x + 49 \)
જ.બા. \( = (3x – 7)^2 \)
\( = (3x)^2 – 2 (3x)(7) + (7)^2 \)
\( = 9x^2 - 42x + 49 \)
આમ, ડા.બા. \( = \) જ.બા.
\( \implies (3x + 7)^2 – 84x = (3x – 7)^2 \)
In simple words: To prove this identity, we first expanded the left-hand side using the \( (a + b)^2 \) identity and simplified it. Then, we expanded the right-hand side using the \( (a - b)^2 \) identity. Since both sides simplified to the same expression, the identity is proven true.

Exam Tip: For "prove" or "verify" questions, always work on the Left-Hand Side (LHS) and Right-Hand Side (RHS) separately until they simplify to the same expression. Do not cross the equality sign until the very end.

Question 5.
(ii) \( (9p – 5q)^2 + 180pq = (9p + 5q)^2 \)
Answer:
ડા.બા. \( = (9p – 5q)^2 + 180pq \)
\( = (9p)^2 – 2(9p)(5q) + (5q)^2 + 180pq \)
\( = 81p^2 – 90pq + 25q^2 + 180pq \)
\( = 81p^2 + 90pq + 25q^2 \)
જ.બા. \( = (9p + 5q)^2 \)
\( = (9p)^2 + 2 (9p)(5q) + (5q)^2 \)
\( = 81p^2 + 90pq + 25q^2 \)
આમ, ડા.બા. \( = \) જ.બા.
\( \implies (9p – 5q)^2 + 180pq = (9p + 5q)^2 \)
In simple words: To prove this identity, we expanded the left-hand side using the \( (a - b)^2 \) identity and simplified by combining like terms. Then, we expanded the right-hand side using the \( (a + b)^2 \) identity. As both sides simplified to the identical expression, the identity is verified.

Exam Tip: When combining terms like \( -90pq \) and \( +180pq \), ensure accurate addition or subtraction to prevent calculation errors. The final expression must match for both sides.

Question 5.
(iii) \( (\frac {4}{3}m - \frac {3}{4}n)^2 + 2mn = \frac {16}{9}m^2 + \frac {9}{16}n^2 \)
Answer:
ડા.બા. \( = [\frac {4}{3}m – \frac {3}{4}n]^2 + 2mn \)
\( = [(\frac {4}{3}m)^2 - 2(\frac {4}{3}m)(\frac {3}{4}n) + (\frac {3}{4}n)^2] + 2mn \)
\( = \frac {16}{9}m^2 - 2mn + \frac {9}{16}n^2 + 2mn \)
\( = \frac {16}{9}m^2 – 2mn + 2mn + \frac {9}{16}n^2 \)
\( = \frac {16}{9}m^2 + \frac {9}{16}n^2 \)
જ.બા. \( = \frac {16}{9}m^2 + \frac {9}{16}n^2 \)
આમ, ડા.બા. \( = \) જ.બા.
\( \implies (\frac {4}{3}m - \frac {3}{4}n)^2 + 2mn = \frac {16}{9}m^2 + \frac {9}{16}n^2 \)
In simple words: To prove this identity, we first expanded the left-hand side using the \( (a - b)^2 \) identity. After expansion, the middle term \( -2mn \) cancelled out with the added \( +2mn \), leaving only the squared fractional terms. This simplified result matched the right-hand side, thus proving the identity.

Exam Tip: When multiplying fractions in the \( 2ab \) term, look for opportunities to cancel common factors (like the '2' and '3' in this case) to simplify calculations.

Question 5.
(iv) \( (4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2 \)
Answer:
ડા.બા. \( = (4pq + 3q)^2 – (4pq – 3q)^2 \)
\( = [(4pq)^2 + 2(4pq)(3q) + (3q)^2] – [(4pq)^2 – 2 (4pq)(3q) + (3q)^2] \)
\( = [16p^2q^2 + 24pq^2 + 9q^2] – [16p^2q^2 – 24pq^2 + 9q^2] \)
\( = 16p^2q^2 + 24pq^2 + 9q^2 - 16p^2q^2 + 24pq^2 – 9q^2 \)
\( = 16p^2q^2 – 16p^2q^2 + 24pq^2 + 24pq^2 + 9q^2 – 9q^2 \)
\( = 0 + 48pq^2 + 0 \)
\( = 48pq^2 \)
જ.બા. \( = 48pq^2 \)
આમ, ડા.બા. \( = \) જ.બા.
\( \implies (4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2 \)
In simple words: We used the identities for squaring a sum and a difference to expand both parts of the left-hand side. After expansion, we subtracted the second expression from the first. Many terms cancelled out, resulting in \( 48pq^2 \), which matched the right-hand side, thus proving the identity.

Exam Tip: This is a classic application of the identity \( (A+B)^2 - (A-B)^2 = 4AB \). Here, \( A = 4pq \) and \( B = 3q \). So, \( 4(4pq)(3q) = 48pq^2 \). Using this shortcut can save significant time.

Question 5.
(v) \( (a – b)(a + b) + (b − c) (b + c) + (c − a) (c + a) = 0 \)
Answer:
ડા.બા. \( = (a – b) (a + b) + (b − c) (b + c) + (c − a) (c + a) \)
\( = (a^2 – b^2) + (b^2 – c^2) + (c^2 – a^2) \)
\( = a^2 – b^2 + b^2 – c^2 + c^2 – a^2 \)
\( = a^2 - a^2 + b^2 – b^2 + c^2 - c^2 \)
\( = 0 + 0 + 0 \)
\( = 0 \)
જ.બા. \( = 0 \)
આમ, ડા.બા. \( = \) જ.બા.
\( \implies (a – b)(a + b) + (b − c) (b + c) + (c − a) (c + a) = 0 \)
In simple words: We applied the difference of squares identity \( (x - y)(x + y) = x^2 - y^2 \) to each of the three products on the left-hand side. This resulted in three terms \( (a^2 - b^2) \), \( (b^2 - c^2) \), and \( (c^2 - a^2) \). When added together, all terms cancelled out, leaving 0, which matches the right-hand side.

Exam Tip: When terms cancel out in an algebraic expression, it often results in zero or a very simple form. This problem shows how repeated application of an identity can lead to significant simplification.

Question 6. નિત્યસમનો ઉપયોગ કરીને ગણતરી કરો:
(i) 71²
Answer:
\( = (70 + 1)^2 \)
\( = (70)^2 + 2(70)(1) + (1)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = 4900 + 140 + 1 \)
\( = 5041 \)
In simple words: To calculate 71 squared, we wrote it as \( (70 + 1)^2 \) and used the \( (a + b)^2 \) identity. We expanded the terms and then added them up to get the final answer.

Exam Tip: For numbers close to multiples of 10, expressing them as a sum or difference (e.g., \( 71 = 70 + 1 \)) allows for easy application of algebraic identities for quicker calculation.

Question 6.
(ii) 99²
Answer:
\( = (100 – 1)^2 \)
\( = (100)^2 – 2(100)(1) + (1)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 10000 - 200 + 1 \)
\( = 9801 \)
In simple words: To calculate 99 squared, we wrote it as \( (100 - 1)^2 \) and used the \( (a - b)^2 \) identity. We expanded the terms, performed the subtraction, and then added to get the correct value.

Exam Tip: Choosing the nearest multiple of 10 or 100 for 'a' in \( (a \pm b)^2 \) makes calculations easier, as squares and products of such numbers are simple to find.

Question 6.
(iii) 102²
Answer:
\( = (100 + 2)^2 \)
\( = (100)^2 + 2(100)(2) + (2)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = 10000 + 400 + 4 \)
\( = 10404 \)
In simple words: To calculate 102 squared, we expressed it as \( (100 + 2)^2 \) and applied the \( (a + b)^2 \) identity. We then expanded the expression and added the resulting terms to find the square.

Exam Tip: Breaking down numbers into simple sums or differences (e.g., \( 102 = 100 + 2 \)) simplifies squaring by allowing the use of identities instead of direct multiplication.

Question 6.
(iv) 998²
Answer:
\( = (1000 – 2)^2 \)
\( = (1000)^2 – 2(1000)(2) + (2)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 1000000 – 4000 + 4 \)
\( = 996004 \)
In simple words: To calculate 998 squared, we wrote it as \( (1000 - 2)^2 \) and used the \( (a - b)^2 \) identity. This allowed us to expand the terms, perform the subtraction, and get the final value quickly.

Exam Tip: For larger numbers, choosing 'a' as a power of 10 (like 1000) greatly simplifies the squaring and multiplication involved in the identity.

Question 6.
(v) 5.2²
Answer:
\( = (5 + 0.2)^2 \)
\( = (5)^2 + 2(5)(0.2) + (0.2)^2 \)
\( [:: (a + b)^2 = a^2 + 2ab + b^2] \)
\( = 25 + 2 + 0.04 \)
\( = 27 + 0.04 \)
\( = 27.04 \)
In simple words: To find the square of 5.2, we expressed it as \( (5 + 0.2)^2 \) and applied the \( (a + b)^2 \) identity. We then expanded the terms, performed the multiplications, and added them to find the correct square.

Exam Tip: For decimals, split the number into an integer and a decimal part (e.g., \( 5.2 = 5 + 0.2 \)). Be careful with decimal arithmetic, especially when squaring the decimal part.

Question 6.
(vi) 297 × 303
Answer:
\( = (300 – 3) \times (300 + 3) \)
\( = (300)^2 – (3)^2 \)
\( [:: (a – b) (a + b) = a^2 – b^2] \)
\( = 90000 - 9 \)
\( = 89991 \)
In simple words: To multiply 297 by 303, we wrote them as \( (300 - 3) \) and \( (300 + 3) \) respectively. Then, we used the difference of squares identity \( (a - b)(a + b) = a^2 - b^2 \). This simplified the calculation significantly, giving us the product.

Exam Tip: Numbers equidistant from a convenient multiple of 10, 100, etc., are perfect candidates for using the difference of squares identity. For example, \( 297 \) is \( 3 \) less than \( 300 \), and \( 303 \) is \( 3 \) more than \( 300 \).

Question 6.
(vii) 78 × 82
Answer:
\( = (80 – 2) \times (80 + 2) \)
\( = (80)^2 – (2)^2 \)
\( [:: (a – b) (a + b) = a^2 – b^2] \)
\( = 6400 - 4 \)
\( = 6396 \)
In simple words: To multiply 78 by 82, we rewrote them as \( (80 - 2) \) and \( (80 + 2) \). Then, we used the difference of squares identity \( (a - b)(a + b) = a^2 - b^2 \). This made the multiplication much easier, providing the final product efficiently.

Exam Tip: Always look for opportunities to express numbers as \( (X-Y)(X+Y) \) to utilize the difference of squares identity for faster mental or written calculation.

Question 6.
(viii) 8.9²
Answer:
\( = (9 – 0.1)^2 \)
\( = (9)^2 – 2(9)(0.1) + (0.1)^2 \)
\( [:: (a – b)^2 = a^2 – 2ab + b^2] \)
\( = 81 – 1.8 + 0.01 \)
\( = 79.2 + 0.01 \)
\( = 79.21 \)
In simple words: To calculate 8.9 squared, we expressed it as \( (9 - 0.1)^2 \) and applied the \( (a - b)^2 \) identity. We then expanded the terms, performed the necessary calculations with decimals, and added them to find the correct square.

Exam Tip: When using identities with decimals, remember to square the decimal part correctly (e.g., \( (0.1)^2 = 0.01 \)) and handle the multiplication in the middle term precisely.

Question 6.
(ix) 10.5 × 9.5
Answer:
\( = (10 + 0.5) \times (10 – 0.5) \)
\( = (10)^2 – (0.5)^2 \)
\( [:: (a + b) (a – b) = a^2 – b^2] \)
\( = 100 - 0.25 \)
\( = 99.75 \)
[નોંધઃ (ix) રકમમાં ભૂલ હોવાથી 1.05ને બદલે 10.5 લીધા છે.]
In simple words: To multiply 10.5 by 9.5, we rewrote them as \( (10 + 0.5) \) and \( (10 - 0.5) \) respectively. Then, we used the difference of squares identity \( (a - b)(a + b) = a^2 - b^2 \). This made the calculation much simpler, giving us the final product efficiently.

Exam Tip: This method is very efficient for multiplying numbers that are equidistant from a whole number or a half-number. Always double-check decimal calculations.

Question 7. નિત્યસમ \( a^2 – b^2 = (a + b)(a – b) \) નો ઉપયોગ કરીને કિંમત શોધો:
(i) 51² - 49²
Answer:
\( = (51 + 49) (51 – 49) \)
\( = (100) \times (2) \)
\( = 200 \)
In simple words: To find the value of \( 51^2 - 49^2 \), we directly applied the difference of squares identity \( a^2 - b^2 = (a + b)(a - b) \). We added 51 and 49, subtracted 49 from 51, and then multiplied the results to get the final answer.

Exam Tip: This identity is extremely useful for simplifying expressions involving the difference of two squares, often turning complex squaring into simple addition, subtraction, and multiplication.

Question 7.
(ii) \( (1.02)^2 – (0.98)^2 \)
Answer:
\( = (1.02 + 0.98) (1.02 – 0.98) \)
\( = (2.0) \times (0.04) \)
\( = 0.08 \)
In simple words: We used the difference of squares identity \( a^2 - b^2 = (a + b)(a - b) \) to simplify this expression. We first added the two decimal numbers and then subtracted them. Multiplying these two results gave us the final answer.

Exam Tip: The difference of squares identity works just as efficiently with decimals as with whole numbers. Practice decimal addition and subtraction for precision.

Question 7.
(iii) 153² - 147²
Answer:
\( = (153 + 147) (153 – 147) \)
\( = (300) \times (6) \)
\( = 1800 \)
In simple words: We applied the difference of squares identity \( a^2 - b^2 = (a + b)(a - b) \) to solve this problem. We summed 153 and 147, then found the difference between 153 and 147, and finally multiplied these two results to get the numerical answer.

Exam Tip: This identity makes calculating the difference of large squares much simpler, avoiding the need to calculate each square separately before subtracting.

Question 7.
(iv) 12.1² - 7.9²
Answer:
\( = (12.1 + 7.9) (12.1 – 7.9) \)
\( = 20 \times 4.2 \)
\( = 84 \)
In simple words: We applied the difference of squares identity \( a^2 - b^2 = (a + b)(a - b) \) to find the value. We added 12.1 and 7.9, then subtracted 7.9 from 12.1, and finally multiplied these two sums to get the result.

Exam Tip: When performing addition and subtraction with decimals, align the decimal points to avoid errors in calculation.

Question 8. (x + a) (x + b) = x² + (a + b)x + abનો ઉપયોગ કરીને કિંમત શોધો:
(i) 103 × 104
Answer:
\( = (100 + 3) \times (100 + 4) \)
\( = (100)^2 + (3 + 4) \times 100 + (3)(4) \)
\( = 10000 + 700 + 12 \)
\( = 10712 \)
In simple words: To multiply 103 by 104, we rewrote them as \( (100 + 3) \) and \( (100 + 4) \). Then, we used the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) with \( x = 100 \), \( a = 3 \), and \( b = 4 \). This simplified the multiplication to simple additions.

Exam Tip: This identity is very useful for multiplying numbers that are close to a power of 10. Breaking them down into \( (\text{base} + a)(\text{base} + b) \) simplifies calculations significantly.

Question 8.
(ii) 5.1 × 5.2
Answer:
\( = (5 + 0.1) (5 + 0.2) \)
\( = (5)^2 + (0.1 + 0.2) \times 5 + (0.1) (0.2) \)
\( = 25 + (0.3) \times 5 + 0.02 \)
\( = 25 + 1.5 + 0.02 \)
\( = 26.52 \)
In simple words: To multiply 5.1 by 5.2, we wrote them as \( (5 + 0.1) \) and \( (5 + 0.2) \). Then, we applied the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) with \( x = 5 \), \( a = 0.1 \), and \( b = 0.2 \). This approach simplified the multiplication of decimals.

Exam Tip: When using this identity with decimals, be careful with decimal placement in addition and multiplication steps to maintain accuracy.

Question 8.
(iii) 103 × 98
Answer:
\( = (100 + 3) (100 – 2) \)
\( = (100)^2 + (3 – 2) 100 + (3)(-2) \)
\( = 10000 + 100 - 6 \)
\( = 10094 \)
In simple words: To multiply 103 by 98, we rewrote them as \( (100 + 3) \) and \( (100 - 2) \). Then, we used the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) with \( x = 100 \), \( a = 3 \), and \( b = -2 \). This simplified the multiplication to basic arithmetic operations.

Exam Tip: Remember that 'a' or 'b' can be negative when expressing numbers as \( (x+a)(x+b) \). Be very careful with signs when calculating \( (a+b)x \) and \( ab \).

Question 8.
(iv) 9.7 × 9.8
Answer:
\( = (10 – 0.3) (10 – 0.2) \)
\( = (10)^2 + [(-0.3) + (-0.2)] 10 + (- 0.3)(- 0.2) \)
\( = 100 + [- 0.5] \times 10 + 0.06 \)
\( = 100 - 5 + 0.06 \)
\( = 95 + 0.06 \)
\( = 95.06 \)
In simple words: To multiply 9.7 by 9.8, we rewrote them as \( (10 - 0.3) \) and \( (10 - 0.2) \). Then, we used the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) with \( x = 10 \), \( a = -0.3 \), and \( b = -0.2 \). This allowed us to calculate the product using simpler decimal arithmetic.

Exam Tip: When both 'a' and 'b' are negative, their sum \( (a+b) \) will be negative, and their product \( ab \) will be positive. Pay close attention to these sign rules in calculations.