RBSE Solutions Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.3

Question 1. Find the products of the following using suitable identity
(i) \( (x + 5) (x + 5) \)
(ii) \( (3x + 2) (3x + 2) \)
(iii) \( (5a - 7) (5a - 7) \)
(iv) \( (3p - \frac { 1 }{ 2 }) (3p - \frac { 1 }{ 2 }) \)
(v) \( (1.2m - 0.3) (1.2m - 0.3) \)
(vi) \( (x^2 + y^2) (x^2 - y^2) \)
(vii) \( (6y + 7) (-6y + 7) \)
(viii) \( (7a + 9b) (7a - 9b) \)
Answer:
(i) We are given \( (x + 5) (x + 5) \), which can be written as \( (x + 5)^2 \). We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = x \) and \( b = 5 \).
\( (x + 5)^2 = (x)^2 + 2(x)(5) + (5)^2 \)
\( = x^2 + 10x + 25 \)
The square of a binomial is always a trinomial.
In simple words: We multiply \( (x+5) \) by itself. We use the rule that \( (a+b) \) squared is \( a \) squared, plus two times \( a \) times \( b \), plus \( b \) squared. So, it becomes \( x^2 + 10x + 25 \).
(ii) We are given \( (3x + 2) (3x + 2) \), which can be written as \( (3x + 2)^2 \). We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = 3x \) and \( b = 2 \).
\( (3x + 2)^2 = (3x)^2 + 2(3x)(2) + (2)^2 \)
\( = 9x^2 + 12x + 4 \)
This identity simplifies the process of squaring expressions.
In simple words: We multiply \( (3x+2) \) by itself. Using the same rule as above, it becomes \( (3x) \) squared, plus two times \( (3x) \) times \( 2 \), plus \( 2 \) squared. This gives \( 9x^2 + 12x + 4 \).
(iii) We are given \( (5a - 7) (5a - 7) \), which can be written as \( (5a - 7)^2 \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 5a \) and \( b = 7 \).
\( (5a - 7)^2 = (5a)^2 - 2(5a)(7) + (7)^2 \)
\( = 25a^2 - 70a + 49 \)
This identity is used when the terms are subtracted inside the parenthesis.
In simple words: We multiply \( (5a-7) \) by itself. We use the rule that \( (a-b) \) squared is \( a \) squared, minus two times \( a \) times \( b \), plus \( b \) squared. So, it becomes \( 25a^2 - 70a + 49 \).
(iv) We are given \( (3p - \frac { 1 }{ 2 }) (3p - \frac { 1 }{ 2 }) \), which can be written as \( (3p - \frac { 1 }{ 2 })^2 \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 3p \) and \( b = \frac { 1 }{ 2 } \).
\( (3p - \frac { 1 }{ 2 })^2 = (3p)^2 - 2(3p)(\frac { 1 }{ 2 }) + (\frac { 1 }{ 2 })^2 \)
\( = 9p^2 - 3p + \frac { 1 }{ 4 } \)
This identity helps simplify squaring binomials involving fractions.
In simple words: We multiply \( (3p - \frac{1}{2}) \) by itself. Using the \( (a-b) \) squared rule, we get \( (3p) \) squared, minus two times \( (3p) \) times \( \frac{1}{2} \), plus \( (\frac{1}{2}) \) squared. This gives \( 9p^2 - 3p + \frac{1}{4} \).
(v) We are given \( (1.2m - 0.3) (1.2m - 0.3) \), which can be written as \( (1.2m - 0.3)^2 \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 1.2m \) and \( b = 0.3 \).
\( (1.2m - 0.3)^2 = (1.2m)^2 - 2(1.2m)(0.3) + (0.3)^2 \)
\( = 1.44m^2 - 0.72m + 0.09 \)
The calculations with decimals follow the same algebraic rules.
In simple words: We multiply \( (1.2m - 0.3) \) by itself. Using the \( (a-b) \) squared rule, we get \( (1.2m) \) squared, minus two times \( (1.2m) \) times \( 0.3 \), plus \( (0.3) \) squared. This results in \( 1.44m^2 - 0.72m + 0.09 \).
(vi) We are given \( (x^2 + y^2) (x^2 - y^2) \). We use the identity \( (a+b)(a-b) = a^2 - b^2 \). Here, \( a = x^2 \) and \( b = y^2 \).
\( (x^2 + y^2) (x^2 - y^2) = (x^2)^2 - (y^2)^2 \)
\( = x^4 - y^4 \)
This identity is particularly useful for quickly multiplying conjugate binomials.
In simple words: We use the rule that \( (a+b) \) times \( (a-b) \) is \( a \) squared minus \( b \) squared. So, for \( (x^2+y^2)(x^2-y^2) \), we get \( (x^2) \) squared minus \( (y^2) \) squared, which is \( x^4 - y^4 \).
(vii) We are given \( (6y + 7) (-6y + 7) \). We can rearrange the second factor as \( (7 - 6y) \). Now the expression is \( (7 + 6y) (7 - 6y) \). We use the identity \( (a+b)(a-b) = a^2 - b^2 \). Here, \( a = 7 \) and \( b = 6y \).
\( (7 + 6y) (7 - 6y) = (7)^2 - (6y)^2 \)
\( = 49 - 36y^2 \)
Rearranging terms to fit an identity is a common strategy in algebra.
In simple words: We rewrite the expression as \( (7+6y)(7-6y) \). Then, using the rule that \( (a+b)(a-b) \) is \( a \) squared minus \( b \) squared, we get \( 7 \) squared minus \( (6y) \) squared, which is \( 49 - 36y^2 \).
(viii) We are given \( (7a + 9b) (7a - 9b) \). We use the identity \( (a+b)(a-b) = a^2 - b^2 \). Here, \( a = 7a \) and \( b = 9b \).
\( (7a + 9b) (7a - 9b) = (7a)^2 - (9b)^2 \)
\( = 49a^2 - 81b^2 \)
This identity is a quick way to find the product of two binomials that are opposites in their second term.
In simple words: We use the rule that \( (a+b) \) times \( (a-b) \) is \( a \) squared minus \( b \) squared. So, for \( (7a+9b)(7a-9b) \), we get \( (7a) \) squared minus \( (9b) \) squared, which is \( 49a^2 - 81b^2 \).

🎯 Exam Tip: Always identify the correct algebraic identity before solving. Look for patterns like identical terms with opposite signs or identical binomials being multiplied.

Question 2. Use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \) to find the following products:
(i) \( (x + 1) (x + 2) \)
(ii) \( (3x + 5) (3x + 1) \)
(iii) \( (4x - 5) (4x - 1) \)
(iv) \( (3a + 5) (3a - 8) \)
(v) \( (xyz - 1) (xyz - 2) \)
Answer:
(i) We are given \( (x + 1) (x + 2) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = x \), \( a = 1 \), and \( b = 2 \).
\( (x + 1) (x + 2) = x^2 + (1 + 2) x + (1)(2) \)
\( = x^2 + 3x + 2 \)
This identity is very useful for multiplying binomials where the first term is common.
In simple words: We use the given rule. Here, `x` is `x`, `a` is `1`, and `b` is `2`. So we get `x` squared, plus `(1+2)` times `x`, plus `1` times `2`. The answer is `x^2 + 3x + 2`.
(ii) We are given \( (3x + 5) (3x + 1) \). We use the identity \( (X + A) (X + B) = X^2 + (A + B) X + AB \). Here, \( X = 3x \), \( A = 5 \), and \( B = 1 \).
\( (3x + 5) (3x + 1) = (3x)^2 + (5 + 1)(3x) + (5)(1) \)
\( = 9x^2 + 6(3x) + 5 \)
\( = 9x^2 + 18x + 5 \)
Remember to treat the common term, like `3x` here, as a single unit `X` in the identity.
In simple words: We use the rule where `X` is `3x`, `A` is `5`, and `B` is `1`. So we square `3x`, add `(5+1)` times `3x`, and then add `5` times `1`. The answer is `9x^2 + 18x + 5`.
(iii) We are given \( (4x - 5) (4x - 1) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 4x \), \( a = -5 \), and \( b = -1 \).
\( (4x - 5) (4x - 1) = (4x)^2 + ((-5) + (-1))(4x) + (-5)(-1) \)
\( = 16x^2 + (-6)(4x) + 5 \)
\( = 16x^2 - 24x + 5 \)
Careful handling of negative values for `a` and `b` is crucial for accuracy.
In simple words: We use the rule. Here, `x` is `4x`, `a` is `-5`, and `b` is `-1`. So we square `4x`, add `(-5` plus `-1)` times `4x`, and then add `(-5)` times `(-1)`. The answer is `16x^2 - 24x + 5`.
(iv) We are given \( (3a + 5) (3a - 8) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 3a \), \( a = 5 \), and \( b = -8 \).
\( (3a + 5) (3a - 8) = (3a)^2 + (5 + (-8))(3a) + (5)(-8) \)
\( = 9a^2 + (-3)(3a) - 40 \)
\( = 9a^2 - 9a - 40 \)
This identity helps multiply binomials where one constant term is positive and the other is negative.
In simple words: We use the rule. Here, `x` is `3a`, `a` is `5`, and `b` is `-8`. So we square `3a`, add `(5` plus `-8)` times `3a`, and then add `5` times `-8`. The answer is `9a^2 - 9a - 40`.
(v) We are given \( (xyz - 1) (xyz - 2) \). We use the identity \( (X + A) (X + B) = X^2 + (A + B) X + AB \). Here, \( X = xyz \), \( A = -1 \), and \( B = -2 \).
\( (xyz - 1) (xyz - 2) = (xyz)^2 + ((-1) + (-2))(xyz) + (-1)(-2) \)
\( = x^2y^2z^2 + (-3)(xyz) + 2 \)
\( = x^2y^2z^2 - 3xyz + 2 \)
This identity works even when the common term is a product of multiple variables.
In simple words: We use the rule. Here, `X` is `xyz`, `A` is `-1`, and `B` is `-2`. So we square `xyz`, add `(-1` plus `-2)` times `xyz`, and then add `(-1)` times `(-2)`. The answer is `x^2y^2z^2 - 3xyz + 2`.

🎯 Exam Tip: When using the \( (x+a)(x+b) \) identity, clearly identify \( x \), \( a \), and \( b \). If \( a \) or \( b \) are negative, include their signs in the substitution.

Question 3. Find the following squares by using the identities.
(i) \( (b - 7)^2 \)
(ii) \( (xy + 3z)^2 \)
(iii) \( (6m^2 - 5n)^2 \)
(iv) \( {\left(\frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 } \)
Answer:
(i) We are given \( (b - 7)^2 \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = b \) and \( b = 7 \).
\( (b - 7)^2 = (b)^2 - 2(b)(7) + (7)^2 \)
\( = b^2 - 14b + 49 \)
This identity helps in quickly expanding squares of binomials with a subtraction sign.
In simple words: To find \( (b-7) \) squared, we square `b`, subtract two times `b` times `7`, and then add `7` squared. This gives `b^2 - 14b + 49`.
(ii) We are given \( (xy + 3z)^2 \). We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = xy \) and \( b = 3z \).
\( (xy + 3z)^2 = (xy)^2 + 2(xy)(3z) + (3z)^2 \)
\( = x^2y^2 + 6xyz + 9z^2 \)
This identity is fundamental for expanding expressions that are sums squared.
In simple words: To find \( (xy+3z) \) squared, we square `xy`, add two times `xy` times `3z`, and then add `3z` squared. This gives `x^2y^2 + 6xyz + 9z^2`.
(iii) We are given \( (6m^2 - 5n)^2 \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 6m^2 \) and \( b = 5n \).
\( (6m^2 - 5n)^2 = (6m^2)^2 - 2(6m^2)(5n) + (5n)^2 \)
\( = 36m^4 - 60m^2n + 25n^2 \)
Remember to apply the exponent rules correctly when squaring terms like \( 6m^2 \).
In simple words: We find \( (6m^2 - 5n) \) squared by squaring `6m^2`, subtracting two times `6m^2` times `5n`, and then adding `5n` squared. The answer is `36m^4 - 60m^2n + 25n^2`.
(iv) We are given \( {\left(\frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 } \). We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = \frac { 3 }{ 2 } x \) and \( b = \frac { 2 }{ 3 } y \).
\( {\left(\frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 } = {\left(\frac { 3 }{ 2 } x\right) }^2 + 2{\left(\frac { 3 }{ 2 } x\right) }{\left(\frac { 2 }{ 3 } y\right) } + {\left(\frac { 2 }{ 3 } y\right) }^2 \)
\( = \frac { 9 }{ 4 } x^2 + 2xy + \frac { 4 }{ 9 } y^2 \)
When working with fractions, simplify the terms carefully after applying the identity.
In simple words: To square this expression, we square the first part \( (\frac{3}{2}x) \), then add two times the first part times the second part \( (\frac{2}{3}y) \), and finally add the square of the second part. This gives \( \frac{9}{4}x^2 + 2xy + \frac{4}{9}y^2 \).

🎯 Exam Tip: Always double-check your calculations, especially with negative signs, fractions, and exponents, after applying the identity.

Question 4. Simplify
(i) \( (a^2 - b^2)^2 \)
(ii) \( (2n + 5)^2 - (2n - 5)^2 \)
(iii) \( (7m - 8n)^2 + (7m + 8n)^2 \)
(iv) \( (m^2 - n^2m)^2 + 2m^3n^2 \)
Answer:
(i) We are given \( (a^2 - b^2)^2 \). We use the identity \( (X-Y)^2 = X^2 - 2XY + Y^2 \). Here, \( X = a^2 \) and \( Y = b^2 \).
\( (a^2 - b^2)^2 = (a^2)^2 - 2(a^2)(b^2) + (b^2)^2 \)
\( = a^4 - 2a^2b^2 + b^4 \)
Squaring terms with exponents means multiplying the exponents, e.g., \( (a^2)^2 = a^{2 \times 2} = a^4 \).
In simple words: To simplify \( (a^2 - b^2) \) squared, we square `a^2`, subtract two times `a^2` times `b^2`, and then add `b^2` squared. This gives `a^4 - 2a^2b^2 + b^4`.
(ii) We are given \( (2n + 5)^2 - (2n - 5)^2 \). We can use the identity \( (a+b)^2 - (a-b)^2 = 4ab \). Here, \( a = 2n \) and \( b = 5 \).
\( (2n + 5)^2 - (2n - 5)^2 = 4(2n)(5) \)
\( = 40n \)
Alternatively, we can use the difference of squares identity, \( A^2 - B^2 = (A+B)(A-B) \), where \( A = (2n+5) \) and \( B = (2n-5) \).
\( = ((2n+5) + (2n-5))((2n+5) - (2n-5)) \)
\( = (2n+5+2n-5)(2n+5-2n+5) \)
\( = (4n)(10) \)
\( = 40n \)
Both methods lead to the same result, showing the versatility of algebraic identities.
In simple words: We can solve this in two ways. One way is to open both squared brackets and then subtract. The other way is to use the rule that `(a+b)` squared minus `(a-b)` squared is simply `4ab`. Here, `a` is `2n` and `b` is `5`, so `4` times `2n` times `5` gives `40n`.
(iii) We are given \( (7m - 8n)^2 + (7m + 8n)^2 \). We use the identities \( (a-b)^2 = a^2 - 2ab + b^2 \) and \( (a+b)^2 = a^2 + 2ab + b^2 \).
\( (7m - 8n)^2 + (7m + 8n)^2 \)
\( = [(7m)^2 - 2(7m)(8n) + (8n)^2] + [(7m)^2 + 2(7m)(8n) + (8n)^2] \)
\( = [49m^2 - 112mn + 64n^2] + [49m^2 + 112mn + 64n^2] \)
\( = 49m^2 - 112mn + 64n^2 + 49m^2 + 112mn + 64n^2 \)
\( = (49m^2 + 49m^2) + (-112mn + 112mn) + (64n^2 + 64n^2) \)
\( = 98m^2 + 0 + 128n^2 \)
\( = 98m^2 + 128n^2 \)
This particular sum simplifies to \( 2(a^2 + b^2) \) when the terms are `(a-b)^2 + (a+b)^2`.
In simple words: We open both squared brackets using their respective rules. When we add them, the middle terms `(-112mn)` and `(+112mn)` cancel each other out. This leaves us with `49m^2` plus `49m^2`, and `64n^2` plus `64n^2`. The total is `98m^2 + 128n^2`.
(iv) We are given \( (m^2 - n^2m)^2 + 2m^3n^2 \). We first apply the identity \( (a-b)^2 = a^2 - 2ab + b^2 \) to the first term. Here, \( a = m^2 \) and \( b = n^2m \).
\( (m^2 - n^2m)^2 + 2m^3n^2 \)
\( = [(m^2)^2 - 2(m^2)(n^2m) + (n^2m)^2] + 2m^3n^2 \)
\( = [m^4 - 2m^3n^2 + n^4m^2] + 2m^3n^2 \)
\( = m^4 - 2m^3n^2 + n^4m^2 + 2m^3n^2 \)
\( = m^4 + n^4m^2 \)
Combining like terms is the final step after expansion to simplify the expression.
In simple words: We first use the \( (a-b) \) squared rule to open the first bracket. This gives us `m^4 - 2m^3n^2 + n^4m^2`. Then we add the last term `+ 2m^3n^2`. The terms `(-2m^3n^2)` and `(+2m^3n^2)` cancel each other out, leaving `m^4 + n^4m^2`.

🎯 Exam Tip: When simplifying expressions, always check if any terms cancel out or can be combined after applying identities. This makes the answer simpler.

Question 5. Show that
(i) \( (2a + 3b)^2 - (2a - 3b)^2 = 24ab \)
(ii) \( (4x + 5)^2 - 80x = (4x - 5)^2 \)
(iii) \( (3x - 2y)^2 + 24xy = (3x + 2y)^2 \)
(iv) \( (a - b) (a + b) + (b - c)(b + c) + (c - a) (c + a) = 0 \)
Answer:
(i) We need to show that \( (2a + 3b)^2 - (2a - 3b)^2 = 24ab \).
Let's take the Left Hand Side (LHS): \( (2a + 3b)^2 - (2a - 3b)^2 \).
We can use the identity \( (A+B)^2 - (A-B)^2 = 4AB \). Here, \( A = 2a \) and \( B = 3b \).
LHS \( = 4(2a)(3b) \)
LHS \( = 24ab \)
This is equal to the Right Hand Side (RHS).
Therefore, \( (2a + 3b)^2 - (2a - 3b)^2 = 24ab \) is shown.
This identity is a direct result of expanding both squares and seeing terms cancel.
In simple words: We start with the left side. We use a special rule that says `(A+B)` squared minus `(A-B)` squared is always equal to `4AB`. Here, `A` is `2a` and `B` is `3b`. So, `4` times `2a` times `3b` gives `24ab`, which is exactly the right side.
(ii) We need to show that \( (4x + 5)^2 - 80x = (4x - 5)^2 \).
Let's start with the Left Hand Side (LHS): \( (4x + 5)^2 - 80x \).
We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \) for \( (4x+5)^2 \). Here, \( a = 4x \) and \( b = 5 \).
\( (4x + 5)^2 = (4x)^2 + 2(4x)(5) + (5)^2 \)
\( = 16x^2 + 40x + 25 \)
Now, substitute this back into the LHS:
LHS \( = (16x^2 + 40x + 25) - 80x \)
LHS \( = 16x^2 + 40x - 80x + 25 \)
LHS \( = 16x^2 - 40x + 25 \)
Now, let's evaluate the Right Hand Side (RHS): \( (4x - 5)^2 \).
We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 4x \) and \( b = 5 \).
RHS \( = (4x)^2 - 2(4x)(5) + (5)^2 \)
RHS \( = 16x^2 - 40x + 25 \)
Since LHS \( = \) RHS, the statement is shown.
Showing both sides are equal confirms the identity.
In simple words: We simplify the left side `(4x+5)` squared minus `80x`. This gives `16x^2 - 40x + 25`. Then we simplify the right side `(4x-5)` squared, which also gives `16x^2 - 40x + 25`. Since both sides are the same, the statement is true.
(iii) We need to show that \( (3x - 2y)^2 + 24xy = (3x + 2y)^2 \).
Let's start with the Left Hand Side (LHS): \( (3x - 2y)^2 + 24xy \).
We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \) for \( (3x-2y)^2 \). Here, \( a = 3x \) and \( b = 2y \).
\( (3x - 2y)^2 = (3x)^2 - 2(3x)(2y) + (2y)^2 \)
\( = 9x^2 - 12xy + 4y^2 \)
Now, substitute this back into the LHS:
LHS \( = (9x^2 - 12xy + 4y^2) + 24xy \)
LHS \( = 9x^2 - 12xy + 24xy + 4y^2 \)
LHS \( = 9x^2 + 12xy + 4y^2 \)
Now, let's evaluate the Right Hand Side (RHS): \( (3x + 2y)^2 \).
We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = 3x \) and \( b = 2y \).
RHS \( = (3x)^2 + 2(3x)(2y) + (2y)^2 \)
RHS \( = 9x^2 + 12xy + 4y^2 \)
Since LHS \( = \) RHS, the statement is shown.
This demonstration illustrates how adding a middle term can transform one identity into another.
In simple words: We simplify the left side `(3x-2y)` squared plus `24xy`. This gives `9x^2 + 12xy + 4y^2`. Then, we simplify the right side `(3x+2y)` squared, which also gives `9x^2 + 12xy + 4y^2`. Since both sides match, the statement is correct.
(iv) We need to show that \( (a - b) (a + b) + (b - c)(b + c) + (c - a) (c + a) = 0 \).
Let's take the Left Hand Side (LHS): \( (a - b) (a + b) + (b - c)(b + c) + (c - a) (c + a) \).
We use the identity \( (x-y)(x+y) = x^2 - y^2 \) for each term:
\( (a - b) (a + b) = a^2 - b^2 \)
\( (b - c) (b + c) = b^2 - c^2 \)
\( (c - a) (c + a) = c^2 - a^2 \)
Now substitute these back into the LHS:
LHS \( = (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) \)
LHS \( = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 \)
LHS \( = (a^2 - a^2) + (-b^2 + b^2) + (-c^2 + c^2) \)
LHS \( = 0 + 0 + 0 \)
LHS \( = 0 \)
This is equal to the Right Hand Side (RHS).
Thus, the statement is shown. This demonstrates a cyclic property where terms cancel out in a sequence.
In simple words: We use the rule that `(x-y)` times `(x+y)` is `x` squared minus `y` squared. We apply this to all three parts. This makes the expression `(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)`. When we add them all up, everything cancels out (like `a^2` and `-a^2`), and we get `0`, which is the right side.

🎯 Exam Tip: When proving identities, clearly separate LHS and RHS. For the LHS, apply identities step-by-step and simplify until it matches the RHS.

Question 6. Using identities, evaluate the following
(i) \( 99^2 \)
(ii) \( 103^2 \)
(iii) \( 297 \times 303 \)
(iv) \( 78 \times 82 \)
Answer:
(i) To evaluate \( 99^2 \), we can write \( 99 \) as \( (100 - 1) \). We use the identity \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = 100 \) and \( b = 1 \).
\( 99^2 = (100 - 1)^2 \)
\( = (100)^2 - 2(100)(1) + (1)^2 \)
\( = 10000 - 200 + 1 \)
\( = 9801 \)
Using identities simplifies calculations involving numbers close to a benchmark like 100.
In simple words: We write `99` as `(100-1)`. Then we use the rule for `(a-b)` squared: `100` squared, minus two times `100` times `1`, plus `1` squared. This calculation gives us `9801`.
(ii) To evaluate \( 103^2 \), we can write \( 103 \) as \( (100 + 3) \). We use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Here, \( a = 100 \) and \( b = 3 \).
\( 103^2 = (100 + 3)^2 \)
\( = (100)^2 + 2(100)(3) + (3)^2 \)
\( = 10000 + 600 + 9 \)
\( = 10609 \)
This identity is efficient for squaring numbers slightly above a round number.
In simple words: We write `103` as `(100+3)`. Then we use the rule for `(a+b)` squared: `100` squared, plus two times `100` times `3`, plus `3` squared. This calculation gives `10609`.
(iii) To evaluate \( 297 \times 303 \), we can write \( 297 \) as \( (300 - 3) \) and \( 303 \) as \( (300 + 3) \). We use the identity \( (a-b)(a+b) = a^2 - b^2 \). Here, \( a = 300 \) and \( b = 3 \).
\( 297 \times 303 = (300 - 3) (300 + 3) \)
\( = (300)^2 - (3)^2 \)
\( = 90000 - 9 \)
\( = 89991 \)
This identity provides a quick way to multiply two numbers that are equidistant from a central value.
In simple words: We write `297` as `(300-3)` and `303` as `(300+3)`. Then we use the rule for `(a-b)(a+b)`: `300` squared minus `3` squared. This calculation gives `89991`.
(iv) To evaluate \( 78 \times 82 \), we can write \( 78 \) as \( (80 - 2) \) and \( 82 \) as \( (80 + 2) \). We use the identity \( (a-b)(a+b) = a^2 - b^2 \). Here, \( a = 80 \) and \( b = 2 \).
\( 78 \times 82 = (80 - 2) (80 + 2) \)
\( = (80)^2 - (2)^2 \)
\( = 6400 - 4 \)
\( = 6396 \)
This method makes mental calculation of such products much easier.
In simple words: We write `78` as `(80-2)` and `82` as `(80+2)`. Then we use the rule for `(a-b)(a+b)`: `80` squared minus `2` squared. This gives `6396`.

🎯 Exam Tip: When evaluating numerical expressions using identities, look for numbers that can be expressed as a sum or difference from a round number (like 10, 100, or 1000).

Question 7. Using \( a^2 - b^2 = (a + b) (a - b) \), find
(i) \( 101^2 - 99^2 \)
(ii) \( (10.3)^2 - (9.7)^2 \)
(iii) \( 153^2 - 147^2 \)
Answer:
(i) We are given \( 101^2 - 99^2 \). We use the identity \( a^2 - b^2 = (a + b) (a - b) \). Here, \( a = 101 \) and \( b = 99 \).
\( 101^2 - 99^2 = (101 + 99) (101 - 99) \)
\( = (200) (2) \)
\( = 400 \)
This identity converts a subtraction of squares into a simpler multiplication.
In simple words: We use the rule that `a` squared minus `b` squared is `(a+b)` times `(a-b)`. So, `(101+99)` times `(101-99)` gives `200` times `2`, which is `400`.
(ii) We are given \( (10.3)^2 - (9.7)^2 \). We use the identity \( a^2 - b^2 = (a + b) (a - b) \). Here, \( a = 10.3 \) and \( b = 9.7 \).
\( (10.3)^2 - (9.7)^2 = (10.3 + 9.7) (10.3 - 9.7) \)
\( = (20.0) (0.6) \)
\( = 12 \)
The identity works just as effectively with decimal numbers.
In simple words: We use the rule `a` squared minus `b` squared is `(a+b)` times `(a-b)`. So, `(10.3+9.7)` times `(10.3-9.7)` gives `20` times `0.6`, which is `12`.
(iii) We are given \( 153^2 - 147^2 \). We use the identity \( a^2 - b^2 = (a + b) (a - b) \). Here, \( a = 153 \) and \( b = 147 \).
\( 153^2 - 147^2 = (153 + 147) (153 - 147) \)
\( = (300) (6) \)
\( = 1800 \)
This method is particularly useful for quickly calculating differences of large squares.
In simple words: We add `153` and `147` to get `300`. We subtract `147` from `153` to get `6`. Then, we multiply `300` by `6` to get `1800`.

🎯 Exam Tip: For problems involving the difference of squares, always remember the formula \( a^2 - b^2 = (a + b) (a - b) \). It makes calculations much faster.

Question 8. Using identities, find the following products:
(i) \( 103 \times 102 \)
(ii) \( 7.1 \times 7.3 \)
(iii) \( 102 \times 99 \)
(iv) \( 9.8 \times 9.6 \)
Answer:
(i) To calculate \( 103 \times 102 \), we can write \( 103 \) as \( (100 + 3) \) and \( 102 \) as \( (100 + 2) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 100 \), \( a = 3 \), and \( b = 2 \).
\( 103 \times 102 = (100 + 3) (100 + 2) \)
\( = (100)^2 + (3 + 2)(100) + (3)(2) \)
\( = 10000 + (5)(100) + 6 \)
\( = 10000 + 500 + 6 \)
\( = 10506 \)
This identity is effective for multiplying numbers slightly larger than a base ten number.
In simple words: We write `103` as `(100+3)` and `102` as `(100+2)`. Using the rule `(x+a)(x+b)`, we get `100` squared, plus `(3+2)` times `100`, plus `3` times `2`. The answer is `10506`.
(ii) To calculate \( 7.1 \times 7.3 \), we can write \( 7.1 \) as \( (7 + 0.1) \) and \( 7.3 \) as \( (7 + 0.3) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 7 \), \( a = 0.1 \), and \( b = 0.3 \).
\( 7.1 \times 7.3 = (7 + 0.1) (7 + 0.3) \)
\( = (7)^2 + (0.1 + 0.3)(7) + (0.1)(0.3) \)
\( = 49 + (0.4)(7) + 0.03 \)
\( = 49 + 2.8 + 0.03 \)
\( = 51.83 \)
This method works well for multiplying decimals close to an integer.
In simple words: We write `7.1` as `(7+0.1)` and `7.3` as `(7+0.3)`. Using the rule `(x+a)(x+b)`, we get `7` squared, plus `(0.1+0.3)` times `7`, plus `0.1` times `0.3`. The answer is `51.83`.
(iii) To calculate \( 102 \times 99 \), we can write \( 102 \) as \( (100 + 2) \) and \( 99 \) as \( (100 - 1) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 100 \), \( a = 2 \), and \( b = -1 \).
\( 102 \times 99 = (100 + 2) (100 - 1) \)
\( = (100)^2 + (2 + (-1))(100) + (2)(-1) \)
\( = 10000 + (1)(100) - 2 \)
\( = 10000 + 100 - 2 \)
\( = 10098 \)
This identity is efficient when one number is slightly above and the other slightly below a round number.
In simple words: We write `102` as `(100+2)` and `99` as `(100-1)`. Using the rule `(x+a)(x+b)`, we get `100` squared, plus `(2 + (-1))` times `100`, plus `2` times `-1`. The answer is `10098`.
(iv) To calculate \( 9.8 \times 9.6 \), we can write \( 9.8 \) as \( (10 - 0.2) \) and \( 9.6 \) as \( (10 - 0.4) \). We use the identity \( (x + a) (x + b) = x^2 + (a + b) x + ab \). Here, \( x = 10 \), \( a = -0.2 \), and \( b = -0.4 \).
\( 9.8 \times 9.6 = (10 - 0.2) (10 - 0.4) \)
\( = (10)^2 + ((-0.2) + (-0.4))(10) + (-0.2)(-0.4) \)
\( = 100 + (-0.6)(10) + 0.08 \)
\( = 100 - 6 + 0.08 \)
\( = 94 + 0.08 \)
\( = 94.08 \)
This method is particularly useful for decimals close to an integer, especially when both are less than the integer.
In simple words: We write `9.8` as `(10-0.2)` and `9.6` as `(10-0.4)`. Using the rule `(x+a)(x+b)`, we get `10` squared, plus `(-0.2 + -0.4)` times `10`, plus `(-0.2)` times `(-0.4)`. The answer is `94.08`.

🎯 Exam Tip: For numerical products, look for common patterns like numbers near 10, 100, or numbers that are equidistant from a central value. Then choose the appropriate identity.