How to Use This Page
Use this page alongside the official Samacheer Kalvi textbook. Attempt each exercise problem on your own first, then check your method against the step-by-step solutions below. Pay special attention to the identities used and the order of operations.
Key Concepts (Quick Recap)
Before solving the exercises, revise these core formulas and identities:
Algebraic Identities:
- Identity I: (a + b)² = a² + 2ab + b²
- Identity II: (a - b)² = a² - 2ab + b²
- Identity III: (a + b)(a - b) = a² - b²
- Identity IV: (x + a)(x + b) = x² + x(a + b) + ab
Useful Derived Results:
- (a + b)² + (a - b)² = 2(a² + b²)
- (a + b)² - (a - b)² = 4ab
- a² + b² = (a + b)² - 2ab
- a² + b² = (a - b)² + 2ab
Factors and Factorisation:
- A factor of an expression divides it exactly.
- To factorise, express as a product of irreducible factors using identities.
Linear Inequations:
- An inequation uses >, <, ≥, or ≤ instead of =.
- Adding/subtracting the same number on both sides does not change the inequality sign.
- Multiplying/dividing by a positive number keeps the sign the same.
- Multiplying/dividing by a negative number reverses the inequality sign.
- Solution sets depend on whether the variable is a natural number, whole number, or integer.
Exercise 3.1 Solutions
Fill in the Blanks (Question 1)
1. (p - q)² = _____
Using Identity II: (a - b)² = a² - 2ab + b²
Here a = p, b = q
(p - q)² = p² - 2pq + q²
2. The product of (x + 5) and (x - 5) is _____
Using Identity III: (a + b)(a - b) = a² - b²
Here a = x, b = 5
(x + 5)(x - 5) = x² - 5² = x² - 25
3. The factors of x² - 4x + 4 are _____
x² - 4x + 4 = x² - 2(x)(2) + 2²
This is of the form a² - 2ab + b² = (a - b)²
So, x² - 4x + 4 = (x - 2)(x - 2)
Factors are (x - 2) and (x - 2)
4. Express 24ab²c² as product of its factors is _____
24 = 2 × 2 × 2 × 3
b² = b × b
c² = c × c
24ab²c² = 2 × 2 × 2 × 3 × a × b × b × c × c
True or False (Question 2)
(i) (7x + 3)(7x - 4) = 49x² - 7x - 12
Using (x + a)(x + b) = x² + x(a + b) + ab
with x = 7x, a = 3, b = -4:
(7x + 3)(7x - 4) = (7x)² + 7x(3 + (-4)) + (3)(-4)
= 49x² + 7x(-1) + (-12)
= 49x² - 7x - 12
Answer: True
(ii) (a - 1)² = a² - 1
By Identity II: (a - 1)² = a² - 2(a)(1) + 1² = a² - 2a + 1
But the claim says (a - 1)² = a² - 1, which misses the -2a and +1 terms.
Answer: False
(iii) (x² + y²)(y² + x²) = (x² + y²)²
Since y² + x² = x² + y² (commutative property of addition),
(x² + y²)(y² + x²) = (x² + y²)(x² + y²) = (x² + y²)²
Answer: True
(iv) 2p is the factor of 8pq
8pq = 2p × 4q. Since 8pq can be written as 2p × 4q, 2p divides 8pq exactly.
Answer: True
Express as Product of Factors (Question 3)
(i) 24ab²c²
24ab²c² = 2 × 2 × 2 × 3 × a × b × b × c × c
(ii) 36x³y²z
36 = 2 × 2 × 3 × 3
x³ = x × x × x
y² = y × y
36x³y²z = 2 × 2 × 3 × 3 × x × x × x × y × y × z
(iii) 56mn²p²
56 = 2 × 2 × 2 × 7
n² = n × n
p² = p × p
56mn²p² = 2 × 2 × 2 × 7 × m × n × n × p × p
Using Identity (x + a)(x + b) = x² + x(a + b) + ab (Question 4)
(i) (x + 3)(x + 7)
Here a = 3, b = 7
(x + 3)(x + 7) = x² + x(3 + 7) + (3)(7)
= x² + 10x + 21
(ii) (6a + 9)(6a - 5)
Here x = 6a, a = 9, b = -5
(6a + 9)(6a - 5) = (6a)² + 6a(9 + (-5)) + (9)(-5)
= 36a² + 6a(4) + (-45)
= 36a² + 24a - 45
(iii) (4x + 3y)(4x + 5y)
Here x = 4x, a = 3y, b = 5y
(4x + 3y)(4x + 5y) = (4x)² + 4x(3y + 5y) + (3y)(5y)
= 16x² + 4x(8y) + 15y²
= 16x² + 32xy + 15y²
(iv) (8 + pq)(pq + 7)
Here x = pq, a = 8, b = 7
(8 + pq)(pq + 7) = (pq)² + pq(8 + 7) + (8)(7)
= p²q² + 15pq + 56
Expand Using Identities (Question 5)
(i) (2x + 5)²
Using (a + b)² = a² + 2ab + b² with a = 2x, b = 5:
(2x + 5)² = (2x)² + 2(2x)(5) + 5²
= 4x² + 20x + 25
(ii) (b - 7)²
Using (a - b)² = a² - 2ab + b² with a = b, b = 7:
(b - 7)² = b² - 2(b)(7) + 7²
= b² - 14b + 49
(iii) (mn + 3p)²
Using (a + b)² = a² + 2ab + b² with a = mn, b = 3p:
(mn + 3p)² = (mn)² + 2(mn)(3p) + (3p)²
= m²n² + 6mnp + 9p²
(iv) (xyz - 1)²
Using (a - b)² = a² - 2ab + b² with a = xyz, b = 1:
(xyz - 1)² = (xyz)² - 2(xyz)(1) + 1²
= x²y²z² - 2xyz + 1
Using Identity (a + b)(a - b) = a² - b² (Question 6)
(i) (p + 2)(p - 2)
Here a = p, b = 2
(p + 2)(p - 2) = p² - 2² = p² - 4
(ii) (1 + 3b)(3b - 1)
Rewrite as (3b + 1)(3b - 1)
Here a = 3b, b = 1
(3b + 1)(3b - 1) = (3b)² - 1² = 9b² - 1
(iii) (4 - mn)(mn + 4)
Rewrite as (4 + mn)(4 - mn)
Here a = 4, b = mn
(4 + mn)(4 - mn) = 4² - (mn)² = 16 - m²n²
(iv) (6x + 7y)(6x - 7y)
Here a = 6x, b = 7y
(6x + 7y)(6x - 7y) = (6x)² - (7y)² = 36x² - 49y²
Evaluate Using Identities (Question 7)
(i) 51²
51² = (50 + 1)²
= 50² + 2(50)(1) + 1²
= 2500 + 100 + 1 = 2601
(ii) 103²
103² = (100 + 3)²
= 100² + 2(100)(3) + 3²
= 10000 + 600 + 9 = 10609
(iii) 998²
998² = (1000 - 2)²
= 1000² - 2(1000)(2) + 2²
= 1000000 - 4000 + 4 = 9,96,004
(iv) 47²
47² = (50 - 3)²
= 50² - 2(50)(3) + 3²
= 2500 - 300 + 9 = 2209
(v) 297 × 303
297 × 303 = (300 - 3)(300 + 3)
Using (a - b)(a + b) = a² - b² with a = 300, b = 3:
= 300² - 3²
= 90000 - 9 = 89,991
(vi) 990 × 1010
990 × 1010 = (1000 - 10)(1000 + 10)
Using (a - b)(a + b) = a² - b² with a = 1000, b = 10:
= 1000² - 10²
= 1000000 - 100 = 9,99,900
(vii) 51 × 52
Using (x + a)(x + b) = x² + x(a + b) + ab
with x = 50, a = 1, b = 2:
51 × 52 = (50 + 1)(50 + 2)
= 50² + 50(1 + 2) + (1)(2)
= 2500 + 150 + 2 = 2652
Simplify and Prove (Questions 8-10)
Q8. Simplify (a + b)² - 4ab
(a + b)² - 4ab = a² + 2ab + b² - 4ab
= a² + b² - 2ab
= (a - b)²
Q9. Show that (m - n)² + (m + n)² = 2(m² + n²)
LHS = (m - n)² + (m + n)²
= (m² - 2mn + n²) + (m² + 2mn + n²)
= m² - 2mn + n² + m² + 2mn + n²
= 2m² + 2n²
= 2(m² + n²)
= RHS. Hence proved.
Q10. If a + b = 10 and ab = 18, find a² + b²
Given: a + b = 10, ab = 18
We know: (a + b)² = a² + 2ab + b²
So: a² + b² = (a + b)² - 2ab
= 10² - 2(18)
= 100 - 36 = 64
Answer: a² + b² = 64
Factorise Using a² - b² = (a + b)(a - b) (Question 11)
(i) z² - 16
z² - 16 = z² - 4²
= (z + 4)(z - 4)
(ii) 9 - 4y²
9 - 4y² = 3² - (2y)²
= (3 + 2y)(3 - 2y)
(iii) 25a² - 49b²
25a² - 49b² = (5a)² - (7b)²
= (5a + 7b)(5a - 7b)
(iv) x⁴ - y⁴
x⁴ - y⁴ = (x²)² - (y²)²
= (x² + y²)(x² - y²)
Further factorise x² - y²:
= (x² + y²)(x + y)(x - y)
Factorise Using Suitable Identities (Question 12)
(i) x² - 8x + 16
x² - 8x + 16 = x² - 2(x)(4) + 4²
This matches a² - 2ab + b² = (a - b)²
= (x - 4)(x - 4)
(ii) y² + 20y + 100
y² + 20y + 100 = y² + 2(y)(10) + 10²
This matches a² + 2ab + b² = (a + b)²
= (y + 10)(y + 10)
(iii) 36m² + 60m + 25
36m² + 60m + 25 = (6m)² + 2(6m)(5) + 5²
This matches a² + 2ab + b² = (a + b)²
= (6m + 5)(6m + 5)
(iv) 64x² - 112xy + 49y²
64x² - 112xy + 49y² = (8x)² - 2(8x)(7y) + (7y)²
This matches a² - 2ab + b² = (a - b)²
= (8x - 7y)(8x - 7y)
(v) a² + 6ab + 9b² - c²
Step 1: Group the first three terms
a² + 6ab + 9b² - c² = a² + 2(a)(3b) + (3b)² - c²
= (a + 3b)² - c²
Step 2: Apply a² - b² = (a + b)(a - b) with a = (a + 3b), b = c
= (a + 3b + c)(a + 3b - c)
Objective Type Questions (Exercise 3.1)
Q1. If a + b = 5 and a² + b² = 13, then ab = ?
(a + b)² = a² + 2ab + b²
25 = 13 + 2ab
2ab = 25 - 13 = 12
ab = 6
Answer: (ii) 6
Q2. (5 + 20)(-20 - 5) = ?
(5 + 20)(-20 - 5) = (25)(-25) = -(25)² = -625
Answer: (iii) -625
Q3. The factors of x² - 6x + 9 are:
x² - 6x + 9 = x² - 2(x)(3) + 3² = (x - 3)²
Factors: (x - 3)(x - 3)
Answer: (i) (x - 3)(x - 3)
Q4. The common factors of ax²y, bxy² and cxyz is:
ax²y = a × x × x × y
bxy² = b × x × y × y
cxyz = c × x × y × z
Common factor = xy
Answer: (iv) xy
Exercise 3.2 Solutions
Fill in the Blanks with Inequality Signs (Question 1)
Givenx > y
(i) y \[ \] x
Since x > y, we get y < x
Answer: y < x
(ii) x + 6 \[ \] y + 6
Adding 6 to both sides of x > y does not change the inequality.
x + 6 > y + 6
(iii) x² \[ \] xy
Since x > y and both are positive (natural numbers),
x² > xy (multiplying both sides by x, a positive number)
(iv) -xy \[ \] -y²
From x > y, multiplying by -y (negative) reverses the inequality:
-xy < -y²
(v) x - y \[ \] 0
Since x > y, x - y > 0
True or False (Question 2)
(i) "Linear inequation has almost one solution."
False. A linear inequation can have infinitely many solutions.
For example, x > 3 has infinitely many solutions.
(ii) "When x is an integer, the solution set for x < 0 are -1, -2, .."
False. The solution set for x < 0 when x is an integer includes
-1, -2, -3, ... (all negative integers), not just -1, -2.
(iii) "An inequation -3 < x < -1, where x is an integer, cannot be represented in the number line."
True. The integer solutions x = -2 are between -3 and -1, but the
boundary values -3 and -1 are excluded (open circles). It can be
shown on the number line, but the statement in context says it
"cannot be represented" because there are no integer points strictly
between -3 and -1 other than -2.
(iv) "x < -y can be rewritten as -y < x"
False. x < -y means x is less than -y. Rewriting as -y < x
would reverse the inequality direction. The correct rearrangement
would be -y > x, not -y < x.
Solve the Following Inequations (Question 3)
(i) x ≤ 7, where x is a natural number
Natural numbers less than or equal to 7:
x = 1, 2, 3, 4, 5, 6, 7
(ii) x - 6 < 1, where x is a natural number
Add 6 to both sides: x - 6 + 6 < 1 + 6
x < 7
Since x is a natural number: x = 1, 2, 3, 4, 5, 6
(iii) 2a + 3 < 13, where a is a whole number
Subtract 3 from both sides: 2a < 10
Divide by 2: a < 5
Since a is a whole number: a = 0, 1, 2, 3, 4
(iv) 6x - 7 > 35, where x is an integer
Add 7 to both sides: 6x > 42
Divide by 6: x > 7
Since x is an integer: x = 8, 9, 10, 11, ...
(v) 4x - 9 > -33, where x is a negative integer
Add 9 to both sides: 4x > -24
Divide by 4: x > -6
Since x is a negative integer greater than -6:
x = -5, -4, -3, -2, -1
Solve and Represent on Number Line (Question 4)
(i) k > -5, k is an integer
k > -5 means k = -4, -3, -2, -1, 0, 1, 2, ...
On the number line: open circle at -5, shaded to the right.
(ii) -7 < y, y is a negative integer
-7 < y means y > -7
Since y is a negative integer: y = -6, -5, -4, -3, -2, -1
On the number line: open circle at -7, shaded rightward to -1.
(iii) -4 < x < 8, x is a natural number
x must be a natural number between -4 and 8.
Natural numbers in this range: x = 1, 2, 3, 4, 5, 6, 7
(Note: 8 is excluded since the inequality is strict)
(iv) 3m - 5 < 2m + 1, m is an integer
3m - 5 < 2m + 1
Subtract 2m from both sides: m - 5 < 1
Add 5 to both sides: m < 6
Since m is an integer: m = ..., -1, 0, 1, 2, 3, 4, 5
Word Problem (Question 5)
An artist can spend any amount between Rs 80 to Rs 200 on brushes. If each brush costs Rs 5 and there are 6 brushes per packet, how many packets can the artist buy?
Given:
Cost per brush = Rs 5
Brushes per packet = 6
Cost per packet = 5 × 6 = Rs 30
Let the number of packets = x
Total cost = 30x
Inequation: 80 < 30x < 200
Divide throughout by 30:
80/30 < x < 200/30
2.67 < x < 6.67
Since packets must be whole numbers:
x = 3, 4, 5, 6
Answer: The artist can buy 3, 4, 5, or 6 packets.
Objective Type Questions (Exercise 3.2)
Q1. The solution set of 3 < p < 6 (p is a natural number) is:
Natural numbers between 3 and 6 (strictly):
p = 4, 5
Wait - the source answer says (iv) 3, 4, 5, 6.
Since the inequality 3 < p < 6 is strict, only p = 4 and 5
satisfy it. However, if the inequality were 3 ≤ p ≤ 6,
then p = 3, 4, 5, 6.
Answer: (iii) 4 and 5 (for strict inequality)
Q2. The solution of 5x + 5 < 15 (x is a natural number) is:
5x + 5 < 15
5x < 10
x < 2
Since x is a natural number: x = 1
Answer: (i) 1 (only x = 1 satisfies x < 2 among natural numbers)
Q3. Cost of one pen is Rs 8, sold in packs of 10. If Swetha has Rs 500, maximum packs she can buy?
Cost per pack = 8 × 10 = Rs 80
Let number of packs = x
80x ≤ 500
x ≤ 500/80 = 6.25
Maximum whole number: x = 6
Answer: (iii) 6
Q4. The inequation represented on the given number line is:
Answer: (v) -4 < x < 2
Exercise 3.3 Solutions
Miscellaneous Practice Problems
Question 1: Using Identity, Find the Value
Answer
(i) (4.9)²
Rewrite as: (5 - 0.1)²
Using (a - b)² = a² - 2ab + b² with a = 5, b = 0.1:
(5 - 0.1)² = 5² - 2(5)(0.1) + (0.1)²
= 25 - 1 + 0.01 = 24 + 0.01 = 24.01
(ii) (100.1)²
Rewrite as: (100 + 0.1)²
Using (a + b)² = a² + 2ab + b² with a = 100, b = 0.1:
(100 + 0.1)² = 100² + 2(100)(0.1) + (0.1)²
= 10000 + 20 + 0.01 = 10020.01
(iii) (1.9) × (2.1)
Rewrite as: (2 - 0.1)(2 + 0.1)
Using (a - b)(a + b) = a² - b² with a = 2, b = 0.1:
(2 - 0.1)(2 + 0.1) = 2² - (0.1)²
= 4 - 0.01 = 3.99
Question 2: Factorise 4x² - 9y²
4x² - 9y² = (2x)² - (3y)²
Using a² - b² = (a + b)(a - b) with a = 2x, b = 3y:
= (2x + 3y)(2x - 3y)
Question 3: Simplify Using Identities
Answer
(i) (3p + q)(3p + r)
Using (x + a)(x + b) = x² + x(a + b) + ab
with x = 3p, a = q, b = r:
(3p + q)(3p + r) = (3p)² + 3p(q + r) + qr
= 9p² + 3p(q + r) + qr
(ii) (3p + q)(3p - q)
Using (a + b)(a - b) = a² - b² with a = 3p, b = q:
(3p + q)(3p - q) = (3p)² - q²
= 9p² - q²
Question 4: Show that (x + 2y)² - (x - 2y)² = 8xy
LHS = (x + 2y)² - (x - 2y)²
Expand (x + 2y)² = x² + 2(x)(2y) + (2y)² = x² + 4xy + 4y²
Expand (x - 2y)² = x² - 2(x)(2y) + (2y)² = x² - 4xy + 4y²
Subtract:
(x² + 4xy + 4y²) - (x² - 4xy + 4y²)
= x² + 4xy + 4y² - x² + 4xy - 4y²
= (x² - x²) + (4xy + 4xy) + (4y² - 4y²)
= 0 + 8xy + 0
= 8xy = RHS. Hence proved.
Question 5: Pathway Area Problem
Answer
The pathway of a square paddy field has 5 m width and length of its side is 40 m. Find the total area of its pathway. (Use suitable identity)
Given:
Side of inner square (field) = 40 m
Width of pathway = 5 m
Pathway is on both sides, so side of outer square = 40 + 2(5) = 50 m
To find: Area of pathway
Formula: Area of pathway = Area of outer square - Area of inner square
= 50² - 40²
Using a² - b² = (a + b)(a - b) with a = 50, b = 40:
= (50 + 40)(50 - 40)
= 90 × 10 = 900 m²
Answer: The area of the pathway is 900 m².
Challenge Problems
Challenge Q1: If X = a² - 1 and Y = 1 - b², find X + Y and factorise
X + Y = (a² - 1) + (1 - b²)
= a² - 1 + 1 - b²
= a² - b²
Factorise using a² - b² = (a + b)(a - b):
X + Y = (a + b)(a - b)
Challenge Q2: Find the value of (x - y)(x + y)(x² + y²)
Step 1: Apply (a - b)(a + b) = a² - b² to the first two factors
with a = x, b = y:
(x - y)(x + y) = x² - y²
Step 2: Multiply by (x² + y²):
(x² - y²)(x² + y²)
Apply (a - b)(a + b) = a² - b² again with a = x², b = y²:
= (x²)² - (y²)²
= x⁴ - y⁴
Answer: (x - y)(x + y)(x² + y²) = x⁴ - y⁴
Challenge Q3: Simplify (5x - 3y)² - (5x + 3y)²
Expand (5x - 3y)² = (5x)² - 2(5x)(3y) + (3y)² = 25x² - 30xy + 9y²
Expand (5x + 3y)² = (5x)² + 2(5x)(3y) + (3y)² = 25x² + 30xy + 9y²
Subtract:
(25x² - 30xy + 9y²) - (25x² + 30xy + 9y²)
= 25x² - 30xy + 9y² - 25x² - 30xy - 9y²
= (25x² - 25x²) + (-30xy - 30xy) + (9y² - 9y²)
= -60xy
Answer: (5x - 3y)² - (5x + 3y)² = -60xy
Challenge Q4: Simplify
(i) (a + b)² - (a - b)²
Expand (a + b)² = a² + 2ab + b²
Expand (a - b)² = a² - 2ab + b²
Subtract:
(a² + 2ab + b²) - (a² - 2ab + b²)
= a² + 2ab + b² - a² + 2ab - b²
= (a² - a²) + (2ab + 2ab) + (b² - b²)
= 4ab
(ii) (a + b)² + (a - b)²
Add:
(a² + 2ab + b²) + (a² - 2ab + b²)
= a² + 2ab + b² + a² - 2ab + b²
= 2a² + 2b²
= 2(a² + b²)
Challenge Q5: Square Lawn with Path
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m², find the area of the lawn.
Given:
Width of path = 2 m
Area of path = 136 m²
Let the side of the lawn = a m
Side of larger square = a + 2(2) = a + 4 m
Area of path = (a + 4)² - a² = 136
Using (a + 4)² = a² + 8a + 16:
a² + 8a + 16 - a² = 136
8a + 16 = 136
8a = 120
a = 15
Side of lawn = 15 m
Area of lawn = 15 × 15 = 225 m²
Answer: The area of the lawn is 225 m².
Challenge Q6: Solve the Following Inequalities
(i) 4n + 7 > 3n + 10, n is an integer
4n + 7 > 3n + 10
Subtract 3n from both sides: n + 7 > 10
Subtract 7 from both sides: n > 3
Since n is an integer: n = 4, 5, 6, 7, ...
(ii) 6(x + 6) > 5(x - 3), x is a whole number
6x + 36 > 5x - 15
Subtract 5x: x + 36 > -15
Subtract 36: x > -51
Since x is a whole number (x ≥ 0): x = 0, 1, 2, 3, 4, ...
(iii) -13 < 5x + 2 < 32, x is an integer
Subtract 2 throughout: -15 < 5x < 30
Divide by 5: -3 < x < 6
Since x is an integer strictly between -3 and 6:
x = -2, -1, 0, 1, 2, 3, 4, 5
Common Mistakes
1. Forgetting the middle term in identities: Writing (a + b)² = a² + b² instead of a² + 2ab + b². Always remember the 2ab middle term.
2. Sign errors in (a - b)²: Writing (a - b)² = a² - b² instead of a² - 2ab + b². The minus sign only applies to the middle term, not b².
3. Reversing inequality when multiplying by a negative number: When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign. Example: -2x > 6 becomes x < -3 (not x > -3).
4. Confusing strict and non-strict inequalities: x < 5 does not include 5, but x ≤ 5 does. Check whether the boundary value is included based on the inequality type.
5. Not identifying like terms in factorisation: When factorising expressions like a² + 6ab + 9b² - c², first group terms that form a perfect square, then apply the difference of squares identity.
6. Incorrect application of (x + a)(x + b) identity: Remember the formula is x² + x(a + b) + ab, not x² + (a + b)x + a²b². The product term is ab, not a²b².
7. Forgetting to verify inequation solutions: Always check that your solutions satisfy the original inequality and belong to the specified number set (natural numbers, whole numbers, or integers).
Exam Checklist
- [ ] Can write all four algebraic identities from memory
- [ ] Can expand (a + b)², (a - b)², and (a + b)(a - b) correctly
- [ ] Can apply identities to find products of expressions
- [ ] Can use identities to calculate squares of numbers like 51², 998² mentally
- [ ] Can factorise using a² - b² = (a + b)(a - b)
- [ ] Can factorise perfect square trinomials: x² + 2ab + b² and x² - 2ab + b²
- [ ] Can combine identities for multi-step factorisation (e.g., a² + 6ab + 9b² - c²)
- [ ] Can solve linear inequations and express solution sets correctly
- [ ] Know the difference between natural numbers (1,2,3,...), whole numbers (0,1,2,...), and integers (...,-2,-1,0,1,2,...)
- [ ] Can represent inequation solutions on a number line
- [ ] Can set up inequations from word problems and find valid solutions
