Exercise 5.2 — Factorisation Method
Solving quadratic equations using factorisation method.
Loading PDF…
What is a Root of a Quadratic Equation?
Core Concept
A root (or solution) of a quadratic equation ax² + bx + c = 0 is a real number α such that substituting x = α makes the equation true — that is, aα² + bα + c = 0.
For example, consider x² − 5x + 6 = 0. Substituting x = 1 gives 1 − 5 + 6 = 2 ≠ 0 (not a root). Substituting x = 2 gives 4 − 10 + 6 = 0 ✔. So x = 2 is a root.
Since a quadratic polynomial can have at most two zeroes, a quadratic equation can have at most two roots.
The Factorisation Method — How It Works
The key idea: if ax² + bx + c = 0 can be written as a product of two linear factors, then setting each factor to zero gives the two roots. The critical step is splitting the middle term (b) into two parts p and q.
Find p and q such that: p + q = b & p × q = a × c
Then rewrite bx as px + qx → group → factor → solve
①
Identify a, b, c from the standard form ax² + bx + c = 0
②
Find p and q: p + q = b and p × q = a × c
③
Split: replace bx with px + qx, then group and take common factors
④
Set each linear factor = 0 and solve for x
Question 1 — Find Roots by Factorisation (All 9 Parts)
(i)
x² − 3x − 10 = 0
Roots: 5 and −2
a=1, b=−3, c=−10
a×c = −10
→
Need p+q=−3, p×q=−10
p = 2, q = −5
→
Split −3x as
+2x − 5x
- 1x² + 2x − 5x − 10 = 0
- 2x(x + 2) − 5(x + 2) = 0
- 3(x + 2)(x − 5) = 0
- 4x + 2 = 0 → x = −2 OR x − 5 = 0 → x = 5
Roots: x = 5 and x = −2(ii)
2x² + x − 6 = 0
Roots: −2 and 3/2
a=2, b=1, c=−6
a×c = −12
→
Need p+q=1, p×q=−12
p = 4, q = −3
→
Split +x as
+4x − 3x
- 12x² + 4x − 3x − 6 = 0
- 22x(x + 2) − 3(x + 2) = 0
- 3(x + 2)(2x − 3) = 0
- 4x = −2 OR x = 3/2
Roots: x = −2 and x = 3/2(iii)
√2 x² + 7x + 5√2 = 0
Roots: −√2 and −5/√2
a=√2, b=7, c=5√2
a×c = √2×5√2 = 10
→
Need p+q=7, p×q=10
p = 2, q = 5
→
Split 7x as
2x + 5x
- 1√2 x² + 2x + 5x + 5√2 = 0
- 2√2 x(x + √2) + 5(x + √2) = 0
- 3(x + √2)(√2 x + 5) = 0
- 4x = −√2 OR x = −5/√2
Roots: x = −√2 and x = −5/√2(iv)
2x² − x + 1/8 = 0 (multiply by 8 → 16x² − 8x + 1 = 0)
Roots: 1/4 and 1/4 (equal roots)
💡 Clearing the fraction first: Multiply entire equation by 8 to get integer coefficients: 16x² − 8x + 1 = 0
a=16, b=−8, c=1
a×c = 16
→
Need p+q=−8, p×q=16
p = −4, q = −4
→
Split −8x as
−4x − 4x
- 116x² − 4x − 4x + 1 = 0
- 24x(4x − 1) − 1(4x − 1) = 0
- 3(4x − 1)(4x − 1) = 0 → (4x−1)² = 0
- 44x = 1 → x = 1/4 (both roots are equal)
Roots: x = 1/4 and x = 1/4 (repeated root)(v)
100x² − 20x + 1 = 0
Roots: 1/10 and 1/10 (equal)
Split −20x as
−10x − 10x
since
(−10)+(−10)=−20
(−10)×(−10)=100
- 1100x² − 10x − 10x + 1 = 0
- 210x(10x − 1) − 1(10x − 1) = 0
- 3(10x − 1)² = 0
- 410x = 1 → x = 1/10 (repeated root)
Roots: x = 1/10 and x = 1/10(vi)
x(x + 4) = 12 → First rearrange!
Roots: −6 and 2
💡 Rearrange first: Expand left and bring 12 to left: x² + 4x − 12 = 0
Need p+q=4, p×q=−12
p = 6, q = −2
→
Split +4x as
+6x − 2x
- 1x² + 6x − 2x − 12 = 0
- 2x(x + 6) − 2(x + 6) = 0
- 3(x + 6)(x − 2) = 0
- 4x = −6 OR x = 2
Roots: x = −6 and x = 2(vii)
3x² − 5x + 2 = 0
Roots: 1 and 2/3
a=3, b=−5, c=2 → a×c=6
p=−3, q=−2
→
−3+(−2)=−5 ✔
−3×(−2)=6 ✔
- 13x² − 3x − 2x + 2 = 0
- 23x(x − 1) − 2(x − 1) = 0
- 3(x − 1)(3x − 2) = 0
- 4x = 1 OR x = 2/3
Roots: x = 1 and x = 2/3(viii)
x − 3/x = 2 → Multiply by x first!
Roots: 3 and −1
💡 Clear the fraction: Multiply both sides by x → x² − 3 = 2x → x² − 2x − 3 = 0
Need p+q=−2, p×q=−3
p = −3, q = +1
→
Split −2x as
−3x + x
- 1x² − 3x + x − 3 = 0
- 2x(x − 3) + 1(x − 3) = 0
- 3(x − 3)(x + 1) = 0
- 4x = 3 OR x = −1
Roots: x = 3 and x = −1(ix)
3(x−4)² − 5(x−4) − 12 = 0 [Substitution method]
Roots: 7 and 8/3
💡 Substitution shortcut: Let a = (x−4). The equation becomes 3a² − 5a − 12 = 0 — much simpler to factor!
- 1Substitute a = x−4: 3a² − 5a − 12 = 0
- 2Split: p+q=−5, p×q=−36 → p=−9, q=4 → 3a² − 9a + 4a − 12 = 0
- 33a(a − 3) + 4(a − 3) = 0 → (a−3)(3a+4) = 0
- 4a = 3 OR a = −4/3
- 5Back-substitute: x−4 = 3 → x = 7 and x−4 = −4/3 → x = 8/3
Roots: x = 7 and x = 8/3
💡
Key Pattern to Spot: Problems (iv) and (v) give repeated roots (both roots equal) — the quadratic is a perfect square. Problems (iii), (viii), and (ix) need extra steps first (clearing surds, clearing fractions, substitution). Always write the equation in standard form before applying the method!
Advertisement
Questions 2–10 — Word Problems Solved Step-by-Step
Q2 — Two Numbers: Sum = 27, Product = 182
Find two numbers whose sum is 27 and product is 182.
Let the first number = x → second number = 27 − x
Product: x(27 − x) = 182 → 27x − x² = 182
Rearrange: x² − 27x + 182 = 0
Split: p+q=−27, p×q=182 → p=−14, q=−13
(x − 14)(x − 13) = 0
(x − 14)(x − 13) = 0
x = 14 or x = 13 → either way, the two numbers are 13 and 14
Answer: The two numbers are 13 and 14Q3 — Consecutive Integers: Sum of Squares = 613
Find two consecutive positive integers whose sum of squares is 613.
Let the integers be x and x+1
x² + (x+1)² = 613 → x² + x² + 2x + 1 = 613
2x² + 2x − 612 = 0 → divide by 2 → x² + x − 306 = 0
Split: p+q=1, p×q=−306 → p=18, q=−17
(x + 18)(x − 17) = 0
(x + 18)(x − 17) = 0
x = −18 (rejected — must be positive) or x = 17
Answer: The consecutive integers are 17 and 18Q4 — Right Triangle: Altitude 7 cm Less than Base, Hypotenuse = 13 cm
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the base and altitude.
Let base = x cm → altitude = x − 7 cm
Pythagoras: x² + (x−7)² = 13²
x² + x² − 14x + 49 = 169 → 2x² − 14x − 120 = 0 → divide by 2 → x² − 7x − 60 = 0
Split: p+q=−7, p×q=−60 → p=−12, q=5
(x − 12)(x + 5) = 0
(x − 12)(x + 5) = 0
x = −5 (rejected) or x = 12 cm. Altitude = 12 − 7 = 5 cm
Answer: Base = 12 cm, Altitude = 5 cmQ5 — Cottage Industry: Total Cost Rs. 90, Cost per Article = 2x + 3
Cost per article = Rs. (3 more than twice the number produced). Total cost = Rs. 90. Find number of articles and cost each.
Let articles produced = x → cost per article = Rs. (2x + 3)
x(2x + 3) = 90 → 2x² + 3x − 90 = 0
Split: p+q=3, p×q=−180 → p=15, q=−12
2x(x − 6) + 15(x − 6) = 0 → (x−6)(2x+15) = 0
2x(x − 6) + 15(x − 6) = 0 → (x−6)(2x+15) = 0
x = −15/2 (rejected) or x = 6. Cost per article = 2(6)+3 = Rs. 15
Answer: 6 articles produced, each costing Rs. 15Q6 — Rectangle: Perimeter = 28 m, Area = 40 m²
Find the dimensions of a rectangle with perimeter 28 m and area 40 m².
Let length = x m. Perimeter: 2(l+b) = 28 → l+b = 14 → breadth = 14−x
Area: x(14−x) = 40 → 14x − x² = 40 → x² − 14x + 40 = 0
Split: p+q=−14, p×q=40 → p=−10, q=−4
(x − 10)(x − 4) = 0
(x − 10)(x − 4) = 0
x = 10 → breadth = 4 | x = 4 → breadth = 10 (same rectangle)
Answer: Dimensions are 10 m × 4 mQ7 — Triangle: Base = Altitude + 4 cm, Area = 48 cm²
The base of a triangle is 4 cm longer than its altitude. Area = 48 cm². Find base and altitude.
Let altitude = x cm → base = x + 4 cm
Area = ½ × base × altitude: ½ × (x+4) × x = 48 → x(x+4) = 96
x² + 4x − 96 = 0
Split: p+q=4, p×q=−96 → p=12, q=−8
(x + 12)(x − 8) = 0
(x + 12)(x − 8) = 0
x = −12 (rejected) or x = 8 cm. Base = 8 + 4 = 12 cm
Answer: Altitude = 8 cm, Base = 12 cmQ8 — Two Trains (West + North), 50 km Apart After 2 Hours
Two trains leave a station simultaneously — one west, one north. First train is 5 km/h faster. After 2 hours they are 50 km apart. Find each train's speed.
| Train | Speed | Time | Distance |
|---|---|---|---|
| First (West) | x+5 km/h | 2 h | 2(x+5) km |
| Second (North) | x km/h | 2 h | 2x km |
Trains travel perpendicular (west & north) → use Pythagoras:
[2(x+5)]² + [2x]² = 50²
[2(x+5)]² + [2x]² = 50²
4(x+5)² + 4x² = 2500 → 4x²+40x+100+4x² = 2500 → 8x²+40x−2400 = 0
Divide by 8: x² + 5x − 300 = 0
Split: p+q=5, p×q=−300 → p=20, q=−15
(x+20)(x−15) = 0
(x+20)(x−15) = 0
x = −20 (rejected) or x = 15. Second train = 15 km/h, First = 20 km/h
Answer: First train = 20 km/h, Second train = 15 km/hQ9 — 60 Students: Boys + Girls Contribute Rs. 1600
In a class of 60 students, each boy contributed ₹ equal to the number of girls, and each girl contributed ₹ equal to the number of boys. Total collected = ₹ 1600. Find the number of boys.
Let boys = x → girls = 60 − x
Boys' total: x(60−x) Girls' total: (60−x)·x
Both contributions are the same expression!
Both contributions are the same expression!
x(60−x) + (60−x)x = 1600 → 2x(60−x) = 1600 → x(60−x) = 800
60x − x² = 800 → x² − 60x + 800 = 0
Split: p+q=−60, p×q=800 → p=−20, q=−40
(x−20)(x−40) = 0
(x−20)(x−40) = 0
x = 20 (20 boys, 40 girls) OR x = 40 (40 boys, 20 girls)
Answer: Number of boys is 20 or 40Q10 — Motorboat: 24 km Upstream + 24 km Downstream in 6 Hours
A motorboat travels 24 km upstream in a river with 3 km/h current. The round trip takes 6 hours. Find the boat's speed in still water.
| Direction | Speed | Distance | Time |
|---|---|---|---|
| Upstream | (x − 3) km/h | 24 km | 24/(x−3) hrs |
| Downstream | (x + 3) km/h | 24 km | 24/(x+3) hrs |
Total time = 6: 24/(x−3) + 24/(x+3) = 6
Factor 24: 24[(x+3 + x−3) / (x²−9)] = 6 → 24 × 2x / (x²−9) = 6
Cross-multiply: 48x = 6(x²−9) → x² − 8x − 9 = 0
Split: p+q=−8, p×q=−9 → p=−9, q=1
(x−9)(x+1) = 0
(x−9)(x+1) = 0
x = −1 (rejected) or x = 9 km/h
Answer: Speed of boat in still water = 9 km/hQuick Reference — All Answers
| # | Equation / Problem | Roots / Answer |
|---|---|---|
| Q1(i) | x² − 3x − 10 = 0 | x = 5, x = −2 |
| Q1(ii) | 2x² + x − 6 = 0 | x = −2, x = 3/2 |
| Q1(iii) | √2 x² + 7x + 5√2 = 0 | x = −√2, x = −5/√2 |
| Q1(iv) | 2x² − x + 1/8 = 0 | x = 1/4, x = 1/4 (equal) |
| Q1(v) | 100x² − 20x + 1 = 0 | x = 1/10, x = 1/10 (equal) |
| Q1(vi) | x(x+4) = 12 | x = −6, x = 2 |
| Q1(vii) | 3x² − 5x + 2 = 0 | x = 1, x = 2/3 |
| Q1(viii) | x − 3/x = 2 | x = 3, x = −1 |
| Q1(ix) | 3(x−4)² − 5(x−4) − 12 = 0 | x = 7, x = 8/3 |
| Q2 | Sum=27, Product=182 | Numbers: 13 and 14 |
| Q3 | Sum of squares = 613 | Integers: 17 and 18 |
| Q4 | Right triangle, hyp=13 cm | Base=12 cm, Alt=5 cm |
| Q5 | Pottery, cost=Rs.90 | 6 articles, Rs.15 each |
| Q6 | Rectangle, P=28 m, A=40 m² | 10 m × 4 m |
| Q7 | Triangle, area=48 cm² | Base=12 cm, Alt=8 cm |
| Q8 | Two trains, 50 km apart | 20 km/h and 15 km/h |
| Q9 | 60 students, Rs.1600 | 20 or 40 boys |
| Q10 | Motorboat round trip | 9 km/h in still water |
Common Mistakes to Avoid
Mistake 1
Splitting the middle term incorrectly — always verify: p + q = b AND p × q = a×c. Both conditions must hold.
Mistake 2
Forgetting to rearrange into ax² + bx + c = 0 first. Problems like x(x+4)=12 must be expanded and rearranged before factorising.
Mistake 3
Accepting negative values for physical quantities — length, speed, number of articles, and number of people cannot be negative.
Mistake 4
In train/boat problems, confusing upstream (speed decreases) and downstream (speed increases) when writing the speed expressions.
- Repeated roots appear when (4x−1)² = 0 or (10x−1)² = 0 — the quadratic is a perfect square trinomial.
- Substitution method (Q1-ix) saves time for expressions like (x−4)² — replace the bracket with a single variable first.
- Pythagoras applies when two directions are perpendicular — right triangles and train-direction problems both use a² + b² = c².
- Time = Distance ÷ Speed is the formula for all speed/distance word problems in this exercise.