Exercise 5.1 — Formation
Formation of quadratic equations.
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What This Exercise Covers
📖 Exercise Overview
Exercise 5.1 has two parts. Question 1 asks you to expand and simplify 8 given equations, then decide whether each is quadratic or not based on the degree of the simplified result. Question 2 asks you to translate 4 real-world word problems into quadratic equation form — a skill tested heavily in board exams.
| Problem | Given Expression | Simplified To | Result |
|---|---|---|---|
| (i) | (x+1)² = 2(x−3) | x² + 7 = 0 | ✔ Quadratic |
| (ii) | x²−2x = −2(3−x) | x²−4x+6 = 0 | ✔ Quadratic |
| (iii) | (x−2)(x+1) = (x−1)(x+3) | −3x+1 = 0 | ✗ Not Quadratic |
| (iv) | (x−3)(2x+1) = x(x+5) | x²−10x−3 = 0 | ✔ Quadratic |
| (v) | (2x−1)(x−3) = (x+5)(x−1) | x²−11x+8 = 0 | ✔ Quadratic |
| (vi) | x²+3x+1 = (x−2)² | 7x−3 = 0 | ✗ Not Quadratic |
| (vii) | (x+2)³ = 2x(x²−1) | −x³+6x²+14x+8 = 0 | ✗ Not Quadratic |
| (viii) | x³−4x²−x+1 = (x−2)³ | 2x²−13x+9 = 0 | ✔ Quadratic |
Question 1 — Step-by-Step Solutions
The strategy for every problem is the same: expand both sides → bring all terms to one side → find the degree → conclude.
(i)
(x + 1)² = 2(x − 3)
✔ Quadratic
- 1Expand the left side using (a+b)² = a²+2ab+b²:
x² + 2x + 1 = 2x − 6 - 2Move all terms to the left side:
x² + 2x + 1 − 2x + 6 = 0 - 3Simplify (the 2x terms cancel!):
x² + 7 = 0
x² + 7 = 0 → a=1, b=0, c=7✔ Degree = 2 → Quadratic Equation
(ii)
x² − 2x = (−2)(3 − x)
✔ Quadratic
- 1Expand the right side:
x² − 2x = −6 + 2x - 2Move all terms to the left:
x² − 2x + 6 − 2x = 0 - 3Simplify:
x² − 4x + 6 = 0
x² − 4x + 6 = 0 → a=1, b=−4, c=6✔ Degree = 2 → Quadratic Equation
(iii)
(x − 2)(x + 1) = (x − 1)(x + 3)
✗ Not Quadratic
- 1Expand left side:
x² + x − 2x − 2 = x² − x − 2 - 2Expand right side:
x² + 3x − x − 3 = x² + 2x − 3 - 3Subtract right from left — the x² terms cancel!
x² − x − 2 − x² − 2x + 3 = 0 - 4Simplify:
−3x + 1 = 0
−3x + 1 = 0✗ Degree = 1 → Linear Equation, NOT Quadratic. The x² terms cancelled each other out!
(iv)
(x − 3)(2x + 1) = x(x + 5)
✔ Quadratic
- 1Expand left side:
2x² + x − 6x − 3 = 2x² − 5x − 3 - 2Expand right side:
x² + 5x - 3Bring all to left:
2x² − 5x − 3 − x² − 5x = 0 - 4Simplify:
x² − 10x − 3 = 0
x² − 10x − 3 = 0 → a=1, b=−10, c=−3✔ Degree = 2 → Quadratic Equation
(v)
(2x − 1)(x − 3) = (x + 5)(x − 1)
✔ Quadratic
- 1Expand left side:
2x² − 6x − x + 3 = 2x² − 7x + 3 - 2Expand right side:
x² − x + 5x − 5 = x² + 4x − 5 - 3Bring all to left:
2x² − 7x + 3 − x² − 4x + 5 = 0 - 4Simplify:
x² − 11x + 8 = 0
x² − 11x + 8 = 0 → a=1, b=−11, c=8✔ Degree = 2 → Quadratic Equation
(vi)
x² + 3x + 1 = (x − 2)²
✗ Not Quadratic
- 1Expand the right side using (a−b)²:
x² + 3x + 1 = x² − 4x + 4 - 2Bring all terms to the left — x² cancels:
x² + 3x + 1 − x² + 4x − 4 = 0 - 3Simplify:
7x − 3 = 0
7x − 3 = 0✗ Degree = 1 → Linear Equation, NOT Quadratic. Both x² terms cancelled.
(vii)
(x + 2)³ = 2x(x² − 1)
✗ Not Quadratic
- 1Expand left using (a+b)³ = a³+3a²b+3ab²+b³:
x³ + 3x²(2) + 3x(4) + 8 = x³ + 6x² + 12x + 8 - 2Expand right side:
2x³ − 2x - 3Bring all to left:
x³ + 6x² + 12x + 8 − 2x³ + 2x = 0 - 4Simplify:
−x³ + 6x² + 14x + 8 = 0
−x³ + 6x² + 14x + 8 = 0✗ Degree = 3 → Cubic Equation, NOT Quadratic
(viii)
x³ − 4x² − x + 1 = (x − 2)³
✔ Quadratic
- 1Expand (x−2)³ using (a−b)³ = a³−3a²b+3ab²−b³:
(x−2)³ = x³ − 6x² + 12x − 8 - 2Bring all terms to the left — x³ cancels!
x³ − 4x² − x + 1 − x³ + 6x² − 12x + 8 = 0 - 3Simplify:
2x² − 13x + 9 = 0
2x² − 13x + 9 = 0 → a=2, b=−13, c=9✔ Degree = 2 → Quadratic Equation (x³ cancelled — a surprise result!)
Board Exam Tip: Problems (iii), (vi), and (viii) are the tricky ones — they look complicated but x² or x³ terms cancel during simplification. Always fully expand and simplify before concluding. Never judge by how the equation looks at first glance!
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Question 2 — Forming Quadratic Equations from Word Problems
This section tests your ability to read a real-world situation, assign a variable, form expressions, and write a quadratic equation. This is one of the most important skills for board exams in CBSE, Telangana, and Andhra Pradesh.
Problem (i) — Rectangular Plot (Area = 528 m²)
The area of a rectangular plot is 528 m². The length of the plot is one metre more than twice its breadth. Find the quadratic equation for the breadth.
Length = 2x + 1
Breadth = x
Area = 528 m²
Let the breadth = x metres
Then length = 2x + 1 metres (one more than twice breadth)
Then length = 2x + 1 metres (one more than twice breadth)
Area of rectangle = length × breadth
(2x + 1) × x = 528
(2x + 1) × x = 528
Expand: 2x² + x = 528
2x² + x − 528 = 0Problem (ii) — Consecutive Positive Integers (Product = 306)
The product of two consecutive positive integers is 306. Find the quadratic equation for the integers.
Let the two consecutive positive integers be x and x + 1
Product = 306:
x(x + 1) = 306
x(x + 1) = 306
Expand: x² + x = 306
x² + x − 306 = 0Problem (iii) — Rohan's Age Problem
Rohan's mother is 26 years older than him. The product of their ages after 3 years will be 360. Find the quadratic equation for Rohan's present age.
Let Rohan's present age = x years
Then his mother's present age = x + 26 years
Then his mother's present age = x + 26 years
After 3 years: Rohan = x + 3, Mother = x + 29
Product of their ages after 3 years = 360:
(x + 3)(x + 29) = 360
(x + 3)(x + 29) = 360
Expand:
x² + 29x + 3x + 87 = 360
x² + 32x + 87 − 360 = 0
x² + 29x + 3x + 87 = 360
x² + 32x + 87 − 360 = 0
x² + 32x − 273 = 0Problem (iv) — Train Speed Problem (Distance = 480 km)
A train travels 480 km at uniform speed. If the speed were 8 km/h less, it would take 3 hours more. Find the quadratic equation for the train's speed.
| Scenario | Speed | Distance | Time = Distance ÷ Speed |
|---|---|---|---|
| Actual | x km/h | 480 km | 480/x hours |
| Reduced speed | (x−8) km/h | 480 km | 480/(x−8) hours |
The reduced-speed journey takes 3 hours more:
480/(x−8) − 480/x = 3
480/(x−8) − 480/x = 3
Factor out 480:
480 × [1/(x−8) − 1/x] = 3
480 × [1/(x−8) − 1/x] = 3
Combine fractions:
480 × [x − (x−8)] / [x(x−8)] = 3
480 × 8 / (x²−8x) = 3
480 × [x − (x−8)] / [x(x−8)] = 3
480 × 8 / (x²−8x) = 3
Cross-multiply:
480 × 8 = 3(x² − 8x)
3840 = 3x² − 24x
x² − 8x = 1280
480 × 8 = 3(x² − 8x)
3840 = 3x² − 24x
x² − 8x = 1280
x² − 8x − 1280 = 0Summary — All Answers at a Glance
| Problem | Given | Standard Form | Verdict |
|---|---|---|---|
| (i) | (x+1)² = 2(x−3) | x² + 7 = 0 | ✔ Quadratic |
| (ii) | x²−2x = −2(3−x) | x²−4x+6 = 0 | ✔ Quadratic |
| (iii) | (x−2)(x+1)=(x−1)(x+3) | −3x+1 = 0 | ✗ Not Quadratic |
| (iv) | (x−3)(2x+1)=x(x+5) | x²−10x−3 = 0 | ✔ Quadratic |
| (v) | (2x−1)(x−3)=(x+5)(x−1) | x²−11x+8 = 0 | ✔ Quadratic |
| (vi) | x²+3x+1=(x−2)² | 7x−3 = 0 | ✗ Not Quadratic |
| (vii) | (x+2)³=2x(x²−1) | −x³+6x²+14x+8 = 0 | ✗ Not Quadratic |
| (viii) | x³−4x²−x+1=(x−2)³ | 2x²−13x+9 = 0 | ✔ Quadratic |
| Q2(i) | Rectangle, area=528 m² | 2x²+x−528 = 0 | ✔ Formed |
| Q2(ii) | Consecutive integers, product=306 | x²+x−306 = 0 | ✔ Formed |
| Q2(iii) | Rohan's age problem | x²+32x−273 = 0 | ✔ Formed |
| Q2(iv) | Train speed, 480 km | x²−8x−1280 = 0 | ✔ Formed |
Common Mistakes in Exercise 5.1
Mistake 1
Not expanding both sides fully before simplifying. Many students skip steps and make sign errors during expansion.
Mistake 2
Assuming (iii) and (vi) are quadratic just because they start with x² terms — always check after simplification!
Mistake 3
In word problems, forgetting to add 3 to both ages in problem (iii) — Rohan's mother becomes x+26+3 = x+29, not x+26.
Mistake 4
In the train problem, setting up the fraction the wrong way (subtracting actual from reduced instead of reduced minus actual).
- Always simplify fully before declaring an equation quadratic — degree can change after cancellation.
- In word problems, clearly define your variable before writing any equation.
- Use Time = Distance ÷ Speed for all speed-related word problems.
- Check your standard form — make sure all terms are on one side and the right side is 0.
- For age problems — write present ages first, then future/past ages from those.