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Class 10 Mathematics Exercise 4.1 NCERT Solutions Olympiad Board Exam

Chapter 4 — QUADRATIC EQUATIONS

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

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Q1
NUMERIC3 marks

Check whether the following are quadratic equations :

Concept Before Solving

A quadratic equation in one variable is of the form:

$$ax^2 + bx + c = 0,\quad a \ne 0$$

where highest power of variable is 2.

Identification Rule
  • Bring all terms to one side
  • Simplify completely
  • Check highest power of \(x\)
  • If degree = 2 → Quadratic
  • If degree ≠ 2 → Not quadratic
Solution Roadmap
  1. Expand brackets carefully
  2. Move all terms to LHS
  3. Combine like terms
  4. Check degree
Visual Insight (Degree Identification)
Quadratic curve (degree = 2)
  1. \((x + 1)^2 = 2(x – 3)\)

    Solution

    $$\begin{aligned} (x+1)^2 &= 2(x-3) \\ x^2 + 2x + 1 &= 2x - 6 \\ x^2 + 2x + 1 - 2x + 6 &= 0 \\ x^2 + 7 &= 0 \end{aligned}$$ Highest power = 2 → Quadratic Equation
  2. \(x^2 – 2x = (–2)(3 – x)\)

    Solution

    $$\begin{aligned} x^2 - 2x &= -2(3 - x) \\ x^2 - 2x &= -6 + 2x \\ x^2 - 2x - 2x + 6 &= 0 \\ x^2 - 4x + 6 &= 0 \end{aligned}$$ Degree = 2 → Quadratic Equation
  3. \((x – 2)(x + 1) = (x – 1)(x + 3)\)

    Solution

    $$\begin{aligned} (x-2)(x+1) &= (x-1)(x+3) \\\ x^2 + x - 2x - 2 &= x^2 + 3x - x - 3 \\\ x^2 - x - 2 &= x^2 + 2x - 3 \\\ x^2 - x - 2 - x^2 - 2x + 3 &= 0 \\\ -3x + 1 &= 0 \end{aligned}$$ Degree = 1 → Not a Quadratic Equation
  4. \((x – 3)(2x +1) = x(x + 5)\)

    Solution

    $$\begin{aligned} (x-3)(2x+1) &= x(x+5) \\ 2x^2 + x - 6x - 3 &= x^2 + 5x \\ 2x^2 - 5x - 3 &= x^2 + 5x \\ 2x^2 - x^2 - 5x - 5x - 3 &= 0 \\ x^2 - 10x - 3 &= 0 \end{aligned}$$ Degree = 2 → Quadratic Equation
  5. \((2x – 1)(x – 3) = (x + 5)(x – 1)\)

    Solution

    $$\begin{aligned} (2x-1)(x-3) &= (x+5)(x-1) \\ 2x^2 - 6x - x + 3 &= x^2 - x + 5x - 5 \\ 2x^2 - 7x + 3 &= x^2 + 4x - 5 \\ 2x^2 - x^2 - 7x - 4x + 3 + 5 &= 0 \\ x^2 - 11x + 8 &= 0 \end{aligned}$$ Degree = 2 → Quadratic Equation
  6. \(x^2 + 3x + 1 = (x – 2)^2\)

    Solution

    $$\begin{aligned} x^2 + 3x + 1 &= (x-2)^2 \\ x^2 + 3x + 1 &= x^2 - 4x + 4 \\ x^2 + 3x + 1 - x^2 + 4x - 4 &= 0 \\ 7x - 3 &= 0 \end{aligned}$$ Degree = 1 → Not a Quadratic Equation
  7. \((x + 2)^3 = 2x (x^2 – 1)\)

    Solution

    $$\begin{aligned} (x+2)^3 &= 2x(x^2 - 1) \\ x^3 + 6x^2 + 12x + 8 &= 2x^3 - 2x \\ x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x &= 0 \\ -x^3 + 6x^2 + 14x + 8 &= 0 \end{aligned}$$ Degree = 3 → Not a Quadratic Equation
  8. \(x^3 – 4x^2 – x + 1 = (x – 2)^3\)

    Solution

    $$\begin{aligned} x^3 - 4x^2 - x + 1 &= (x-2)^3 \\ x^3 - 4x^2 - x + 1 &= x^3 - 6x^2 + 12x - 8 \\ x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 &= 0 \\ 2x^2 - 13x + 9 &= 0 \end{aligned}$$ Degree = 2 → Quadratic Equation

Exam Significance

  • This concept is directly asked in CBSE Board exams (1–2 mark questions)
  • Foundation for solving quadratics using factorisation, formula, graph
  • Frequently tested in NTSE, Olympiads, and JEE foundation level
  • Critical for understanding discriminant and nature of roots
Common Mistakes to Avoid
  • Not bringing all terms to one side
  • Wrong expansion of identities
  • Ignoring cancellation of highest degree terms
  • Misidentifying degree after simplification
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Q2 →
Q2
NUMERIC3 marks

Represent the following situations in the form of quadratic equations :

Concept: Converting Real-Life Problems into Quadratic Equations

Many real-life situations can be translated into quadratic equations using variables.

Standard Modelling Framework
  1. Define variable clearly
  2. Translate words into algebraic expressions
  3. Use given condition (area, product, time etc.)
  4. Form equation
  5. Simplify to standard form \(ax^2 + bx + c = 0\)
Visual Idea (Word → Equation Mapping)
Words Equation Solution

Represent the following situations in the form of quadratic equations :

  1. The area of a rectangular plot is 528 \(\mathrm{m^2}\). The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

    Solution

    Let breadth = \(x\) m
    Length = \(2x + 1\)

    \[\text{Area} = \text{Length} \times \text{Breadth}\] $$\begin{aligned} x(2x+1) &= 528 \\ 2x^2 + x - 528 &= 0 \end{aligned}$$

    Factorisation:

    \[2 \times (-528) = -1056\] \[33 - 32 = 1\] $$\begin{aligned} 2x^2 + 33x - 32x - 528 &= 0 \\ x(2x + 33) -16(2x + 33) &= 0 \\ (2x + 33)(x - 16) &= 0 \end{aligned}$$ \[x = 16 \quad \text{or} \quad x = -\frac{33}{2}\]

    Breadth cannot be negative → \(x = 16\)

    Breadth = 16 m
    Length = \(2(16) + 1 = 33\) m

  2. The product of two consecutive positive integers is 306. We need to find the integers

    Solution

    Let first integer = \(x\)
    Next integer = \(x + 1\)

    $$\begin{aligned} x(x+1) &= 306 \\ x^2 + x - 306 &= 0 \end{aligned}$$ \[306 = 18 \times 17\] $$\begin{aligned} x^2 + 18x - 17x - 306 &= 0 \\ x(x + 18) -17(x + 18) &= 0 \\ (x + 18)(x - 17) &= 0 \end{aligned}$$ \[x = 17 \quad \text{or} \quad x = -18\]

    Only positive values valid:

    First number = 17
    Second number = 18
  3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

    Solution

    Let Rohan’s present age = \(x\)
    Mother’s age = \(x + 26\)

    After 3 years:
    Rohan = \(x + 3\)
    Mother = \(x + 29\)

    $$\begin{aligned} (x+3)(x+29) &= 360 \\ x^2 + 29x + 3x + 87 &= 360 \\ x^2 + 32x - 273 &= 0 \end{aligned}$$ \[273 = 39 \times 7\] $$\begin{aligned} x^2 + 39x - 7x - 273 &= 0 \\ x(x+39) -7(x+39) &= 0 \\ (x+39)(x-7) &= 0 \end{aligned}$$ \[x = 7 \quad \text{or} \quad x = -39\]

    Age cannot be negative → \(x = 7\)

    Rohan’s age = 7 years
  4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

    Solution

    Let speed = \(v\) km/h

    $$\text{Time} = \frac{480}{v}$$

    Reduced speed = \(v - 8\)

    \[\text{New time} = \frac{480}{v - 8}\]

    Given:

    \[\frac{480}{v-8} = \frac{480}{v} + 3\] $$\begin{aligned} \frac{480}{v-8} - \frac{480}{v} &= 3 \\ \frac{480v - 480(v-8)}{v(v-8)} &= 3 \\ \frac{480v - 480v + 3840}{v(v-8)} &= 3 \\ \frac{3840}{v(v-8)} &= 3 \end{aligned}$$ $$\begin{aligned} 3840 &= 3v(v-8) \\ 3840 &= 3v^2 - 24v \\ 3v^2 - 24v - 3840 &= 0 \\ v^2 - 8v - 1280 &= 0 \end{aligned}$$ \[1280 = 40 \times 32\] $$\begin{aligned} v^2 - 40v + 32v - 1280 &= 0 \\ v(v-40) +32(v-40) &= 0 \\ (v-40)(v+32) &= 0 \end{aligned}$$ \[v = 40 \quad \text{or} \quad v = -32\]

    Speed cannot be negative → \(v = 40\)

    Speed of train = 40 km/h

Exam Significance

  • Very frequently asked in CBSE Board (3–4 mark questions)
  • Core application of quadratic equations in real-life modelling
  • Important for NTSE, Olympiads, and foundation of higher algebra
  • Strengthens word → equation conversion (high-weight skill)
Common Mistakes
  • Wrong variable assumption
  • Incorrect translation of statements
  • Errors in fraction handling (especially time problems)
  • Not rejecting negative values in real-life context
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All 2 solutions for QUADRATIC EQUATIONS covered.

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Advanced Concept Framework

Quadratic equations arise in multiple real-world domains such as geometry, motion, number systems, and optimization.

Extended Modelling Strategy
  1. Define variable with unit
  2. Translate relational statements carefully
  3. Use domain formulas:
    • Area → \(l \times b\), \( \pi r^2 \)
    • Speed → \( \frac{distance}{time} \)
    • Numbers → digit representation
  4. Form equation
  5. Simplify → quadratic form
  6. Solve and validate (reject non-physical roots)

Additional Word Problems

  1. The sum of two numbers is 27 and their product is 182.
    Solution
    Let first number = \(x\)
    Second number = \(27 - x\) $$\begin{aligned} x(27-x) &= 182 \\ 27x - x^2 &= 182 \\ x^2 - 27x + 182 &= 0 \end{aligned}$$ $$182 = 13 \times 14$$ $$\begin{aligned} (x-13)(x-14) &= 0 \\ x = 13, 14 \end{aligned}$$
  2. The square of a number exceeds its double by 15.
    Solution
    Let number = \(x\) $$\begin{aligned} x^2 &= 2x + 15 \\ x^2 - 2x - 15 &= 0 \end{aligned}$$ $$15 = 5 \times 3$$ $$\begin{aligned} (x-5)(x+3)=0 \\ x=5, -3 \end{aligned}$$
  3. A rectangular garden length exceeds breadth by 5 m and area is 104 m².
    Solution
    Let breadth = \(x\)
    Length = \(x+5\) $$\begin{aligned} x(x+5)=104 \\ x^2+5x-104=0 \end{aligned}$$ $$104=13 \times 8$$ $$\begin{aligned} (x-8)(x+13)=0 \\ x=8 \end{aligned}$$ Breadth = 8 m, Length = 13 m
  4. Product of two consecutive even numbers is 168.
    Solution
    Let first even = \(x\)
    Next = \(x+2\) $$\begin{aligned} x(x+2)=168 \\ x^2+2x-168=0 \end{aligned}$$ $$168=14 \times 12$$ $$\begin{aligned} (x-12)(x+14)=0 \\ x=12 \end{aligned}$$ Numbers: 12, 14
  5. Sum of squares of two consecutive integers is 85.
    Solution
    Let numbers = \(x, x+1\) $$\begin{aligned} x^2+(x+1)^2=85 \\ x^2+x^2+2x+1=85 \\ 2x^2+2x-84=0 \\ x^2+x-42=0 \end{aligned}$$ $$42=7 \times 6$$ $$\begin{aligned} (x-6)(x+7)=0 \\ x=6 \end{aligned}$$
  6. A number and its reciprocal differ by \( \frac{3}{2} \).
    Solution
    Let number = \(x\) $$\begin{aligned} x - \frac{1}{x} = \frac{3}{2} \\ 2x^2 - 2 = 3x \\ 2x^2 - 3x - 2 = 0 \end{aligned}$$ $$\begin{aligned} (2x+1)(x-2)=0 \\ x=2, -\frac{1}{2} \end{aligned}$$
  7. Area of square equals area of rectangle (length 8 m more than breadth 6 m).
    Solution
    Let square side = \(x\) Rectangle: breadth = \(x-6\), length = \(x+2\) $$\begin{aligned} x^2 = (x-6)(x+2) \\ x^2 = x^2 -4x -12 \\ 4x +12 = 0 \\ x = -3 \end{aligned}$$ No valid physical solution → check modelling (important insight)
  8. Speed increased by 10 km/h reduces time from 6 hrs to 5 hrs.
    Solution
    Distance same $$\begin{aligned} 6v = 5(v+10) \\ 6v = 5v + 50 \\ v = 50 \end{aligned}$$ Linear → not quadratic (important classification insight)
  9. Two-digit number: sum of digits = 10, product = 21.
    Solution
    Let digits = \(x, y\) $$\begin{aligned} x+y=10 \\ xy=21 \end{aligned}$$ Substitute \(y=10-x\) $$\begin{aligned} x(10-x)=21 \\ x^2-10x+21=0 \\ (x-3)(x-7)=0 \end{aligned}$$ Digits: 3, 7 → numbers: 37, 73
  10. A ball is thrown upward with velocity such that height equation is \(h=20t-5t^2\). Find time when height = 15 m.
    Solution
    $$\begin{aligned} 20t - 5t^2 = 15 \\ 5t^2 - 20t + 15 = 0 \\ t^2 - 4t + 3 = 0 \\ (t-1)(t-3)=0 \end{aligned}$$ $$t=1s, 3s$$

Higher-Level Insights

Competitive Edge

What Students Must Master (Quick Revision)

Most Searched Concepts (Exam Focus)

Common Mistakes Students Make

Formula Toolbox

These formulas are repeatedly used in word problems:

\[\text{Area of rectangle} = l \times b\] \[\text{Distance} = \text{Speed} \times \text{Time}\] \[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\] \[\text{Consecutive numbers} = x, x+1, x+2\] \[\text{Even numbers} = x, x+2\]

CBSE Board Question Pattern Insight

Case Study (CBSE Latest Pattern)

A rectangular park has area 600 m². Its length is 5 m more than twice its breadth.

Let breadth = \(x\)
Length = \(2x + 5\)

$$\begin{aligned} x(2x+5) &= 600 \\ 2x^2 + 5x - 600 &= 0 \end{aligned}$$

This type directly appears in CBSE case-study questions.

HOTS (Higher Order Thinking Questions)

  1. A number when increased by its reciprocal gives 5. Form the equation.
  2. The difference of squares of two consecutive numbers is 25.
  3. A train covers same distance in two speeds, find relation equation.

Exam Strategy (High Scoring Approach)

Micro Learning Summary

Word problems in quadratic equations test your ability to translate real-life situations into algebra. Mastery of this skill ensures success in board exams as well as competitive exams like NTSE and Olympiads.

SEO Rich FAQs (Highly Rankable)

Q1. How do you identify a quadratic word problem?
If the situation leads to an equation where highest power of variable is 2.

Q2. Why do we reject negative solutions?
Because quantities like age, length, time cannot be negative.

Q3. Which chapter is most important in Class 10 Maths?
Quadratic equations is one of the highest weightage and most applied chapters.

Q4. How to improve word problem solving?
Practice translating sentences into algebra step by step.

Q5. Are word problems asked in board exams?
Yes, almost every year in 3–4 mark questions.

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