If you’re searching for NCERT solutions class 10 maths chapter 4, you’re at the right place. Quadratic Equations is one of the most scoring chapters in the CBSE Class 10 Maths board exam — it usually carries around 7 marks, and unlike some chapters, the question patterns barely change from year to year. That means mastering this one chapter properly can lock in marks you don’t have to “hope” for on exam day.
At Convex Classes, Jaipur, this is one of the first chapters we cover in depth with our Class 10 batches, because it builds directly into Chapter 5 (Arithmetic Progressions) and comes back again in Class 11. Below, our faculty has solved every single question from Exercise 4.1, 4.2, and 4.3 of the NCERT textbook — with the working shown, not just the final answer, so you actually understand the method behind each solution.
Chat with Convex Classes Jaipur Get Class 10 Maths batch details on WhatsApp Chat Now →What Is a Quadratic Equation? (Quick Recap)
Before jumping into the class 10 maths chapter 4 exercise 4.1 solutions, let’s recap the one definition this entire chapter is built on:
Any equation that can be rearranged into the form ax² + bx + c = 0 (where a ≠ 0) is called a quadratic equation.
The key thing examiners test again and again: expand and simplify fully before deciding whether something is quadratic. An equation can look cubic or linear before simplification and turn out to be exactly ax² + bx + c = 0 afterward — or the other way round.
Once you know a, b, and c, there’s one number that tells you everything about the roots without solving anything further — the discriminant:
D = b² − 4ac
| Value of D | Nature of the roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots (repeated root) |
| D < 0 | No real roots |
Keep this table close — a good chunk of Exercise 4.3, and several board-exam MCQs, are really just this table applied to a specific equation.
NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1
Q1. Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x – 3)
Expanding: x² + 2x + 1 = 2x – 6 → x² + 7 = 0. This fits ax² + bx + c = 0, so it is a quadratic equation.
(ii) x² – 2x = (–2)(3 – x)
Expanding: x² – 2x = –6 + 2x → x² – 4x + 6 = 0. Quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
Expanding both sides, the x² terms cancel out, leaving 3x – 1 = 0. Since there’s no x² term left, this is not a quadratic equation (it’s linear).
(iv) (x – 3)(2x + 1) = x(x + 5)
Expanding: 2x² – 5x – 3 = x² + 5x → x² – 10x – 3 = 0. Quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Expanding: 2x² – 7x + 3 = x² + 4x – 5 → x² – 11x + 8 = 0. Quadratic equation.
(vi) x² + 3x + 1 = (x – 2)²
Expanding the RHS and simplifying, the x² terms cancel, leaving 7x – 3 = 0 — not a quadratic equation.
(vii) (x + 2)³ = 2x(x² – 1)
Expanding fully leaves a genuine x³ term that doesn’t cancel — this is a cubic equation, not quadratic.
(viii) x³ – 4x² – x + 1 = (x – 2)³
Here the x³ terms on both sides cancel each other out, leaving 2x² – 13x + 9 = 0 — which is a quadratic equation, even though the question started out looking cubic. This one trips up a lot of students who stop simplifying too early.
Q2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length is one more than twice its breadth. Find the length and breadth.
Let breadth = x m, so length = (2x + 1) m. Area: x(2x + 1) = 528 → 2x² + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. Find the integers.
Let the first integer = x, so the next is (x + 1). x(x + 1) = 306 → x² + x – 306 = 0
(iii) Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan’s present age.
Let Rohan’s age = x. Mother’s age = x + 26. Three years later: (x + 3)(x + 29) = 360 → x² + 32x – 273 = 0
(iv) A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would take 3 hours more. Find the speed.
Let speed = x km/h. Working through the time relation: (x – 8)(480/x + 3) = 480, which simplifies to x² – 8x – 1280 = 0
NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2
Q1. Find the roots of the following by factorisation:
(i) x² – 3x – 10 = 0
Split the middle term: x² – 5x + 2x – 10 = 0 → x(x – 5) + 2(x – 5) = 0 → (x – 5)(x + 2) = 0 x = 5 or x = –2
(ii) 2x² + x – 6 = 0
2x² + 4x – 3x – 6 = 0 → 2x(x + 2) – 3(x + 2) = 0 → (x + 2)(2x – 3) = 0 x = –2 or x = 3/2
(iii) √2x² + 7x + 5√2 = 0
√2x² + 5x + 2x + 5√2 = 0 → x(√2x + 5) + √2(√2x + 5) = 0 → (√2x + 5)(x + √2) = 0 x = –5/√2 or x = –√2
(iv) 2x² – x + 1/8 = 0
Multiply through by 8: 16x² – 8x + 1 = 0 → this is a perfect square: (4x – 1)² = 0 x = 1/4 (a repeated root)
(v) 100x² – 20x + 1 = 0
This too is a perfect square: (10x – 1)² = 0 x = 1/10 (a repeated root)
Q2. Word problems (marbles and toy production):
(i) John and Jivanti together have 45 marbles. After both lose 5 marbles each, the product of their remaining marbles is 124. Find how many each had.
Let John’s marbles = x, so Jivanti’s = 45 – x. After losing 5 each: (x – 5)(40 – x) = 124 This simplifies to (x – 36)(x – 9) = 0 → x = 36 or x = 9 So either John had 36 and Jivanti had 9, or John had 9 and Jivanti had 36.
(ii) A cottage industry’s cost per toy = ₹(55 – number of toys produced). Total cost on a particular day was ₹750. Find the number of toys.
Let toys produced = x. Then x(55 – x) = 750 → (x – 25)(x – 30) = 0 Number of toys produced was either 25 or 30.
Q3. Find two numbers whose sum is 27 and product is 182.
Let the numbers be x and (27 – x). Then x(27 – x) = 182 → (x – 13)(x – 14) = 0 The two numbers are 13 and 14.
Q4. Find two consecutive positive integers whose squares sum to 365.
Let the integers be x and (x + 1). Then x² + (x + 1)² = 365 → (x + 14)(x – 13) = 0 Since integers must be positive, reject x = –14. The integers are 13 and 14.
Q5. A right triangle’s altitude is 7 cm less than its base; the hypotenuse is 13 cm. Find the other two sides.
Let base = x cm, altitude = (x – 7) cm. By Pythagoras: x² + (x – 7)² = 13² → (x – 12)(x + 5) = 0 A side can’t be negative, so x = 12. Base = 12 cm, altitude = 5 cm.
Q6. A cottage industry produces pottery articles. Cost per article = ₹(2× number produced + 3). Total cost that day was ₹90. Find the number produced and cost per article.
Let articles produced = x. Then x(2x + 3) = 90 → (2x + 15)(x – 6) = 0 Since the count must be a positive integer, x = 6. 6 articles produced, at ₹15 each.
NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3
Q1. Find the nature of the roots, and the real roots where they exist:
(i) 2x² – 3x + 5 = 0 a = 2, b = –3, c = 5 → D = 9 – 40 = –31 < 0
→ No real roots.
(ii) 3x² – 4√3x + 4 = 0 a = 3, b = –4√3, c = 4 → D = 48 – 48 = 0
→ Two equal real roots, x = 4√3 / 6 = 2/√3 (both roots equal).
(iii) 2x² – 6x + 3 = 0 a = 2, b = –6, c = 3 → D = 36 – 24 = 12 > 0
→ Two distinct real roots, found using the quadratic formula: x = (6 ± √12) / 4.
Q2. Find k so that each equation has two equal roots:
(i) 2x² + kx + 3 = 0
For equal roots, D = 0: k² – 24 = 0 → k = ±2√6
(ii) kx(x – 2) + 6 = 0
Rewriting: kx² – 2kx + 6 = 0. Setting D = 0: 4k² – 24k = 0 → k(k – 6) = 0. k = 0 is rejected (the equation would stop being quadratic), so k = 6.
Q3. Can a rectangular mango grove with length = 2 × breadth and area 800 m² be designed? Find the dimensions.
Let breadth = x m, length = 2x m. Area: 2x² = 800 → x² = 400 → x = ±20. Negative length is rejected. Yes, it’s possible — breadth = 20 m, length = 40 m.
Q4. Two friends’ ages sum to 20. Four years ago, the product of their ages was 48. Is this situation possible?
Setting up the equation gives x² – 20x + 112 = 0, with D = 400 – 448 = –48 < 0. Since D is negative, there are no real roots — this situation is not possible. (A good example of using the discriminant to answer a “possible or not” question without solving for x at all.)
Q5. Can a rectangular park with perimeter 80 m and area 400 m² be designed? Find the dimensions.
l + b = 40, and l(40 – l) = 400 gives l² – 40l + 400 = 0, with D = 1600 – 1600 = 0. Since D = 0, this situation is possible — and because the roots are equal, the park turns out to be a square: length = breadth = 20 m.
Frequently Asked Questions — Class 10 Maths Chapter 4
Q. What is the weightage of Quadratic Equations in the CBSE Class 10 board exam?
This chapter typically carries around 6–7 marks and is considered a scoring chapter because the question patterns are fairly predictable year to year.
Q. What is the easiest method to solve a quadratic equation?
Factorisation (splitting the middle term) is fastest when the roots are simple integers or fractions. When factorisation isn’t obvious, the quadratic formula always works, even for irrational roots.
Q. Can a quadratic equation have only one root?
Technically, every quadratic equation has two roots. When the discriminant equals zero, both roots are equal in value — so it’s common to say the equation has “one repeated root,” but there are still two roots mathematically.
Q. What’s the most common mistake students make in Exercise 4.2?
Getting the sign wrong while splitting the middle term. Always double-check that your two numbers multiply to give exactly a × c, signs included — not just b.
Q. How is the discriminant useful outside of finding roots?
It lets you answer “is this situation even possible?” word problems (like Q4 and Q5 in Exercise 4.3) without solving the full equation — just check the sign of D.
Convex Classes Jaipur — How to Score Full Marks in This Chapter
- Fully expand before you judge. Several questions in Exercise 4.1 are designed to look cubic or linear until you simplify completely — don’t decide “quadratic or not” halfway through.
- Learn the discriminant table by heart. D > 0, D = 0, D < 0 — this single table answers a large share of Exercise 4.3 and recurring board MCQs.
- Reject the impossible root. In almost every word problem here (ages, sides, counts), one algebraic root has to be thrown out because it’s negative or non-physical. Examiners specifically check whether you did this.
- Drill middle-term splitting daily until it’s instant — slow factorisation is the biggest reason students run out of time on this chapter in the exam.
- Know the quadratic formula without hesitation: x = (–b ± √(b² – 4ac)) / 2a — your fallback whenever factorisation isn’t obvious.
If you’d like guided practice on this chapter with live doubt-solving, our faculty at Convex Classes, Jaipur runs regular chapter tests and one-on-one sessions for Class 10 CBSE Maths students
