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NCERT Class 10 Maths Chapter 4, Quadratic Equations, explained simply

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Quadratic Equations is Chapter 4 of the NCERT Class 10 Mathematics textbook (Reprint 2026-27), printed pages 38 to 47. It teaches one kind of equation, the one with an x², and two ways to solve it: by factorisation and by the quadratic formula. It also shows how one number, the discriminant, tells you whether real roots exist before you solve. This guide explains each idea in plain words, works through the question types the CBSE sample paper uses, walks through Exercises 4.1, 4.2 and 4.3, and ends with a self-check and a way to test an AI's answer.

Keep the textbook open next to it. Our explanations and examples are our own; the textbook is the authority.

How much it counts in the board exam

CBSE's Class 10 Mathematics curriculum for 2026-27 puts this chapter in Unit II, Algebra, worth 20 of the 80 theory marks. The unit has four chapters (Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations and Arithmetic Progressions), and the curriculum does not split the 20 marks between them. For this chapter it lists:

The 2026-27 sample paper (Mathematics Standard) asks the chapter in two places:

QuestionMarksWhat it asks
Q41A multiple-choice question: a given expression in p is a root of a quadratic; find both roots
Q325A rectangular park with a swimming pool in the middle and a grass strip of equal width all round; given the strip's area, find the pool's length and breadth

That is 6 marks in this paper, and one of only four 5-mark questions. Both question types are worked below.

1. What is a quadratic equation? (section 4.2)

A quadratic equation in x is one that can be written as ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. Written with the powers going down (x², then x, then the number), that is its standard form. The rule a ≠ 0 matters: without the x² term it is just a linear equation.

The book's warning is the useful part: simplify before you decide. An equation can look quadratic and not be one, or look cubic and turn out quadratic. Two of our own:

Turning words into an equation

Exercise 4.1 also asks you to write situations as quadratic equations, without solving them yet. The method: name the unknown x, write every other quantity in terms of x, then use the one fact the question gives.

Worked, our own. The product of two consecutive even numbers is 168. Let the smaller be x; the next even number is x + 2. So x(x + 2) = 168, which in standard form is x² + 2x − 168 = 0. (Solving it later gives x = 12 or x = −14, so the numbers are 12 and 14, or −14 and −12. Both pairs multiply to 168.)

2. Roots, and solving by factorisation (section 4.3)

A number α is a root of ax² + bx + c = 0 if putting x = α makes the left side exactly 0. The roots of the equation are the same as the zeroes of the polynomial ax² + bx + c from Chapter 2, so a quadratic equation has at most two roots.

To solve by factorisation, split the middle term, as you learnt in Class 9:

  1. Multiply a and c.
  2. Find two numbers whose product is a × c and whose sum is b.
  3. Use them to split bx into two terms, group, and take out the common factor.
  4. Set each factor equal to 0.

Worked, our own: 3x² − 10x + 8 = 0. Here a × c = 24, and the two numbers that multiply to 24 and add to −10 are −4 and −6.

3x² − 4x − 6x + 8 = x(3x − 4) − 2(3x − 4) = (x − 2)(3x − 4)

So x − 2 = 0 or 3x − 4 = 0, and the roots are x = 2 and x = 4/3. Check x = 2: 12 − 20 + 8 = 0.

When both roots are the same. 4x² − 12x + 9 = (2x − 3)(2x − 3). Both factors give x = 3/2, so the equation has two equal roots, 3/2 and 3/2. The book writes a repeated root twice, once for each factor.

3. The quadratic formula and the discriminant (section 4.4)

When b² − 4ac ≥ 0, the roots of ax² + bx + c = 0 are

x = (−b ± √(b² − 4ac)) / 2a

The current book states this formula at the start of section 4.4 and uses it; it does not derive it. Its introduction adds the history: Sridharacharya (C.E. 1025) derived the formula by completing the square, as quoted by Bhaskara II, and Brahmagupta had earlier given an explicit formula for one form of the equation.

The quantity under the square root, b² − 4ac, is the discriminant. It decides what kind of roots you get:

Discriminant b² − 4acNature of the roots
greater than 0two distinct real roots
equal to 0two equal real roots, both −b/2a
less than 0no real roots

If it is negative, there is no real number whose square it could be, so stop and write "no real roots". The curriculum asks for real roots only.

Worked, our own: x² − 4x + 1 = 0. a = 1, b = −4, c = 1. The discriminant is 16 − 4 = 12, which is positive, so there are two distinct real roots: x = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 + √3 and 2 − √3. This one does not factorise with whole numbers, which is when the formula earns its place.

Worked, our own: 3x² + 2x + 1 = 0. The discriminant is 4 − 12 = −8, which is negative: no real roots.

The "find k" question. For what values of k does x² + kx + 16 = 0 have two equal roots? Set the discriminant to 0: k² − 64 = 0, so k = 8 or k = −8. Check k = 8: x² + 8x + 16 = (x + 4)², a repeated root.

4. Word problems: the 5-mark question

In the 2026-27 sample paper, this chapter's 5-mark question is a story (Q32). Work it in the same six steps every time:

  1. Say what x stands for, with its unit.
  2. Write the other quantities in terms of x.
  3. Form the equation from the one fact given.
  4. Bring it to standard form and simplify (divide through by any common number).
  5. Solve, by factorisation or the formula.
  6. Reject a root that cannot happen, and say why (a negative length, a negative speed, a side that comes out negative). Then answer the question that was asked, with units, and check.

Worked, our own (the Q32 type, turned inside out). A photo 20 cm by 15 cm is put on a card that leaves a border of the same width, x cm, all round. The card's area is 500 cm². Find x.

Worked, our own (the speed type, like Exercise 4.1). A bus covers 300 km at a steady speed. Had it gone 10 km/h faster, it would have taken 1 hour less. Find its speed.

The 1-mark "a root is given" question (the Q4 type)

If a number is a root, it satisfies the equation, so put it in for x. That gives an equation for the unknown coefficient; then solve the quadratic.

Worked, our own. x = 2 is a root of x² + kx − 10 = 0. Find k and the other root. Put x = 2: 4 + 2k − 10 = 0, so k = 3. The equation is x² + 3x − 10 = 0, that is (x + 5)(x − 2) = 0, so the other root is −5.

How the marks are given in Q32

CBSE's marking scheme for the sample paper's pool question splits the 5 marks like this:

So the final answer is worth 1 of the 5 marks. Most of the marks are for the working, and the 1 mark for x = 8 comes with saying which root you rejected and why.

5. Exercises 4.1, 4.2 and 4.3: what each asks, with one worked item

ExercisePrinted page (PDF page)What it asks
4.141-42 (4-5)Q1: decide whether 8 equations are quadratic. Q2: write 4 situations as quadratic equations
4.244 (7)Q1: solve 5 equations by factorisation. Q2: finish Example 1. Q3-Q6: word problems (two numbers, consecutive integers, a right triangle, a cottage industry)
4.347 (10)Q1: nature of the roots, and the roots if real. Q2: values of k for equal roots. Q3-Q5: "is this situation possible?"

Exercise 4.1, Q1 (iii): is (x − 2)(x + 1) = (x − 1)(x + 3) quadratic? Expand both sides: x² − x − 2 = x² + 2x − 3. The x² terms cancel, leaving −3x + 1 = 0. Not quadratic.

Exercise 4.2, Q5: a right triangle whose altitude is 7 cm less than its base, hypotenuse 13 cm. Let the base be x cm; the altitude is x − 7. By Pythagoras, x² + (x − 7)² = 13² = 169, so 2x² − 14x − 120 = 0, that is x² − 7x − 60 = 0, so (x − 12)(x + 5) = 0. Reject −5. Base 12 cm, altitude 5 cm. Check: 144 + 25 = 169.

Exercise 4.3, Q2 (ii): find k so that kx(x − 2) + 6 = 0 has two equal roots. Standard form: kx² − 2kx + 6 = 0. Discriminant: (−2k)² − 4 × k × 6 = 4k² − 24k = 4k(k − 6). Setting it to 0 gives k = 0 or k = 6. But k = 0 is not allowed: then there is no x² term and the equation is not quadratic (it becomes 6 = 0). So k = 6, and the equation is 6(x − 1)² = 0, with the repeated root 1.

Exercise 4.3, Q5: a rectangular park with perimeter 80 m and area 400 m²? Length + breadth = 40, so if the length is x, the breadth is 40 − x, and x(40 − x) = 400, that is x² − 40x + 400 = 0. The discriminant is 1600 − 1600 = 0: two equal roots, x = 20. Yes, it is possible: the park is a 20 m by 20 m square.

6. Five things that cost marks

  1. Dividing both sides by x. From x² = 3x, dividing by x gives only x = 3 and loses x = 0. Bring everything to one side instead: x² − 3x = 0, so x(x − 3) = 0, and x = 0 or 3.
  2. Not reaching standard form first. The discriminant needs the right a, b and c, signs included. In 2x² − 7x + 3 = 0, b is −7, not 7.
  3. Forgetting the minus in −b. The formula starts with −b. With b = −4, −b is +4.
  4. Keeping k = 0 in a "find k" question when k is the coefficient of x². The equation must stay quadratic.
  5. Not rejecting a root, or rejecting it without a reason. "x = 37 is rejected because the pool's sides would be negative" is a sentence the marking scheme pays for.

7. Self-check: 6 questions with answers

Try them on paper first. Answers are below.

  1. Is (x − 1)² = x² + 4 a quadratic equation?
  2. Solve 2x² − 7x + 3 = 0 by factorisation.
  3. What is the nature of the roots of 5x² − 3x + 2 = 0?
  4. For what value of k does x² − 10x + k = 0 have two equal roots?
  5. Solve x² − 6x + 4 = 0 by the quadratic formula.
  6. A girl is 5 years older than her brother, and the product of their ages is 84. Find their ages.

Answers: 1) No. It becomes x² − 2x + 1 = x² + 4, so −2x − 3 = 0, which is linear. 2) a × c = 6; −1 and −6 multiply to 6 and add to −7. 2x² − x − 6x + 3 = x(2x − 1) − 3(2x − 1) = (x − 3)(2x − 1), so x = 3 or x = 1/2. 3) Discriminant 9 − 40 = −31, negative: no real roots. 4) 100 − 4k = 0, so k = 25 (the equation is then (x − 5)² = 0). 5) Discriminant 36 − 16 = 20, so x = (6 ± √20)/2 = (6 ± 2√5)/2 = 3 + √5 or 3 − √5. 6) Let the brother be x years old: x(x + 5) = 84, so x² + 5x − 84 = 0, that is (x + 12)(x − 7) = 0. Reject −12. The brother is 7 and the girl is 12 (7 × 12 = 84).

8. How to check an AI's answer for this chapter

Chatbots solve quadratics quickly and still slip, usually on a sign. These checks take a minute:

More on using AI well: how to use ChatGPT for homework without cheating yourself.

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Sources

All checked on 21 October 2026.

The definitions, the formula, the section and exercise numbers, the page numbers and the board question types are from the documents above. All explanations, worked examples and self-check questions on this page are our own, and every number was checked by calculation (235 checks). Our working for the four exercise items is ours, not NCERT's.

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