Class 10 Mathematics - NCERT Mathematics

NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations

Chapter 4 covers Quadratic Equations in the standard form ax^2 + bx + c = 0. Students learn to check whether a given equation is quadratic, solve quadratic equations by factorisation and by using the quadratic formula, and determine the nature of roots using the discriminant D = b^2 - 4ac. Word problems involving areas, speeds, ages, and consecutive integers are a key focus. This chapter carries 6-8 marks in the CBSE Class 10 board exam 2027.

3 Exercises13 QuestionsStep-by-Step SolutionsUpdated 2026-27
Practice Quadratic Equations with AI

Syllabus Update 2026-27

As per the CBSE 2026-27 syllabus, Chapter 4 Quadratic Equations is FULLY included with 3 exercises - Exercise 4.1 (2 questions on checking if equations are quadratic), Exercise 4.2 (6 questions on solving by factorisation and word problems), and Exercise 4.3 (5 questions on nature of roots using discriminant). No content has been deleted from this chapter. The method of completing the square has been removed from the syllabus but the quadratic formula is retained.

Unit: Algebra
20 Marks1 MCQ on discriminant/nature of roots (1 mark) + 1 short answer on solving by factorisation (2-3 marks) + 1 word problem (3-4 marks)

Table of Contents

4.1Exercise 4.1 - Introduction to Quadratic Equations

Q1Check whether the following are quadratic equations: (i) (x + 1)^2 = 2(x - 3) (ii) x^2 - 2x = (-2)(3 - x) (iii) (x - 2)(x + 1) = (x - 1)(x + 3) (iv) (x - 3)(2x + 1) = x(x + 5) (v) (2x - 1)(x - 3) = (x + 5)(x - 1) (vi) x^2 + 3x + 1 = (x - 2)^2 (vii) (x + 2)^3 = 2x(x^2 - 1) (viii) x^3 - 4x^2 - x + 1 = (x - 2)^3

(i) (x + 1)^2 = 2(x - 3)
Expanding: x^2 + 2x + 1 = 2x - 6
x^2 + 2x + 1 - 2x + 6 = 0
x^2 + 7 = 0
This is of the form ax^2 + bx + c = 0 (a=1, b=0, c=7). Yes, it is a quadratic equation.

(ii) x^2 - 2x = (-2)(3 - x)
x^2 - 2x = -6 + 2x
x^2 - 2x - 2x + 6 = 0
x^2 - 4x + 6 = 0
This is of the form ax^2 + bx + c = 0 (a=1, b=-4, c=6). Yes, it is a quadratic equation.

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
LHS: x^2 + x - 2x - 2 = x^2 - x - 2
RHS: x^2 + 3x - x - 3 = x^2 + 2x - 3
x^2 - x - 2 = x^2 + 2x - 3
-x - 2 - 2x + 3 = 0
-3x + 1 = 0
This is a linear equation (no x^2 term). No, it is not a quadratic equation.

(iv) (x - 3)(2x + 1) = x(x + 5)
LHS: 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3
RHS: x^2 + 5x
2x^2 - 5x - 3 - x^2 - 5x = 0
x^2 - 10x - 3 = 0
This is of the form ax^2 + bx + c = 0. Yes, it is a quadratic equation.

(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
LHS: 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3
RHS: x^2 - x + 5x - 5 = x^2 + 4x - 5
2x^2 - 7x + 3 - x^2 - 4x + 5 = 0
x^2 - 11x + 8 = 0
This is of the form ax^2 + bx + c = 0. Yes, it is a quadratic equation.

(vi) x^2 + 3x + 1 = (x - 2)^2
x^2 + 3x + 1 = x^2 - 4x + 4
3x + 1 + 4x - 4 = 0
7x - 3 = 0
This is a linear equation. No, it is not a quadratic equation.

(vii) (x + 2)^3 = 2x(x^2 - 1)
LHS: x^3 + 6x^2 + 12x + 8
RHS: 2x^3 - 2x
x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0
-x^3 + 6x^2 + 14x + 8 = 0
This has an x^3 term, so the degree is 3. No, it is not a quadratic equation.

(viii) x^3 - 4x^2 - x + 1 = (x - 2)^3
RHS: x^3 - 6x^2 + 12x - 8
x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0
2x^2 - 13x + 9 = 0
This is of the form ax^2 + bx + c = 0. Yes, it is a quadratic equation.

Q2Represent the following situations in the form of quadratic equations: (i) The area of a rectangular plot is 528 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. (ii) The product of two consecutive positive integers is 306. We need to find the integers. (iii) 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. (iv) 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.

(i) Rectangular plot:
Let breadth = x metres.
Length = 2x + 1 metres.
Area = length x breadth = 528
(2x + 1)(x) = 528
2x^2 + x = 528
2x^2 + x - 528 = 0

(ii) Consecutive positive integers:
Let the integers be x and x + 1.
Product = 306
x(x + 1) = 306
x^2 + x = 306
x^2 + x - 306 = 0

(iii) Rohan's age:
Let Rohan's present age = x years.
Mother's age = x + 26 years.
After 3 years: Rohan = x + 3, Mother = x + 29.
Product = 360
(x + 3)(x + 29) = 360
x^2 + 29x + 3x + 87 = 360
x^2 + 32x + 87 - 360 = 0
x^2 + 32x - 273 = 0

(iv) Train speed:
Let speed = x km/h.
Time at speed x = 480/x hours.
Time at speed (x - 8) = 480/(x - 8) hours.
Difference = 3 hours
480/(x - 8) - 480/x = 3
480x - 480(x - 8) = 3x(x - 8)
480x - 480x + 3840 = 3x^2 - 24x
3x^2 - 24x - 3840 = 0
Dividing by 3:
x^2 - 8x - 1280 = 0

4.2Exercise 4.2 - Solution of a Quadratic Equation by Factorisation

Q1Find the roots of the following quadratic equations by factorisation: (i) x^2 - 3x - 10 = 0 (ii) 2x^2 + x - 6 = 0 (iii) sqrt(2)x^2 + 7x + 5sqrt(2) = 0 (iv) 2x^2 - x + 1/8 = 0 (v) 100x^2 - 20x + 1 = 0

(i) x^2 - 3x - 10 = 0
We need two numbers whose product is -10 and sum is -3: (-5) and 2.
x^2 - 5x + 2x - 10 = 0
x(x - 5) + 2(x - 5) = 0
(x + 2)(x - 5) = 0
x = -2 or x = 5

(ii) 2x^2 + x - 6 = 0
Product = 2 x (-6) = -12. Sum = 1. Numbers: 4 and -3.
2x^2 + 4x - 3x - 6 = 0
2x(x + 2) - 3(x + 2) = 0
(2x - 3)(x + 2) = 0
x = 3/2 or x = -2

(iii) sqrt(2)x^2 + 7x + 5sqrt(2) = 0
Product = sqrt(2) x 5sqrt(2) = 10. Sum = 7. Numbers: 5 and 2.
sqrt(2)x^2 + 5x + 2x + 5sqrt(2) = 0
x(sqrt(2)x + 5) + sqrt(2)(sqrt(2)x + 5) = 0
(x + sqrt(2))(sqrt(2)x + 5) = 0
x = -sqrt(2) or x = -5/sqrt(2) = -5sqrt(2)/2

(iv) 2x^2 - x + 1/8 = 0
Multiply by 8: 16x^2 - 8x + 1 = 0
Product = 16. Sum = -8. Numbers: -4 and -4.
16x^2 - 4x - 4x + 1 = 0
4x(4x - 1) - 1(4x - 1) = 0
(4x - 1)(4x - 1) = 0
x = 1/4, 1/4 (equal roots)

(v) 100x^2 - 20x + 1 = 0
Product = 100. Sum = -20. Numbers: -10 and -10.
100x^2 - 10x - 10x + 1 = 0
10x(10x - 1) - 1(10x - 1) = 0
(10x - 1)(10x - 1) = 0
x = 1/10, 1/10 (equal roots)

Q2Solve the problems given in Q2 of Exercise 4.1.

(i) Area of rectangular plot: 2x^2 + x - 528 = 0
Product = 2 x (-528) = -1056. Sum = 1. Numbers: 33 and -32.
2x^2 + 33x - 32x - 528 = 0
x(2x + 33) - 16(2x + 33) = 0
(x - 16)(2x + 33) = 0
x = 16 or x = -33/2
Since breadth cannot be negative, x = 16.
Breadth = 16 m, Length = 2(16) + 1 = 33 m.

(ii) Consecutive integers: x^2 + x - 306 = 0
Product = -306. Sum = 1. Numbers: 18 and -17.
x^2 + 18x - 17x - 306 = 0
x(x + 18) - 17(x + 18) = 0
(x - 17)(x + 18) = 0
x = 17 or x = -18
Since positive integers, x = 17.
The integers are 17 and 18.

(iii) Rohan's age: x^2 + 32x - 273 = 0
Product = -273. Sum = 32. Numbers: 39 and -7.
x^2 + 39x - 7x - 273 = 0
x(x + 39) - 7(x + 39) = 0
(x - 7)(x + 39) = 0
x = 7 or x = -39
Age cannot be negative, so Rohan's present age = 7 years.

(iv) Train speed: x^2 - 8x - 1280 = 0
Product = -1280. Sum = -8. Numbers: -40 and 32.
x^2 - 40x + 32x - 1280 = 0
x(x - 40) + 32(x - 40) = 0
(x + 32)(x - 40) = 0
x = 40 or x = -32
Speed cannot be negative, so speed of the train = 40 km/h.

Q3Find two numbers whose sum is 27 and product is 182.

Let one number be x. Other number = 27 - x.
Product: x(27 - x) = 182
27x - x^2 = 182
x^2 - 27x + 182 = 0

Product = 182. Sum = -27. Numbers: -13 and -14.
x^2 - 13x - 14x + 182 = 0
x(x - 13) - 14(x - 13) = 0
(x - 13)(x - 14) = 0
x = 13 or x = 14

The two numbers are 13 and 14.

Q4Find two consecutive positive integers, sum of whose squares is 365.

Let the integers be x and x + 1.
x^2 + (x + 1)^2 = 365
x^2 + x^2 + 2x + 1 = 365
2x^2 + 2x + 1 - 365 = 0
2x^2 + 2x - 364 = 0
x^2 + x - 182 = 0

Product = -182. Sum = 1. Numbers: 14 and -13.
x^2 + 14x - 13x - 182 = 0
x(x + 14) - 13(x + 14) = 0
(x - 13)(x + 14) = 0
x = 13 or x = -14

Since positive integers, x = 13.
The two consecutive positive integers are 13 and 14.

Q5The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let base = x cm. Then altitude = (x - 7) cm.
By Pythagoras theorem:
x^2 + (x - 7)^2 = 13^2
x^2 + x^2 - 14x + 49 = 169
2x^2 - 14x + 49 - 169 = 0
2x^2 - 14x - 120 = 0
x^2 - 7x - 60 = 0

Product = -60. Sum = -7. Numbers: -12 and 5.
x^2 - 12x + 5x - 60 = 0
x(x - 12) + 5(x - 12) = 0
(x + 5)(x - 12) = 0
x = 12 or x = -5

Since length cannot be negative, x = 12.
Base = 12 cm, Altitude = 12 - 7 = 5 cm.

Q6A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Let number of articles = x.
Cost of each article = 2x + 3 rupees.
Total cost = x(2x + 3) = 90
2x^2 + 3x = 90
2x^2 + 3x - 90 = 0

Product = 2 x (-90) = -180. Sum = 3. Numbers: 15 and -12.
2x^2 + 15x - 12x - 90 = 0
x(2x + 15) - 6(2x + 15) = 0
(x - 6)(2x + 15) = 0
x = 6 or x = -15/2

Since number of articles cannot be negative, x = 6.
Cost per article = 2(6) + 3 = 15 rupees.

Number of articles = 6, Cost of each article = Rs 15.

4.3Exercise 4.3 - Nature of Roots

Q1Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x^2 - 3x + 5 = 0 (ii) 3x^2 - 4sqrt(3)x + 4 = 0 (iii) 2x^2 - 6x + 3 = 0

(i) 2x^2 - 3x + 5 = 0
a = 2, b = -3, c = 5
D = b^2 - 4ac = 9 - 40 = -31
Since D < 0, no real roots exist.

(ii) 3x^2 - 4sqrt(3)x + 4 = 0
a = 3, b = -4sqrt(3), c = 4
D = b^2 - 4ac = 48 - 48 = 0
Since D = 0, two equal real roots exist.
x = -b/(2a) = 4sqrt(3)/(2 x 3) = 4sqrt(3)/6 = 2sqrt(3)/3
x = 2sqrt(3)/3, 2sqrt(3)/3

(iii) 2x^2 - 6x + 3 = 0
a = 2, b = -6, c = 3
D = b^2 - 4ac = 36 - 24 = 12
Since D > 0, two distinct real roots exist.
x = [-b +- sqrt(D)]/(2a) = [6 +- sqrt(12)]/4 = [6 +- 2sqrt(3)]/4
x = (3 + sqrt(3))/2 or x = (3 - sqrt(3))/2

Q2Find the values of k for each of the following quadratic equations, so that they have two equal roots: (i) 2x^2 + kx + 3 = 0 (ii) kx(x - 2) + 6 = 0

For equal roots, discriminant D = 0.

(i) 2x^2 + kx + 3 = 0
a = 2, b = k, c = 3
D = k^2 - 4(2)(3) = 0
k^2 - 24 = 0
k^2 = 24
k = +-2sqrt(6)

(ii) kx(x - 2) + 6 = 0
kx^2 - 2kx + 6 = 0
a = k, b = -2k, c = 6
D = (-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
4k(k - 6) = 0
k = 0 or k = 6

But if k = 0, the equation becomes 6 = 0, which is not valid. So k cannot be 0.
k = 6

Q3Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^2? If so, find its length and breadth.

Let breadth = x m. Then length = 2x m.
Area = x(2x) = 800
2x^2 = 800
x^2 = 400
x = +-20

Since breadth cannot be negative, x = 20.

Check: This equation x^2 = 400 has D = 0 - 4(1)(-400) = 1600 > 0, so real roots exist.

Yes, it is possible. Breadth = 20 m, Length = 40 m.

Q4Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Let one friend's age = x years. Other friend's age = (20 - x) years.

Four years ago: ages were (x - 4) and (20 - x - 4) = (16 - x).
Product = 48
(x - 4)(16 - x) = 48
16x - x^2 - 64 + 4x = 48
-x^2 + 20x - 64 - 48 = 0
-x^2 + 20x - 112 = 0
x^2 - 20x + 112 = 0

D = b^2 - 4ac = 400 - 448 = -48

Since D < 0, no real roots exist.

This situation is not possible.

Q5Is it possible to design a rectangular park of perimeter 80 m and area 400 m^2? If so, find its length and breadth.

Perimeter = 2(length + breadth) = 80
length + breadth = 40
Let breadth = x m. Then length = (40 - x) m.

Area = x(40 - x) = 400
40x - x^2 = 400
x^2 - 40x + 400 = 0

D = b^2 - 4ac = 1600 - 1600 = 0

Since D = 0, real (equal) roots exist.
x = -b/(2a) = 40/2 = 20

Breadth = 20 m, Length = 40 - 20 = 20 m.

Yes, it is possible. The park is a square with length = breadth = 20 m.

Stuck on a question? AI explains any step in detail - in English or Hindi.

Ask AI Tutor

Previous Year Board Exam Questions

CBSE 20262 marks

If the quadratic equation x^2 + 2(k + 1)x + k^2 = 0 has equal roots, then find the value of k.

For equal roots, D = 0. Here a = 1, b = 2(k+1), c = k^2. D = [2(k+1)]^2 - 4(1)(k^2) = 4(k^2 + 2k + 1) - 4k^2 = 4k^2 + 8k + 4 - 4k^2 = 8k + 4 = 0. Solving: 8k = -4, so k = -1/2. Therefore the value of k is -1/2.

CBSE 20252 marks

Solve for x: 2x^2 - 5x + 3 = 0 by factorisation method.

We have 2x^2 - 5x + 3 = 0. Product = 2 x 3 = 6, sum = -5. Numbers: -2 and -3. Splitting: 2x^2 - 2x - 3x + 3 = 0. Factoring: 2x(x - 1) - 3(x - 1) = 0, so (2x - 3)(x - 1) = 0. Therefore x = 3/2 or x = 1. The roots are x = 1 and x = 3/2.

CBSE 20252 marks

A rectangular field has an area of 300 sq m. If its length is 5 m more than its breadth, form the quadratic equation to find the breadth of the field.

Let breadth = x m, then length = (x + 5) m. Area = x(x + 5) = 300. Expanding: x^2 + 5x = 300. Rearranging: x^2 + 5x - 300 = 0. This is the required quadratic equation. Solving: (x + 20)(x - 15) = 0, so x = 15 (rejecting x = -20). The breadth is 15 m and the quadratic equation is x^2 + 5x - 300 = 0.

CBSE 20242 marks

Find the discriminant of the quadratic equation 3x^2 - 2x + 1/3 = 0 and hence find the nature of its roots.

Multiply by 3: 9x^2 - 6x + 1 = 0. Here a = 9, b = -6, c = 1. Discriminant D = b^2 - 4ac = 36 - 36 = 0. Since D = 0, the equation has two equal real roots. Alternatively, using original equation: a = 3, b = -2, c = 1/3. D = 4 - 4(3)(1/3) = 4 - 4 = 0. The discriminant is 0 and the equation has two equal real roots.

CBSE 20242 marks

Solve for x: 6x^2 + x - 2 = 0 by factorisation.

We have 6x^2 + x - 2 = 0. Product = 6 x (-2) = -12, sum = 1. Numbers: 4 and -3. Splitting: 6x^2 + 4x - 3x - 2 = 0. Factoring: 2x(3x + 2) - 1(3x + 2) = 0, so (2x - 1)(3x + 2) = 0. Therefore x = 1/2 or x = -2/3. The roots are x = 1/2 and x = -2/3.

CBSE 20243 marks

The sum of two numbers is 15 and the sum of their reciprocals is 3/10. Find the two numbers.

Let one number = x. Other number = 15 - x. Sum of reciprocals: 1/x + 1/(15-x) = 3/10. Taking LCM: (15-x+x)/[x(15-x)] = 3/10. Simplifying: 15/[x(15-x)] = 3/10. Cross-multiplying: 150 = 3x(15-x). So 50 = 15x - x^2. Rearranging: x^2 - 15x + 50 = 0. Factoring: (x-5)(x-10) = 0. So x = 5 or x = 10. The two numbers are 5 and 10.

CBSE 20232 marks

For what values of k does the quadratic equation 4x^2 - 12x - k = 0 have no real roots?

For no real roots, D < 0. Here a = 4, b = -12, c = -k. D = (-12)^2 - 4(4)(-k) = 144 + 16k. For no real roots: 144 + 16k < 0. So 16k < -144, which gives k < -9. The equation has no real roots when k < -9.

CBSE 20232 marks

Find the value of p so that the quadratic equation px(x - 3) + 9 = 0 has two equal roots.

Expanding: px^2 - 3px + 9 = 0. Here a = p, b = -3p, c = 9. For equal roots: D = 0. D = (-3p)^2 - 4(p)(9) = 9p^2 - 36p = 0. Factoring: 9p(p - 4) = 0. So p = 0 or p = 4. Since p = 0 would make it not a quadratic equation, p = 4.

CBSE 20233 marks

If one root of the quadratic equation 2x^2 + kx - 6 = 0 is 2, find the value of k. Also, find the other root.

Since x = 2 is a root, substitute in 2x^2 + kx - 6 = 0: 2(4) + 2k - 6 = 0. So 8 + 2k - 6 = 0, giving 2k = -2, hence k = -1. The equation becomes 2x^2 - x - 6 = 0. Factoring: 2x^2 - 4x + 3x - 6 = 0, so 2x(x - 2) + 3(x - 2) = 0, giving (2x + 3)(x - 2) = 0. Therefore x = -3/2 or x = 2. k = -1 and the other root is -3/2.

CBSE 20222 marks

Find the roots of the quadratic equation x^2 - 3x - 10 = 0.

We have x^2 - 3x - 10 = 0. We need two numbers whose product is -10 and sum is -3: these are -5 and 2. Splitting: x^2 - 5x + 2x - 10 = 0. Factoring: x(x - 5) + 2(x - 5) = 0, so (x + 2)(x - 5) = 0. Therefore x = -2 or x = 5. The roots are -2 and 5.

CBSE 20224 marks

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Let the speed of the train be x km/h. Time = 360/x hours. If speed were (x+5), time = 360/(x+5). According to the condition: 360/x - 360/(x+5) = 1. Taking LCM: [360(x+5) - 360x] / [x(x+5)] = 1. Simplifying: 360x + 1800 - 360x = x^2 + 5x, so 1800 = x^2 + 5x. Rearranging: x^2 + 5x - 1800 = 0. Factoring: (x + 45)(x - 40) = 0. So x = 40 or x = -45. Since speed cannot be negative, the speed of the train is 40 km/h.

CBSE 20213 marks

The product of two consecutive positive integers is 240. Form a quadratic equation and find the integers.

Let the two consecutive positive integers be x and x + 1. Their product: x(x + 1) = 240. Expanding: x^2 + x - 240 = 0. Factoring: (x + 16)(x - 15) = 0. So x = 15 or x = -16. Since the integers are positive, x = 15. The quadratic equation is x^2 + x - 240 = 0 and the integers are 15 and 16.

CBSE 2021 (Term 1)1 mark

If one root of the equation (k - 1)x^2 - 10x + 3 = 0 is the reciprocal of the other, then find the value of k.

If one root is alpha and the other is 1/alpha, then product of roots = alpha x (1/alpha) = 1. For the equation (k-1)x^2 - 10x + 3 = 0, product of roots = c/a = 3/(k-1). So 3/(k-1) = 1, giving k - 1 = 3, hence k = 4.

CBSE 20201 mark

Find the nature of roots of the quadratic equation 2x^2 - 4x + 3 = 0.

Here a = 2, b = -4, c = 3. Discriminant D = b^2 - 4ac = 16 - 24 = -8. Since D < 0, the quadratic equation 2x^2 - 4x + 3 = 0 has no real roots.

CBSE 20193 marks

If the quadratic equation (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 has equal roots, prove that c^2 = a^2(1 + m^2).

For equal roots, D = 0. Here A = 1+m^2, B = 2mc, C = c^2 - a^2. D = (2mc)^2 - 4(1+m^2)(c^2-a^2) = 0. Expanding: 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0. Simplifying: 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = 0. So -4c^2 + 4a^2 + 4m^2a^2 = 0, giving 4c^2 = 4a^2(1 + m^2). Hence c^2 = a^2(1 + m^2).

CBSE 20193 marks

Solve for x: x^2 + 5x - (a^2 + a - 6) = 0.

Using the quadratic formula: x = [-5 +- sqrt(25 + 4(a^2 + a - 6))] / 2. Simplifying the discriminant: 25 + 4a^2 + 4a - 24 = 4a^2 + 4a + 1 = (2a + 1)^2. So x = [-5 +- (2a + 1)] / 2. Taking + sign: x = (-5 + 2a + 1)/2 = (2a - 4)/2 = a - 2. Taking - sign: x = (-5 - 2a - 1)/2 = -(2a + 6)/2 = -(a + 3). The roots are x = (a - 2) and x = -(a + 3).

Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.

Practice PYQ Patterns

Assertion-Reason Questions

(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true

AR1

Assertion (A): The equation x^2 + 3x + 1 = (x - 2)^2 is not a quadratic equation.

Reason (R): Any equation of the form ax^2 + bx + c = 0 where a is not equal to 0 is called a quadratic equation.

Answer: (A)

Assertion is TRUE: Expanding gives x^2 + 3x + 1 = x^2 - 4x + 4, which simplifies to 7x - 3 = 0 (linear, not quadratic since the x^2 terms cancel out). Reason is TRUE and correctly explains the assertion - since the simplified form has no x^2 term (a = 0), it fails the definition of a quadratic equation.

AR2

Assertion (A): The quadratic equation 2x^2 - 8x + 6 = 0 has two distinct real roots.

Reason (R): For the equation ax^2 + bx + c = 0, if discriminant D = b^2 - 4ac > 0, then the roots are real and distinct.

Answer: (A)

Assertion is TRUE: D = (-8)^2 - 4(2)(6) = 64 - 48 = 16 > 0, so two distinct real roots exist (x = 1 and x = 3). Reason is TRUE and correctly explains the assertion - since D > 0, the nature of roots is two distinct real roots.

AR3

Assertion (A): The quadratic equation x^2 + x + 1 = 0 has no real roots.

Reason (R): The discriminant D = b^2 - 4ac of the equation x^2 + x + 1 = 0 is negative.

Answer: (A)

Assertion is TRUE: the equation has no real roots. Reason is TRUE: D = 1 - 4(1)(1) = 1 - 4 = -3 < 0. Since D < 0, no real roots exist. The reason correctly explains the assertion.

AR4

Assertion (A): If the discriminant of a quadratic equation is zero, then the equation has only one root.

Reason (R): When D = 0, the quadratic formula gives x = -b/(2a), which is a single value.

Answer: (D)

Assertion is FALSE: When D = 0, the equation has two equal real roots (repeated root), not "only one root." A quadratic always has exactly two roots counting multiplicity. Reason is TRUE: when D = 0, x = -b/(2a) is indeed a single value, and both roots are equal to this value. Since assertion is false and reason is true, the answer is D.

AR5

Assertion (A): The roots of the equation 9x^2 - 6x + 1 = 0 are equal.

Reason (R): The quadratic formula x = [-b +- sqrt(b^2-4ac)]/(2a) gives real roots only when b^2 - 4ac >= 0.

Answer: (B)

Assertion is TRUE: D = 36 - 36 = 0, so roots are equal (x = 1/3, 1/3). Reason is TRUE: the quadratic formula indeed gives real roots only when D >= 0. However, the reason does not specifically explain why the roots are equal (it would need D = 0 for that). Both are true but the reason does not correctly explain the assertion.

AR6

Assertion (A): For k = 4, the equation kx(x - 3) + 9 = 0 has two equal roots.

Reason (R): The discriminant of the equation 4x^2 - 12x + 9 = 0 is zero.

Answer: (A)

Assertion is TRUE: Substituting k = 4 gives 4x^2 - 12x + 9 = 0. D = (-12)^2 - 4(4)(9) = 144 - 144 = 0, so the equation has two equal roots. Reason is TRUE: D = 0 is correctly computed. The reason correctly explains the assertion, as equal roots occur precisely when the discriminant is zero.

AR7

Assertion (A): The quadratic equation 2x^2 - 3x + 5 = 0 has two real roots.

Reason (R): Every quadratic equation has exactly two roots.

Answer: (D)

Assertion is FALSE: D = 9 - 40 = -31 < 0, so the equation has no real roots. Reason is FALSE as stated - while every quadratic equation has exactly two roots in the complex number system, it does not always have two real roots. Since both are false, the answer is D.

Case Study Questions

CS1Swimming Pool DesignCBSE 2023-24

A school is planning to build a rectangular swimming pool in its compound. The length of the pool is 2 metres more than twice its breadth. The area of the rectangular pool is 24 sq metres. Based on this information, answer the following questions.

(i) If the breadth of the pool is x metres, which of the following quadratic equations represents the given situation? (1 mark)

Breadth = x m. Length = 2x + 2 m. Area = x(2x + 2) = 24. So 2x^2 + 2x - 24 = 0, or x^2 + x - 12 = 0 (dividing by 2). This is the required quadratic equation.

(ii) What is the breadth of the pool? (1 mark)

Solving x^2 + x - 12 = 0: (x + 4)(x - 3) = 0. So x = 3 or x = -4. Since breadth cannot be negative, the breadth of the pool is 3 metres.

(iii)(a) What is the length of the pool? (2 marks)

Length = 2x + 2 = 2(3) + 2 = 6 + 2 = 8 metres. Verification: Area = 3 x 8 = 24 sq m, which matches the given area.

(iii)(b) Find the value of the discriminant of the quadratic equation formed. (2 marks)

The equation is x^2 + x - 12 = 0. Here a = 1, b = 1, c = -12. Discriminant D = b^2 - 4ac = 1 - 4(1)(-12) = 1 + 48 = 49. Since D > 0, the equation has two distinct real roots.

CS2Train Journey ProblemCBSE 2023-24

Raj and Ajay are very close friends. Both the families decided to go to Ranikhet by their own cars. Raj's car travels at a speed of x km/h while Ajay's car travels 5 km/h faster than Raj's car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

(i) What is the speed of Ajay's car? (1 mark)

Ajay's speed = (x + 5) km/h. We need to first find x. Time for Raj = 400/x. Time for Ajay = 400/(x+5). Given: 400/x - 400/(x+5) = 4. Solving: 400(x+5) - 400x = 4x(x+5). So 2000 = 4x^2 + 20x, giving x^2 + 5x - 500 = 0. Factoring: (x+25)(x-20) = 0. So x = 20 (rejecting -25). Ajay's speed = 20 + 5 = 25 km/h.

(ii) Form the quadratic equation from the given situation. (1 mark)

From the condition 400/x - 400/(x+5) = 4, simplifying gives 2000 = 4x^2 + 20x. Dividing by 4: x^2 + 5x - 500 = 0. The required quadratic equation is x^2 + 5x - 500 = 0.

(iii)(a) How much time did Raj take to complete the journey? (2 marks)

Raj's speed = 20 km/h. Time = Distance/Speed = 400/20 = 20 hours.

(iii)(b) What is the discriminant of the quadratic equation x^2 + 5x - 500 = 0? (2 marks)

Here a = 1, b = 5, c = -500. Discriminant D = b^2 - 4ac = 25 - 4(1)(-500) = 25 + 2000 = 2025. Note: sqrt(2025) = 45, confirming that the roots are rational.

MCQ Practice

MCQ1. Which of the following is a quadratic equation?

(A) x^2 + 2x + 1 = (4 - x)^2 + 3

(B) -2x^2 = (5 - x)(2x - 2/5)

(C) (k + 1)x^2 + 3/2 x = 7, where k = -1

(D) x^3 - x^2 = (x - 1)^3

Answer: (D) - Check each: (A) x^2+2x+1 = x^2-8x+16+3 gives 10x-18=0 (linear). (B) -2x^2 = 10x - 2x^2 - 2 + 2x/5 gives 10x+2x/5-2=0 (linear). (C) With k=-1, the equation becomes 0.x^2 + 3x/2 = 7 (linear). (D) x^3-x^2 = x^3-3x^2+3x-1 gives 2x^2-3x+1 = 0, which is quadratic.

MCQ2. The roots of the quadratic equation x^2 - 0.04 = 0 are:

(A) +- 0.2

(B) +- 0.02

(C) 0.4

(D) 2

Answer: (A) - x^2 = 0.04. So x = +- sqrt(0.04) = +- 0.2.

MCQ3. If the discriminant of the equation 6x^2 - bx + 2 = 0 is 1, then the value of b is:

(A) 7

(B) -7

(C) +- 7

(D) 49

Answer: (C) - D = b^2 - 4ac = b^2 - 4(6)(2) = b^2 - 48 = 1. So b^2 = 49, giving b = +- 7.

MCQ4. The quadratic equation 2x^2 - sqrt(5)x + 1 = 0 has:

(A) two distinct real roots

(B) two equal real roots

(C) no real roots

(D) more than 2 real roots

Answer: (C) - D = (-sqrt(5))^2 - 4(2)(1) = 5 - 8 = -3. Since D < 0, the equation has no real roots.

MCQ5. If x = 3 is one root of x^2 - 2kx - 6 = 0, then the value of k is:

(A) -1/2

(B) 1/2

(C) 3

(D) 1

Answer: (B) - Substituting x = 3: 9 - 6k - 6 = 0. So 3 - 6k = 0, giving 6k = 3, hence k = 1/2.

MCQ6. The sum of the roots of the quadratic equation 3x^2 - 9x + 5 = 0 is:

(A) 3

(B) -3

(C) 9

(D) 5/3

Answer: (A) - Sum of roots = -b/a = -(-9)/3 = 9/3 = 3.

MCQ7. A quadratic equation whose roots are 2 and -5 is:

(A) x^2 + 3x - 10 = 0

(B) x^2 - 3x + 10 = 0

(C) x^2 - 3x - 10 = 0

(D) x^2 + 3x + 10 = 0

Answer: (A) - Sum of roots = 2 + (-5) = -3. Product = 2 x (-5) = -10. Equation: x^2 - (sum)x + product = 0, so x^2 - (-3)x + (-10) = 0, i.e., x^2 + 3x - 10 = 0.

MCQ8. The value of k for which the equation x^2 + 2(k+1)x + k^2 = 0 has equal roots is:

(A) -1/2

(B) 1/2

(C) 1

(D) -1

Answer: (A) - For equal roots, D = 0. D = [2(k+1)]^2 - 4(k^2) = 4(k^2+2k+1) - 4k^2 = 4k^2 + 8k + 4 - 4k^2 = 8k + 4 = 0. So k = -1/2.

MCQ9. If the equation (a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 has equal roots, then:

(A) a/b = c/d

(B) a/c = b/d

(C) a/d = b/c

(D) ab = cd

Answer: (B) - For equal roots, D = 0. D = 4(ac+bd)^2 - 4(a^2+b^2)(c^2+d^2) = 0. Expanding: a^2c^2 + 2abcd + b^2d^2 = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2. Simplifying: 2abcd = a^2d^2 + b^2c^2, so (ad - bc)^2 = 0, giving ad = bc, i.e., a/c = b/d.

MCQ10. If one root of 5x^2 + 13x + k = 0 is the reciprocal of the other, then k equals:

(A) 0

(B) 5

(C) 1/6

(D) 6

Answer: (B) - If one root is alpha and other is 1/alpha, then product of roots = alpha x (1/alpha) = 1. Product of roots = c/a = k/5 = 1. So k = 5.

MCQ11. The nature of roots of the equation x^2 + 4x + 5 = 0 is:

(A) real and distinct

(B) real and equal

(C) not real

(D) one real, one imaginary

Answer: (C) - D = 16 - 20 = -4 < 0. Since the discriminant is negative, the equation has no real roots (roots are not real/imaginary).

MCQ12. If the roots of the equation kx^2 - 5x + 6 = 0 are in the ratio 2:3, then k equals:

(A) 1

(B) 2

(C) 6

(D) -1

Answer: (A) - Let roots be 2m and 3m. Sum of roots = 2m + 3m = 5m = 5/k, so m = 1/k. Product of roots = 2m x 3m = 6m^2 = 6/k. Substituting m = 1/k: 6/k^2 = 6/k. So k^2 = k, giving k(k-1) = 0. Since k cannot be 0 (equation would not be quadratic), k = 1.

Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.

Start Quiz Mode

Important Questions - Mark-wise

1 Mark Questions

Write the discriminant of the quadratic equation 3x^2 - 4x + 1 = 0.

D = b^2 - 4ac = (-4)^2 - 4(3)(1) = 16 - 12 = **4.**

State the nature of roots of 2x^2 + 3x + 5 = 0.

D = 9 - 40 = -31. Since D < 0, the equation has **no real roots.**

For what value of k does x^2 + 4x + k = 0 have equal roots?

For equal roots D = 0: 16 - 4k = 0, so **k = 4.**

Is x = -2 a root of 2x^2 + 5x - 3 = 0?

Substituting: 2(4) + 5(-2) - 3 = 8 - 10 - 3 = -5, which is not 0. So **x = -2 is not a root.**

Check whether (x - 2)^2 + 1 = 2x - 3 is a quadratic equation.

Expanding: x^2 - 4x + 4 + 1 = 2x - 3, so x^2 - 6x + 8 = 0. Since it is of the form ax^2 + bx + c = 0 with a = 1, **yes, it is a quadratic equation.**

2-3 Mark Questions

Solve by factorisation: 3x^2 - 2sqrt(6)x + 2 = 0.

Product = 3 x 2 = 6. Sum = -2sqrt(6). Numbers: -sqrt(6) and -sqrt(6).
3x^2 - sqrt(6)x - sqrt(6)x + 2 = 0
sqrt(3)x(sqrt(3)x - sqrt(2)) - sqrt(2)(sqrt(3)x - sqrt(2)) = 0
(sqrt(3)x - sqrt(2))(sqrt(3)x - sqrt(2)) = 0
x = sqrt(2)/sqrt(3) = sqrt(6)/3 (equal roots).

Find the value of k for which the equation 2x^2 + kx + 3 = 0 has two equal roots.

For equal roots, D = 0: k^2 - 4(2)(3) = 0. k^2 = 24. k = +- 2sqrt(6).

Find the roots of the quadratic equation 4x^2 + 4sqrt(3)x + 3 = 0 using the quadratic formula.

a = 4, b = 4sqrt(3), c = 3. D = 48 - 48 = 0 (equal roots).
x = -b/(2a) = -4sqrt(3)/8 = -sqrt(3)/2 (repeated root).

Find two consecutive odd positive integers whose sum of squares is 290.

Let the integers be x and x+2. x^2 + (x+2)^2 = 290. 2x^2 + 4x + 4 = 290. x^2 + 2x - 143 = 0. (x+13)(x-11) = 0. x = 11 (positive). The integers are 11 and 13.

3-4 Mark Questions

A train travels 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. Find the speed of the train.

Let speed = x km/h. 480/x - 480/(x) is the original time. Setting up: 480/(x-8) - 480/x = 3. Simplifying: 480x - 480(x-8) = 3x(x-8). So 3840 = 3x^2 - 24x. Dividing by 3: x^2 - 8x - 1280 = 0. Factoring: (x-40)(x+32) = 0. x = 40. Speed = 40 km/h.

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let larger number = x. Smaller number's square = 8x. Difference of squares: x^2 - 8x = 180. x^2 - 8x - 180 = 0. (x-18)(x+10) = 0. x = 18 or x = -10. If x = 18: smaller^2 = 144, smaller = +-12. If x = -10: smaller^2 = -80 (impossible). The numbers are 18 and 12 (or 18 and -12).

A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park.

Let length = x m. Breadth = (x-3) m. Area of rectangle = x(x-3). Area of triangle = (1/2)(x-3)(12) = 6(x-3). Given: x(x-3) = 6(x-3) + 4. Expanding: x^2 - 3x = 6x - 18 + 4. So x^2 - 9x + 14 = 0. Factoring: (x-7)(x-2) = 0. x = 7 or x = 2. If x = 2, breadth = -1 (invalid). Length = 7 m, Breadth = 4 m.

Key Concepts

Standard form of a quadratic equation: ax^2 + bx + c = 0, where a, b, c are real numbers and a is not equal to 0Roots (or zeroes) of a quadratic equation are the values of x that satisfy the equationSolving by factorisation: split the middle term and use the zero product propertyQuadratic formula: x = [-b +- sqrt(b^2 - 4ac)] / (2a)Discriminant D = b^2 - 4ac determines the nature of rootsIf D > 0: two distinct real roots; if D = 0: two equal real roots (repeated root); if D < 0: no real roots

Important Formulas

Standard form: ax^2 + bx + c = 0, a is not equal to 0
Quadratic formula: x = [-b +- sqrt(b^2 - 4ac)] / (2a)
Discriminant: D = b^2 - 4ac
Sum of roots = -b/a
Product of roots = c/a
For equal roots: D = 0, i.e., b^2 = 4ac

Study Plan - Quadratic Equations

1

Understanding Quadratic Equations

Read NCERT section 4.1 on the standard form ax^2 + bx + c = 0. Learn to identify whether a given equation is quadratic by simplifying and checking the degree. Solve all questions of Exercise 4.1 (Q1 and Q2). Practice converting word problems into quadratic equations

2

Solving by Factorisation - Part 1

Learn the splitting the middle term method for factorisation. Practise finding two numbers with given sum and product. Solve Exercise 4.2 Q1 (all five parts). Revise zero product property: if ab = 0, then a = 0 or b = 0

3

Solving by Factorisation - Word Problems

Solve Exercise 4.2 Q2 (word problems from Q2 of Ex 4.1). Solve Exercise 4.2 Q3 to Q6 (number problems, ages, areas, speeds). Focus on setting up equations correctly before solving. Always verify answers by substituting back into the original problem context

4

Quadratic Formula and Discriminant

Memorise the quadratic formula x = [-b +- sqrt(b^2-4ac)]/(2a). Understand the discriminant D = b^2 - 4ac and its three cases. Solve Exercise 4.3 Q1 (find nature of roots and solve if roots exist). Practise computing discriminant for various equations

5

Nature of Roots and Finding k

Solve Exercise 4.3 Q2 (find k for equal roots). Solve Exercise 4.3 Q3, Q4, Q5 (real-life applications using discriminant). Practise previous year questions on finding k for specific root conditions. Review common PYQ patterns: equal roots, no real roots, distinct roots

6

Revision and Exam Practice

Solve 10-12 previous year questions from CBSE 2019-2026. Practise 2 case study questions and 5-6 MCQs on discriminant and nature of roots. Revise all formulas and write them from memory. Time yourself on word problems - aim for 5-7 minutes per 3-mark question

Common Mistakes to Avoid

  • ✗Forgetting to check if the equation is actually quadratic after simplification - Always simplify both sides first. If x^2 terms cancel out, the equation is linear, not quadratic. For example, (x+1)^2 = x^2 + 2x + 3 simplifies to a linear equation.
  • ✗Sign errors when computing the discriminant D = b^2 - 4ac - Be careful with negative values of b and c. For example, in 3x^2 - 5x - 2 = 0, b = -5 (not 5), so b^2 = 25. And c = -2, so -4ac = -4(3)(-2) = +24. D = 25 + 24 = 49.
  • ✗Accepting negative roots for real-world quantities like length, age, speed, or number of items - Always reject negative roots when the variable represents a physical quantity that must be positive. State this explicitly in your answer: 'Since speed cannot be negative, x = -32 is rejected.'
  • ✗Incorrectly splitting the middle term - choosing wrong pair of numbers - When splitting bx in ax^2 + bx + c = 0, find two numbers whose product equals ac (not just c) and whose sum equals b. For 2x^2 + x - 6 = 0: product = 2 x (-6) = -12, sum = 1. The numbers are 4 and -3, not 2 and -3.
  • ✗Setting k = 0 when finding conditions for equal roots, even though k = 0 makes the equation non-quadratic - When the leading coefficient contains k (e.g., kx^2 - 3kx + 9 = 0), always check that k is not equal to 0 after solving. If k = 0, the equation is no longer quadratic, so reject that value.
  • ✗Confusing 'no real roots' with 'no roots at all' and writing D = 0 means 'one root' - When D < 0, the equation has no real roots (it has complex roots, but that is beyond Class 10 scope). When D = 0, the equation has two equal real roots (not 'one root'). A quadratic always has exactly 2 roots counting multiplicity.

Tips for Board Exam

  • ✓Memorise the quadratic formula and discriminant conditions: Write x = [-b +- sqrt(b^2-4ac)]/(2a) at the top of your rough sheet. Know the three cases by heart: D > 0 (two distinct), D = 0 (two equal), D < 0 (no real roots). This covers MCQs and 1-mark questions instantly.
  • ✓For word problems, define the variable clearly before forming the equation: Write 'Let the speed = x km/h' or 'Let the breadth = x m' as the first line. Then express all other quantities in terms of x. This makes it easy for the examiner to follow your logic and awards you method marks even if the final answer is wrong.
  • ✓Always verify your answer by substituting back into the original equation: After finding the roots, substitute each root into the original equation to check. In the board exam, this verification step often carries 1/2 to 1 mark. For word problems, check that the answer makes sense in context (positive lengths, integer counts, etc.).
  • ✓For 'find k' questions, show the discriminant condition clearly: Write D = b^2 - 4ac = 0 (for equal roots) or D < 0 (for no real roots) explicitly. Then substitute values and solve for k. Do not skip steps - the method carries marks. Always check if k = 0 is valid when k appears in the coefficient of x^2.
  • ✓Attempt the MCQ on discriminant/nature of roots first in the exam: This is typically the easiest mark in the chapter. Just compute D = b^2 - 4ac and check the sign. It takes under 30 seconds and builds confidence. Common trap: the question gives the equation in non-standard form, so always rearrange to ax^2 + bx + c = 0 first.

Practice Quadratic Equations with AI Tutor

Stuck on a question? Ask our AI tutor to explain step-by-step in English or Hindi. Choose from 10 learning modes - Lecture, Quiz, Flashcards, and more. Free for 3 chapters per day.

Start Quadratic Equations - Choose Your Mode

About Quadratic Equations - Class 10 Mathematics

What Are Quadratic Equations?

A quadratic equation is a polynomial equation of degree 2. Its standard form is ax^2 + bx + c = 0, where a, b, and c are real numbers and a is not equal to 0. The condition a is not 0 is critical because if a were 0, the equation would become linear (bx + c = 0), not quadratic. The values of x that satisfy the equation are called its roots or solutions. A quadratic equation always has exactly two roots, which may be real or complex, equal or distinct.

Examples of quadratic equations include 2x^2 + 3x - 5 = 0, x^2 - 4 = 0, and x^2 + 1 = 0. Not every equation that looks quadratic actually is one. For instance, (x + 1)^2 = x^2 + 2x + 3 simplifies to 2x - 2 = 0, which is linear because the x^2 terms cancel out. Exercise 4.1 of the NCERT textbook focuses on this skill of checking whether a given equation is quadratic after simplification. Students must expand both sides, rearrange to one side, and check if the highest power of x is 2.

Methods of Solving Quadratic Equations

The CBSE 2026-27 syllabus covers two methods for solving quadratic equations: factorisation and the quadratic formula.

Factorisation (also called the splitting the middle term method) works by expressing the quadratic as a product of two linear factors. For ax^2 + bx + c = 0, find two numbers p and q such that p + q = b and p x q = a x c. Then split the middle term bx into px + qx and factor by grouping. For example, to solve x^2 - 3x - 10 = 0, we need two numbers with product -10 and sum -3, which are -5 and 2. Splitting: x^2 - 5x + 2x - 10 = x(x-5) + 2(x-5) = (x+2)(x-5) = 0, giving x = -2 or x = 5.

The zero product property is fundamental here: if the product of two expressions is zero, then at least one of them must be zero. This is why we can set each factor equal to zero.

The Quadratic Formula

The quadratic formula provides a universal method to find the roots of any quadratic equation ax^2 + bx + c = 0:

x = [-b +- sqrt(b^2 - 4ac)] / (2a)

This formula is derived by the method of completing the square (which has been removed from the CBSE syllabus but the formula itself is retained). The formula works for every quadratic equation, whether or not it can be factorised easily. For example, 2x^2 - 6x + 3 = 0 cannot be neatly factorised, but using the formula with a = 2, b = -6, c = 3 gives x = [6 +- sqrt(36-24)]/4 = [6 +- 2sqrt(3)]/4 = (3 +- sqrt(3))/2.

Students should memorise this formula thoroughly, as it is used directly in board exam questions and also as a conceptual foundation for understanding the discriminant.

Discriminant and Nature of Roots

The discriminant of a quadratic equation ax^2 + bx + c = 0 is D = b^2 - 4ac. It determines the nature of roots without actually solving the equation:

  • If D > 0: The equation has two distinct real roots. The roots are x = (-b + sqrt(D))/(2a) and x = (-b - sqrt(D))/(2a). If D is a perfect square, the roots are rational; otherwise, they are irrational.
  • If D = 0: The equation has two equal real roots (also called repeated or coincident roots). Both roots equal x = -b/(2a).
  • If D < 0: The equation has no real roots. The roots are complex numbers (beyond the scope of Class 10).

This concept is heavily tested in CBSE board exams. The most common question type is: "Find the value of k for which the equation has equal roots." To solve these, set D = 0 and solve for k. Always check that the value of k does not make the leading coefficient zero, as that would make the equation non-quadratic.

For example, if kx(x - 2) + 6 = 0 must have equal roots: kx^2 - 2kx + 6 = 0. D = 4k^2 - 24k = 0, giving k(k-6) = 0. Since k = 0 would eliminate the x^2 term, only k = 6 is valid.

Word Problems

Word problems are a major component of this chapter, contributing 3-4 marks in most board exams. The key is to translate the English statement into a mathematical equation. Common types include:

  1. Number problems: Find two consecutive integers whose product is 306. Let them be x and x+1: x(x+1) = 306 gives x^2 + x - 306 = 0.

  2. Geometry problems: The altitude of a right triangle is 7 cm less than its base, and the hypotenuse is 13 cm. Using Pythagoras theorem: base^2 + altitude^2 = 13^2 leads to a quadratic equation.

  3. Speed-distance-time problems: A train travels 480 km at uniform speed. If speed were 8 km/h less, it would take 3 hours more. Setting up: 480/(x-8) - 480/x = 3 leads to x^2 - 8x - 1280 = 0.

  4. Age problems: Rohan's mother is 26 years older. Product of their ages 3 years hence is 360. This gives (x+3)(x+29) = 360, leading to a quadratic in x.

  5. Area problems: A rectangular park has perimeter 80 m and area 400 sq m. This leads to x^2 - 40x + 400 = 0, where D = 0 means the park is actually a square.

For each word problem, always define the variable clearly, form the equation, solve it, and then interpret the result in the context of the problem. Reject any root that does not make physical sense (negative length, speed, age, etc.).

How Tested in Board Exams

Quadratic equations is one of the most important chapters for CBSE Class 10, appearing in every board exam with 6-8 marks. The typical question distribution is:

  • 1 MCQ (1 mark): Usually on discriminant, nature of roots, or a direct computation like finding sum/product of roots.
  • 1 Short Answer (2-3 marks): Solving by factorisation, finding k for equal roots, or finding the discriminant.
  • 1 Long Answer/Word Problem (3-4 marks): Speed-distance, area, or age problem that requires forming and solving a quadratic equation.

Since CBSE 2023, case study questions involving quadratic equations have become common. These present a real-life scenario (swimming pool dimensions, train journey) and ask 4 sub-questions worth 1+1+2 marks. The questions test equation formation, solving by factorisation, and discriminant computation.

Assertion-reason questions test conceptual understanding of discriminant conditions and the definition of quadratic equations. Common traps include: D = 0 means "one root" (wrong - it means two equal roots), and forgetting to reject k = 0.

Tips for Full Marks

  1. Always write the standard form ax^2 + bx + c = 0 before applying any formula. Identify a, b, and c explicitly.
  2. For factorisation questions, show all steps: splitting, grouping, factoring, and stating both roots.
  3. For word problems, the first line should define the variable: "Let the speed = x km/h." The last line should state the answer in words: "Therefore, the speed of the train is 40 km/h."
  4. For discriminant questions, write D = b^2 - 4ac with values substituted, then simplify step by step.
  5. For "find k" questions, clearly state the condition (D = 0 for equal roots, D > 0 for real roots, etc.) before solving.
  6. Verify your answers: substitute roots back into the original equation, and check that word problem answers make physical sense.

Connection to Other Chapters

Quadratic equations connect to several other chapters in Class 10 Mathematics:

  • Chapter 2 (Polynomials): The roots of ax^2 + bx + c = 0 are the zeroes of the polynomial p(x) = ax^2 + bx + c. The relationships sum of roots = -b/a and product of roots = c/a are directly from the polynomial chapter.
  • Chapter 3 (Pair of Linear Equations): Some problems can be reduced to quadratic equations after substitution. Understanding both linear and quadratic solving methods gives a complete toolkit.
  • Chapter 5 (Arithmetic Progressions): Problems like "the sum of first n terms equals a certain value" can lead to quadratic equations in n.
  • Chapter 6 (Triangles) and Chapter 8 (Trigonometry): Geometry problems involving Pythagoras theorem or trigonometric identities sometimes reduce to quadratic equations.

Mastering this chapter builds algebraic fluency that is essential for Class 11 and 12 Mathematics, where quadratic expressions appear in coordinate geometry, calculus, and complex numbers.

Frequently Asked Questions

What is a quadratic equation in standard form?

A quadratic equation in standard form is ax^2 + bx + c = 0, where a, b, and c are real numbers and a is not equal to 0. The term ax^2 is the quadratic term, bx is the linear term, and c is the constant term. The condition a is not 0 ensures the equation is of degree 2.

How do you solve a quadratic equation by factorisation?

To solve by factorisation: (1) Write the equation in standard form ax^2 + bx + c = 0. (2) Find two numbers whose product is a x c and whose sum is b. (3) Split the middle term using these two numbers. (4) Group and factor. (5) Use the zero product property to find the roots. For example, x^2 - 7x + 12 = 0: numbers with product 12 and sum -7 are -3 and -4. So (x-3)(x-4) = 0, giving x = 3 or x = 4.

What is the quadratic formula and when should you use it?

The quadratic formula is x = [-b +- sqrt(b^2 - 4ac)]/(2a). Use it when the quadratic equation cannot be easily factorised, or when the question specifically asks for it. It works for any quadratic equation. First identify a, b, c from the standard form, compute the discriminant D = b^2 - 4ac, and substitute into the formula.

What is the discriminant and how does it determine the nature of roots?

The discriminant D = b^2 - 4ac tells you about the roots without solving. If D > 0, there are two distinct real roots. If D = 0, there are two equal real roots (repeated root). If D < 0, there are no real roots. In CBSE board exams, questions asking for the 'nature of roots' expect you to compute D and state one of these three cases.

How do you find the value of k for which a quadratic equation has equal roots?

For equal roots, set the discriminant equal to zero: D = b^2 - 4ac = 0. Substitute the values of a, b, c (which will contain k) into this equation and solve for k. Important: if k appears in the coefficient of x^2, check that the value of k does not make that coefficient zero, as the equation would then not be quadratic.

How many marks does Chapter 4 Quadratic Equations carry in the CBSE Class 10 board exam?

Quadratic Equations is part of the Algebra unit, which carries 20 marks total in the CBSE 2026-27 exam. Chapter 4 typically contributes 6-8 marks through a combination of 1 MCQ (1 mark), 1 short answer on factorisation or discriminant (2-3 marks), and 1 word problem (3-4 marks). Case study questions from this chapter are also common.

What are the most common types of word problems from this chapter in CBSE exams?

The most common word problem types are: (1) Speed-distance-time problems (train/car journey where reduced speed increases time), (2) Area problems (rectangular field/park with given area and a length-breadth relationship), (3) Number problems (consecutive integers with given product or sum of squares), (4) Age problems (product of ages after certain years). In recent years (2023-2026), speed-distance and area problems have been the most frequently asked.

What is the difference between roots of a quadratic equation and zeroes of a quadratic polynomial?

They are the same values but viewed differently. The roots of ax^2 + bx + c = 0 are the values of x that satisfy the equation. The zeroes of p(x) = ax^2 + bx + c are the values of x for which p(x) = 0. Both are found using the same methods. Chapter 2 (Polynomials) discusses the relationship between zeroes and coefficients: sum of zeroes = -b/a and product of zeroes = c/a.

More Maths NCERT Solutions - Class 10