This chapter introduces the students to quadratic equations by showing how to solve the equations, along with factoring, use of the formula for quadratic equations and determining the nature of roots. The objective is of vast importance as these very ideas form a basis for advanced mathematics and problem-solving in science and engineering. The step-by-step solutions with detailed explanations in ALLEN Study Materials make concepts easy to grasp and interpret in a clear and convenient way.
These NCERT Solutions for Class 10 Maths Chapter 4 strictly adhere to the instructions of the CBSE board. They also help you score well in exams and tests. The NCERT Solutions for Class 10 quadratic equations in are, therefore, of utmost importance for a student who wants to excel at their exams.
Students can download NCERT Solutions Class 10 Mathematics Chapter 4 from the below table in PDF format.
Chapter 4, Quadratic Equations, focuses on equations of the form , where a ≠ 0 . Students learn various methods to solve these equations, including factorization, completing the square, and the quadratic formula. The chapter also covers the discriminant to determine the nature of roots and their real-life applications. Mastering quadratic equations is essential for solving advanced mathematical problems and understanding key concepts in algebra.
Definition of a Quadratic Equation: A quadratic equation is a second-degree polynomial equation in the form: where a, b, and c are constants, and a ≠ 0.
Examples of Quadratic Equations:
The standard form is , where:
Important Point: a ≠ 0, because if a = 0, the equation becomes linear, not quadratic.
Roots of a Quadratic Equation: The solutions of the quadratic equation are called roots of a quadratic equation.
1. Check whether the following are quadratic equations :
(i) (x+1)² = 2(x-3)
(ii) x² - 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) x² + 3x + 1 = (x-2)²
(vii) (x+2)³ = 2x(x²-1)
(viii) x³ - 4x² - x + 1 = (x-2)³
Solution:
(i) (x+1)² = 2(x-3)
=> x² + 2x + 1 = 2x - 6
=> x² + 2x - 2x + 1 + 6 = 0
=> x² + 0x + 7 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
(ii) x² - 2x = (-2)(3-x)
=> x² - 2x = -6 + 2x
=> x² - 4x + 6 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
(iii) (x-2)(x+1) = (x-1)(x+3)
=> x² + x - 2x - 2 = x² + 3x - x - 3
=> x² - x - 2 = x² + 2x - 3
=> -x - 2x - 2 + 3 = 0
=> -3x + 1 = 0 or 3x - 1 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(iv) (x-3)(2x+1) = x(x+5)
=> 2x² - 5x - 3 = x² + 5x
=> x² - 10x - 3 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
(v) (2x-1)(x-3) = (x+5)(x-1)
=> 2x² - 7x + 3 = x² + 4x - 5
=> x² - 11x + 8 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
(vi) x² + 3x + 1 = (x-2)²
=> x² + 3x + 1 = x² + 4 - 4x
=> 7x - 3 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(vii) (x+2)³ = 2x(x²-1)
=> x³ + 3 × x × 2 × (x+2) + 2³ = 2x(x²-1)
=> x³ + 6x(x+2) + 8 = 2x³ - 2x
=> x³ + 6x² + 12x + 8 = 2x³ - 2x
=> -x³ + 6x² + 14x + 8 = 0
=> x³ - 6x² - 14x - 8 = 0
It is not of the form ax² + bx + c = 0
Hence, the given equation is not a quadratic equation.
(viii) x³ - 4x² - x + 1 = (x-2)³
=> x³ - 4x² - x + 1 = x³ - 8 - 6x² + 12x
=> 2x² - 13x + 9 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². 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.
Solution:
(i) Let the breadth of the rectangular plot be x meters.
Then, the length of the plot is (2x + 1) meters.
The area of the plot is given as 528 m².
Therefore, x × (2x + 1) = 528
=> 2x² + x = 528
=> 2x² + x - 528 = 0
(ii) Let the two consecutive positive integers be x and x + 1.
It is given that their product is 306.
Therefore, x(x + 1) = 306
=> x² + x = 306
=> x² + x - 306 = 0
(iii) Let Rohan's present age be x years.
Then, the present age of Rohan's mother is (x + 26) years.
After 3 years, Rohan's age will be (x + 3) years, and his mother's age will be (x + 26 + 3) = (x + 29) years.
It is given that the product of their ages after 3 years is 360.
Therefore, (x + 3)(x + 29) = 360
=> x² + 29x + 3x + 87 = 360
=> x² + 32x + 87 = 360
=> x² + 32x + 87 - 360 = 0
=> x² + 32x - 273 = 0
(iv) Let the speed of the train be x km/h.
Time taken to travel 480 km = 480/x hours.
In the second condition, the speed of the train is (x - 8) km/h.
It is given that the train takes 3 hours more to cover the same distance.
Therefore, the time taken to travel 480 km is (480/x + 3) hours.
Speed × Time = Distance
(x - 8)(480/x + 3) = 480
=> 480 + 3x - 3840/x - 24 = 480
=> 3x - 3840/x = 24
=> 3x² - 3840 = 24x
=> 3x² - 24x - 3840 = 0
=> x² - 8x - 1280 = 0
3. Find the roots of the following quadratic equations by factorisation :
(i) x² - 3x - 10 = 0
(ii) 2x² + x - 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2x² - x + 1/8 = 0
(v) 100x² - 20x + 1 = 0
Solution:
(i) x² - 3x - 10 = 0
=> x² - 5x + 2x - 10 = 0
=> x(x - 5) + 2(x - 5) = 0
=> (x + 2)(x - 5) = 0
=> x + 2 = 0 or x - 5 = 0
=> x = -2 or x = 5
Hence, the roots are -2 and 5.
(ii) 2x² + x - 6 = 0
=> 2x² + 4x - 3x - 6 = 0
=> 2x(x + 2) - 3(x + 2) = 0
=> (x + 2)(2x - 3) = 0
=> x + 2 = 0 or 2x - 3 = 0
=> x = -2 or x = 3/2
(v) 100x² - 20x + 1 = 0
=> (10x)² - 2 × 10x × 1 + 1² = 0
=> (10x - 1)² = 0
=> Both roots are given by 10x - 1 = 0, i.e. x = 1/10.
Hence, the roots are 1/10, 1/10.
4. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² - 3x + 5 = 0
(ii) 3x² - 4√3x + 4 = 0
(iii) 2x² - 6x + 3 = 0
Solution:
(i) 2x² - 3x + 5 = 0
Here a = 2, b = -3 and c = 5
Therefore, discriminant D = b² - 4ac
= (-3)² - 4 × 2 × 5
= 9 - 40 = -31
As D < 0
Hence, no real root.
(ii) 3x² - 4√3x + 4 = 0
Here a = 3, b = -4√3 and c = 4
Therefore, discriminant D = b² - 4ac
= (-4√3)² - 4(3)(4) = 48 - 48 = 0
As D = 0
Hence, two equal real roots.
Now, the roots are
= (-b ± √D) / 2a = (4√3 ± 0) / (2 × 3) = 2/√3
Hence, the roots are 2/√3 and 2/√3.
(iii) 2x² - 6x + 3 = 0
a = 2, b = -6 and c = 3
Discriminant D = b² - 4ac
= (-6)² - 4(2)(3)
= 36 - 24 = 12
=> D > 0
Hence, roots are distinct and real.
The roots are
x = (-b ± √D) / 2a
= (6 ± √12) / (2 × 2) = (6 ± 2√3) / 4 = (3 ± √3) / 2
Therefore, the roots are (3 + √3) / 2 and (3 - √3) / 2.
5. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x - 2) + 6 = 0
Solution:
(i) 2x² + kx + 3 = 0
Here a = 2, b = k and c = 3
Therefore, D = b² - 4ac
= k² - 4 × 2 × 3 = k² - 24
Two roots will be equal if D = 0
=> k² - 24 = 0
=> k = ±√24
=> k = ±2√6
(ii) kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Here, a = k, b = -2k and c = 6
Therefore, D = b² - 4ac
= (-2k)² - 4(k)(6) = 4k² - 24k
Two roots will be equal if D = 0
=> 4k² - 24k = 0
=> 4k(k - 6) = 0
Either 4k = 0 or k - 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'. Hence, k = 6.
Quadratic equations have many applications in real life, including:
(Session 2025 - 26)