NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1 will significantly improve the problem-solving skills of students on the topic of quadratic equations. The exercise gives a basic introduction to the quadratic equations and their formation, which is crucial for understanding this chapter. Understanding this chapter will significantly help students build a strong base for their future mathematics or other fields of study. This is why, here, you can find the brief explanation of key concepts of the exercise, along with NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1 PDF.
Unlike linear equations, quadratic equations are those with the degree or the highest power of the variable “2 or square”. The equations have a magnificent history with numerous contributions from famous mathematicians of the time. To mention some, Babylonians were the first to solve a quadratic equation, and then came other mathematicians like Euclid, Brahmagupta, and many more. The equations have only one variable (mostly x) with two solutions. A quadratic equation is written in a standard form as:
Here, a and b are the real number coefficients of the single variable, such that a0, while c is the constant term of the quadratic equation. These equations are used not only to solve the chapter's numerical problems but also to solve real-world scenarios.
Exercise 4.1 gives a resourceful intro to the interesting realm of solving quadratic equations. Let’s explore some valuable concepts used to solve this exercise:
Formation of quadratic equations from a given word problem is the first and the most basic thing a student needs to learn to thoroughly understand these equations. This can be done by following the below-mentioned approach:
And now you have your required equation.
Another important concept of the exercise is to verify the quadratic behaviour of the given equation. In some cases, you will be provided with non-simplified quadratic equations. For these questions, you need to first simplify these equations. If, after the simplification, the highest power of the equation becomes two, then the given equation will be a quadratic one.
In Exercise 4.1, students will stumble across the following questions, each solved with a different approach, which includes:
Begin practising the concepts today from NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1, and learn a deeper understanding of quadratic equations that will be helpful for both exams and future mathematical challenges.
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
(Session 2025 - 26)