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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
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.
1.0Download NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations : Free PDF
Students can download NCERT Solutions Class 10 Mathematics Chapter 4 from the below table in PDF format.
2.0NCERT Exercise-wise Solutions for Class 10 Maths Chapter 4 Quadratic Equations :Overview
- NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1: Exercise 4.1 of NCERT Class 10 Maths Chapter 4 contains basic concepts about quadratic equations. Questions have been provided to the students to identify quadratic equations and word problems associated with creating the quadratic equations.
- NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2: Exercise 4.2 of NCERT Class 10 Maths Chapter 4 deals with solving quadratic equations by factoring. In this exercise, students are asked to factorize quadratic equations and find their roots.
- NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3: Exercise 4.3 of NCERT Class 10 Maths Chapter 4 deals with the nature of the roots of any quadratic equations. Students are asked to determine the discriminant of quadratic equations.
3.0Important Topics - Class 10 Maths Chapter 4 Quadratic Equations Solutions
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.
4.0Class 10 Maths Chapter 4: Quadratic Equations - Key Topics
- Quadratic Equations: Equations of the form (where a ≠ 0).
- Solving by Factorization: Factoring the quadratic equation to find x.
- Solving by Completing the Square: Converting the equation into a perfect square form and solving.
- Quadratic Formula: to find the roots.
- Nature of Roots: Using the discriminant to determine the number and type of roots.
- Applications: Solving real-life problems using quadratic equations.
5.0General Outline for Class 10 Maths Chapter 4 Quadratic Equation
Introduction
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:
Standard Form of a Quadratic Equation
The standard form is , where:
- a is the coefficient of ,
- b is the coefficient of x,
- c is the constant term.
Important Point: a ≠ 0, because if a = 0, the equation becomes linear, not quadratic.
Solutions of a Quadratic Equation
Roots of a Quadratic Equation: The solutions of the quadratic equation are called roots of a quadratic equation.
Methods to find the roots:
- Factorization Method:
- Factor the quadratic expression and solve for x.
- Example: factors to (x − 2)(x − 3) = 0, so the roots are x = 2 and x = 3.
- Quadratic Formula:
- The general solution is given by the quadratic formula:
- This formula can be used for any quadratic equation to find the roots.
- Completing the Square Method: Rewrite the equation in the form , then solve for x.
Discriminant
- Discriminant Formula: The discriminant of a quadratic equation is
- Interpretation of the Discriminant:
- If , there are two real and distinct roots.
- If , there is one real and repeated root (also called a double root).
- If , there are no real roots (the roots are complex).
- Nature of Roots
- Real and Distinct Roots: Occur when the discriminant is positive .
- Real and Equal Roots: Occur when the discriminant is zero
- Complex Roots: Occur when the discriminant is negative
Applications of Quadratic Equations
- Word Problems: Quadratic equations are used to solve problems involving areas, speed, time, profit-loss, etc.
- Example: A rectangle’s area is given by a quadratic equation, or solving projectile motion problems.
- Real-Life Applications:
- Physics (motion, trajectories),
- Economics (profit-maximization problems),
- Engineering (structural calculations), and more.
Sum and Product of Roots
- For a quadratic equation , the sum and product of the roots ( and ) are given by:
- Sum of roots:
- Product of roots:
6.0Sample NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
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.
7.0Key Benefits of Quadratic Equations
- Foundation for Advanced Math: Builds skills for higher algebra and calculus.
- Problem-Solving: Enhances critical thinking and analytical skills.
- Real-World Applications: Used in physics, economics, engineering, and finance.
- Understanding Relationships: Models non-linear relationships in various fields.
- Graphical Interpretation: Helps understand parabolic curves and optimization.
- Competitive Exams: Important for preparing for various entrance exams.
- Practical Problem Solving: Solves real-life problems involving areas, motion, and more.
- Improves Algebra: Strengthens skills like factoring and solving equations.
- Multiple Methods: Solving by different approaches deepens understanding.
- Boosts Confidence: Increases confidence in math and its practical uses.
8.0Applications of Quadratic Equations
Quadratic equations have many applications in real life, including:
- Projectile Motion: Finding the trajectory of moving objects, such as a ball or rocket, along with maximum height and range.
- Business Profit and Loss: This tool helps find the optimum production levels which can maximize profit or minimize the cost.
- Area Calculations: In all such problems where the area is given, dimensions of geometric shapes such as rectangles and squares are found.
- Speed, Distance, and Time Problems: Solve for the meeting point or overtaking time of two objects in motion at different speeds.
- Optimization Problems: It optimizes maximization or minimization problems in some quantity such as crop yield or revenue in industrial applications.