Chapter 4 Class 10 Maths Quadratic Equations represented in the form of ax² +bx+c=0 where x is the unknown variable which can be anything from a, b, and c are constants, and a≠0. However, the right side is zero. In ax² +bx+c=0, ax2 is the non-zero constant, bx is constant and c is constant. Since ax² is a bigger term, it is known to be the quadratic equation. However, the term bx is represented as the linear term, and c is the constant term whose value is 0. 

 

Let’s understand this through an example:

Check whether the following are quadratic equations: 

(x – 2) (x + 1) = (x – 1) (x + 3)

 

By multiplication

⇒ x² – x – 2 = x² + 2x – 3

⇒ 3x – 1 = 0

Since the above equation is not in the form of ax² + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

 

Chapter 4 Class 10 Maths exercise 4.2 discusses the solution of quadratic equations by factorization. 

• • • Transform the equation to bring one side equal to zero 
    • Start the factorization process of the non-zero side 
    • Solve the resulting equation

Let’s understand the factorization of quadratic equations through an example. 

Solve the equation,

 

 

 

In order to convert the quadratic equation in the form of ax2 + bx + c, the most common method used is by completing the square. It can be done by rearranging the expression obtained after completing the square. Let’s understand this via an example-

Example - Find the roots of the following quadratic equations by Completing the Square:

Nature of roots is a combination of real roots, imaginary or unreal roots, and irrational roots. To find out the nature of roots of a quadratic equation, the first step is to calculate the roots of an equation. Further roots is classified into three major parts: 

1. Real and distinct roots, where the discriminant is positive. 

2. Real and equal roots, where the discriminant is equal to zero. 

3. Complex roots, where the discriminant is negative.

Let’s understand calculating nature of roots through an example: 

Find the nature of the roots of quadratic equations-

 

1- Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x – 3)

(ii) x² – 2x = (–2) (3 – x)

Solutions:

(i) Given,

(x + 1)² = 2(x – 3)

By using the formula for (a+b)² = a²+2ab+b²

⇒ x² + 2x + 1 = 2x – 6

⇒ x² + 7 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(ii) Given, x² – 2x = (–2) (3 – x)

⇒ x² – 2x = -6 + 2x

⇒ x² – 4x + 6 = 0

Since the above equation is in the form of ax² + bx + c = 0.

Therefore, the given equation is a quadratic equation.

 

2- Find two numbers whose sum is 27 and the product is 182.

Solution:

Let us say, the first number is x and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x² – 27x – 182 = 0

⇒ x² – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

 

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

Solution:

Let us say, the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base2 + Altitude2 = Hypotenuse2

∴ x² + (x – 7)² = 132

⇒ x² + x² + 49 – 14x = 169

⇒ 2x² – 14x – 120 = 0

⇒ x² – 7x – 60 = 0

⇒ x² – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

4- 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.

Solution:

Let us say the speed of the train is x km/hr.

Time taken to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360/x – 1) = 360

⇒ 360 – x + 1800/x-5 = 360

⇒ x² + 5x – 1800 = 0

⇒ x² + 45x – 40x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

As we know, the value of speed cannot be negative.

Therefore, the speed of the train is 40 km/h.

5- Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:

Let the breadth of the mango grove be l.

Length of the mango grove will be 2l.

Area of mango grove = (2l) (l)= 2l2

2l2 = 800

l2 = 800/2 = 400

l2 – 400 =0

Comparing the given equation with ax² + bx + c = 0, we get

a = 1, b = 0, c = 400

As we know, Discriminant = b² – 4ac

=> (0)² – 4 × (1) × ( – 400) = 1600

Here, b² – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

l = ±20

As we know, the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

 

1. How many exercises are there in chapter 4 Class 10? 

Chapter 4 Class 10 Maths Quadratic Equation have 4 exercises- 

• • Quadratic Equations (Introduction)  

  • Solution of Quadratic Equations by Factorization

  • Solution of a Quadratic Equation by Completing the Square 

  • Nature of Roots  

 

2. What is a quadratic equation for class 10th?

A Quadratic Equation in the variable x is an equation of the form ax²+ bx + c = 0, where a, b, c are real numbers, a ≠ 0. That is, ax²+ bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation. 

 

3. List the important concepts of class 10 Quadratic Equations. 

The major concepts explained in Chapter 4 Class 10 Maths Quadratic Equations are the meaning of quadratic equations, finding its roots through factorization and completing squares, and the nature of the roots.