Class 10 NCERT Solutions- Chapter 4 Quadratic Equations - Exercise 4.1

Chapter 4 of the Class 10 NCERT Mathematics textbook covers Quadratic Equations a fundamental topic in algebra. This chapter helps students understand how to solve quadratic equations using various methods and apply these solutions to real-world problems. Exercise 4.1 focuses on solving basic quadratic equations reinforcing the concepts introduced in the chapter.

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree 2 which can be written in the standard form:

ax₂+bx+c=0

where a, b, and c are constants, and 𝑎≠0. The solutions to the quadratic equation are the values of the x that satisfy the equation. These solutions can be found using various methods including factoring completing the square and using the quadratic formula:

Question 1. Check whether the following are quadratic equations :

(i) (x + 1)² = 2(x – 3) 

Solution:

Here,

LHS = (x + 1)²

= x² + 2x + 1        (Using identity (a+b)² = a² + 2ab + b²)

and, RHS = 2(x–3)

= 2x – 6

As, LHS = RHS

x² + 2x + 1 = 2x – 6

x² + 7 = 0 .............(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

(ii) x² – 2x = (–2) (3 – x)

Solution:

Here,

LHS = x² – 2x

and, RHS = (–2) (3 – x)

= 2x–6

As, LHS = RHS

x² – 2x = 2x – 6

x² – 4x + 6 = 0 .............(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3) 

Solution:

Here,

LHS = (x – 2)(x + 1)

= x² + (–2+1)x + (–2)(1)    (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² - x - 2

and, RHS = (x – 1)(x + 3) 

= x² + (–1+3)x + (–1)(3)   (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² + 2x - 3

As, LHS = RHS

x² – x – 2 = x² + 2x - 3

3x – 1 = 0 .............(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 because (a=0).

Hence, the equation is equation because highest power of x is 1.

(iv) (x – 3)(2x +1) = x(x + 5)

Solution:

Here,

LHS = (x – 3)(2x +1)

= 2x² + x +(–3)(2x) + (–3)(1)

= 2x² – 5x - 3

and, RHS = x(x + 5)

= x² + 5x

As, LHS = RHS

2x² – 5x - 3 = x² + 5x

x² – 10x - 3 = 0 .............(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1) 

Solution:

Here,

LHS = (2x – 1)(x – 3)

= 2x² + (2x)(–3) +(–1)(x) + (–1)(–3)

= 2x² – 7x + 3

and, RHS = (x + 5)(x – 1) 

= x² + 5(x) + (–1)(x) + (5)(–1)     (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² + 4(x) – 5

As, LHS = RHS

2x² – 7x + 3 = x² + 4(x) – 5

x² – 11x + 8 = 0 .............(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

(vi) x² + 3x + 1 = (x – 2)²

Solution:

Here,

LHS = x² + 3x + 1

and, RHS = (x – 2)²

= x² – 4x + 4      (Using identity (a–b)² = a² – 2ab + b²)

As, LHS = RHS

x² + 3x + 1 = x² – 4x + 4

7x – 3 = 0 .............(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 because (a=0).

Hence, the equation is equation because highest power of x is 1.

(vii) (x + 2)³ = 2x (x² – 1) 

Solution:

Here,

LHS = (x + 2)³

= x³ + 2³ + 3x(2)(x+2)   (Using identity (x+y)³ = x³ + y³ + 3xy(x+y))

= x³ + 8 + 6x² +12x

and, RHS = 2x (x² – 1) 

= 2x³  – 2x      

As, LHS = RHS

x³ + 8 + 6x² +12x = 2x³  – 2x 

x³ – 6x² – 14x – 8 = 0 .............(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 3.

(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution:

Here,

LHS = x³ – 4x² – x + 1

and, RHS = (x – 2)³

= x³ – 2³ – 3x(2)(x–2)     (Using identity (x–y)³ = x³ – y³ – 3xy(x-y))

= x³ – 8 – 6x² +12x

As, LHS = RHS

x³ – 4x² – x + 1 = x³ – 8 – 6x² +12x

2x² – 13x + 9 = 0 .............(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

Question 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 meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:

Let's consider,

Breadth of the rectangular plot = b m

Then, length of the plot = (2b + 1) m.

As, Area of rectangle = length × breadth

528 m² = (2x + 1) × x

2x² + x =528

2x² + x – 528 = 0 ...................(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x² + x – 528 = 0, which is the required representation of the problem mathematically.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solution:

Let's consider,

The first integer number = x

Next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1)

x² + x = 306

x² + x – 306 = 0 ...................(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

Therefore, the two integers x and x+1, satisfies the quadratic equation, x² + x – 306 = 0, which is the required representation of the problem mathematically.

(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.

Solution:

Let's consider,

Age of Rohan’s = x  years

So, Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 – 360 = 0

x² + 32x – 273 = 0 ...................(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is equation because highest power of x is 2.

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x² + 32x – 273 = 0, which is the required representation of the problem mathematically.

(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:

Let's consider,

The speed of train = x  km/hr

Time taken to travel 480 km = (480/x) hr

Time = Distance / Speed

Here, According to the given condition,

The speed of train = (x – 8) km/hr

As, the train will take 3 hours more to cover the same distance.

Hence, Time taken to travel 480 km = 480/(x+3) km/hr

As,

Speed × Time = Distance

(x – 8)(480/(x + 3) = 480

480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

3x² – 8x – 1280 = 0 ...................(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the speed of the train, satisfies the quadratic equation, 3x² – 8x – 1280 = 0, which is the required representation of the problem mathematically.

Read More:

• Quadratic Equations| Methods for solving Quadratic Equation and Examples
• Quadratic Formula

Summary

Exercise 4.1 focuses on identifying quadratic equations and finding their solutions. It covers the standard form of quadratic equations (ax² + bx + c = 0, where a ≠ 0), distinguishing between quadratic and non-quadratic equations, and solving simple quadratic equations by factorization. Students learn to recognize the coefficients a, b, and c in quadratic equations and practice solving equations where one solution is provided to find the other solution.