Class 10 Math Ch 4 Quadratic Equations Ex 4.1 NCERT Solution

In Exercise 4.1 of Chapter 4 – Quadratic Equations, students learn how to recognize and form quadratic equations from given word problems. This exercise helps build the foundation by understanding the standard form of a quadratic equation and identifying quadratic expressions. It focuses on converting real-life situations into quadratic equations.

Class 10 Math Ch 4 Quadratic Equations Ex 4.1 – Textbook Answers

Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)²=2(x-3)
(ii) x – 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)

Expand both sides:
On the left side:
(x + 1)² = (x + 1)(x + 1) = x² + 2x + 1

On the right side:
2(x – 3) = 2x – 6

Now, the equation becomes:
x² + 2x + 1 = 2x – 6

Rearrange it into standard form:
x² + 2x + 1 – 2x + 6 = 0
Simplify:
x² + 7 = 0

This is a quadratic equation because it has the highest degree of 2.

(ii) x – 2x = (-2)(3 – x)

Simplify both sides:
On the left side:
x – 2x = -x

On the right side:
(-2)(3 – x) = -6 + 2x

Now, the equation becomes:
-x = -6 + 2x

Rearrange it into standard form:
-x – 2x = -6
Simplify:
-3x = -6

This is a linear equation (degree 1), not a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

Expand both sides:
On the left side:
(x – 2)(x + 1) = x² + x – 2x – 2 = x² – x – 2

On the right side:
(x – 1)(x + 3) = x² + 3x – x – 3 = x² + 2x – 3

Now, the equation becomes:
x² – x – 2 = x² + 2x – 3

Rearrange it into standard form:
x² – x – 2 – x² – 2x + 3 = 0
Simplify:
-3x + 1 = 0

This is a linear equation (degree 1), not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)

Expand both sides:
On the left side:
(x – 3)(2x + 1) = 2x² + x – 6x – 3 = 2x² – 5x – 3

On the right side:
x(x + 5) = x² + 5x

Now, the equation becomes:
2x² – 5x – 3 = x² + 5x

Rearrange it into standard form:
2x² – 5x – 3 – x² – 5x = 0
Simplify:
x² – 10x – 3 = 0

This is a quadratic equation because the highest degree is 2.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Expand both sides:
On the left side:
(2x – 1)(x – 3) = 2x² – 6x – x + 3 = 2x² – 7x + 3

On the right side:
(x + 5)(x – 1) = x² – x + 5x – 5 = x² + 4x – 5

Now, the equation becomes:
2x² – 7x + 3 = x² + 4x – 5

Rearrange it into standard form:
2x² – 7x + 3 – x² – 4x + 5 = 0
Simplify:
x² – 11x + 8 = 0

This is a quadratic equation because the highest degree is 2.

(vi) x² + 3x + 1 = (x – 2)²

Expand the right side:
(x – 2)² = (x – 2)(x – 2) = x² – 4x + 4

Now, the equation becomes:
x² + 3x + 1 = x² – 4x + 4

Rearrange it into standard form:
x² + 3x + 1 – x² + 4x – 4 = 0
Simplify:
7x – 3 = 0

This is a linear equation (degree 1), not a quadratic equation.

(vii) (x + 2)³ = 2x(x² – 1)

Expand both sides:
On the left side:
(x + 2)³ = (x + 2)(x + 2)(x + 2) = x³ + 6x² + 12x + 8

On the right side:
2x(x² – 1) = 2x³ – 2x

Now, the equation becomes:
x³ + 6x² + 12x + 8 = 2x³ – 2x

Rearrange it into standard form:
x³ + 6x² + 12x + 8 – 2x³ + 2x = 0
Simplify:
-x³ + 6x² + 14x + 8 = 0

This is a cubic equation (degree 3), not a quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

Expand the right side:
(x – 2)³ = (x – 2)(x – 2)(x – 2) = x³ – 6x² + 12x – 8

Now, the equation becomes:
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8

Rearrange it into standard form:
x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
Simplify:
2x² – 13x + 9 = 0

This is a quadratic equation because the highest degree is 2.

Summary:

Not quadratic: (ii), (iii), (vii)

Quadratic equations: (i), (iv), (v), (vi), (viii)

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 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) Rectangular Plot:

Let the breadth of the plot be x meters.
Then the length of the plot is 2x + 1 meters (one more than twice the breadth).
The area of the rectangle is given as 528 square meters.

So, we can write the equation for the area as:
528 = (2x + 1) × x

Expanding the equation:
528 = 2x² + x

This can be rearranged to:
2x² + x – 528 = 0

(ii) Product of Two Consecutive Integers:

Let the two consecutive integers be x and x + 1.
The product of these two integers is given as 306.

So, we can write the equation as:
x × (x + 1) = 306

Expanding the equation:
x² + x = 306

Rearranging it into a standard quadratic form:
x² + x – 306 = 0

(iii) Rohan’s Age:

Let Rohan’s present age be x.
His mother’s present age is x + 26 (26 years older than Rohan).

In 3 years, Rohan’s age will be x + 3, and his mother’s age will be x + 29.
The product of their ages 3 years from now is given as 360.

So, we can write the equation as:
(x + 3) × (x + 29) = 360

Expanding the equation:
x² + 32x + 87 = 360

Rearranging it into a standard quadratic form:
x² + 32x – 273 = 0

(iv) Train Speed:

Let the speed of the train be x km/h.
The time taken to cover the distance of 480 km at speed x is 480 / x hours.

If the speed had been 8 km/h less, the time taken would have been 480 / (x – 8) hours.
According to the problem, this time is 3 hours more, so we have:

480 / (x – 8) = 480 / x + 3

Multiplying both sides by x(x – 8) to clear the denominators:
480x = 480(x – 8) + 3x(x – 8)

Expanding the equation:
480x = 480x – 3840 + 3x² – 24x

Cancelling out the 480x terms:
0 = -3840 + 3x² – 24x

Rearranging it into standard quadratic form:
3x² – 24x – 3840 = 0

New Syllabus – Class 10 Math Ch 4 Quadratic Equations Ex 4.1

Question 1: Check whether the following are quadratic equations

(i) (x + 1)² = 2(x – 3)

LHS: (x + 1)² = x² + 2x + 1
RHS: 2(x – 3) = 2x – 6
x² + 2x + 1 = 2x – 6
⇒ x² + 7 = 0
Yes, it’s a quadratic equation.

(ii) x² – 2x = (–2)(3 – x)

RHS: (–2)(3 – x) = –6 + 2x
x² – 2x = –6 + 2x
⇒ x² – 4x + 6 = 0
Yes, it’s a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

LHS: (x – 2)(x + 1) = x² – x – 2x + 3 = x² – 3x + 3
RHS: (x – 1)(x + 3) = x² + 3x – x – 3 = x² + 2x – 3
x² – 3x + 3 = x² + 2x – 3
⇒ –5x + 6 = 0
No, it’s a linear equation, not quadratic.

(iv) (x – 3)(2x + 1) = x(x + 5)

LHS: (x – 3)(2x + 1) = 2x² – 6x + x – 3 = 2x² – 5x – 3
RHS: x(x + 5) = x² + 5x
2x² – 5x – 3 = x² + 5x
⇒ x² – 10x – 3 = 0
Yes, it’s a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

LHS: (2x – 1)(x – 3) = 2x² – 6x – x + 3 = 2x² – 7x + 3
RHS: (x + 5)(x – 1) = x² – x + 5x – 5 = x² + 4x – 5
2x² – 7x + 3 = x² + 4x – 5
⇒ x² – 11x + 8 = 0
Yes, it’s a quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

RHS: (x – 2)² = x² – 4x + 4
x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0
No, it’s a linear equation.

(vii) (x + 2)³ = 2x(x² – 1)

LHS: (x + 2)³ = x³ + 6x² + 12x + 8
RHS: 2x(x² – 1) = 2x³ – 2x
x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ –x³ + 6x² + 14x + 8 = 0
No, it’s a cubic equation (degree 3), not quadratic.

(viii) x³ – 4x² – x + 1 = (x – 2)³

RHS: (x – 2)³ = x³ – 6x² + 12x – 8
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ 2x² – 13x + 9 = 0
Yes, it’s a quadratic equation.

Question 2: Represent the following situations as quadratic equations

(i) Area = 528 m², Length = 2×Breadth + 1

Let breadth = x m
Then length = 2x + 1
Area = x(2x + 1) = 528
⇒ 2x² + x – 528 = 0
This is a quadratic equation.

(ii) Product of two consecutive positive integers = 306

Let first number = x
Next = x + 1
Product = x(x + 1) = 306
⇒ x² + x – 306 = 0
This is a quadratic equation.

(iii) Rohan’s mother is 26 years older. After 3 years, product = 360

Let Rohan’s age = x
Mother’s age = x + 26
After 3 years:
(x + 3)(x + 29) = 360
⇒ x² + 32x + 87 = 360
⇒ x² + 32x – 273 = 0
This is a quadratic equation.

(iv) Distance = 480 km, speed reduced by 8 km/h takes 3 more hours

Let speed = x km/h
Then time = 480/x
Reduced speed = x – 8 ⇒ time = 480/(x – 8)
Given:
480/(x – 8) – 480/x = 3
Multiply both sides by x(x – 8):
480x – 480(x – 8) = 3x(x – 8)
⇒ 480x – 480x + 3840 = 3x² – 24x
⇒ 3x² – 24x – 3840 = 0
⇒ x² – 8x – 1280 = 0
This is a quadratic equation.

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions

This page will guide you to the textbook and solutions, as provided by the National Council of Educational Research and Training (NCERT).

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In conclusion, Class 10 Math Ch 4 Quadratic Equations Ex 4.1 provides a solid foundation for understanding the basic concepts of quadratic equations. The exercise involves solving simple quadratic equations using factorization, which helps students develop a step-by-step approach to solving problems. By practicing these problems, students not only learn how to solve equations but also improve their ability to recognize and work with quadratic equations in different forms. The solutions to these equations offer valuable insight into the nature of roots, whether they are real or imaginary. Overall, Class 10 Math Ch 4 Quadratic Equations Ex 4.1 plays a crucial role in building confidence and skills needed for more complex problems in the future.

Class 10 Math Ch 4 Quadratic Equations Ex 4.1 helps in solving quadratic equations by factorization, laying a strong foundation for understanding their nature and solutions.”
Class 10 Math Ch 4 Quadratic Equations Ex 4.1 helps in solving quadratic equations by factorization, laying a strong foundation for understanding their nature and solutions.”
Class 10 Math Ch 4 Quadratic Equations Ex 4.1 helps in solving quadratic equations by factorization, laying a strong foundation for understanding their nature and solutions.”