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

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1 - FREE PDF Download

Before talking about the NCERT solutions for Chapter 4 Ex 4.1 Class 10 Maths, let’s do a quick recapitulation- from chapter Polynomials, we already know what quadratic polynomials are. Isn’t it? When a polynomial of quadratic order of the form as  ax2 + bx + c = 0 (a ≠ 0) becomes zero, we termed it as a quadratic equation. Now, ax2 + bx + c = 0 (a ≠ 0) is known as the general form of the quadratic equation where ‘a’ and ‘b’ are called variables of the first order and c is a constant. Class 10 maths chapter 4 exercise puts students into the practice of solving quadratic equations. Ex 4.1 class 10 solutions lend them a helping hand and make them understand the right method of solving those problems. Students can now download Exercise 4.1 class 10  NCERT Book Solutions PDF to boost their exam preparation. 

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Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 4 Exercise 4.1 Class 10 | Vedantu
3. Access PDF for Maths NCERT Chapter 4 Quadratic Equations Exercise 4.1 Class 10
    3.1Class 10 Maths Chapter 4 Exercise 4.1
4. Class 10 Maths Chapter 4: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 4 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
7. NCERT Study Resources for Class 10 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 4 Exercise 4.1 Class 10 | Vedantu

  • Quadratic Equation: An equation of the form ax2 + bx + c = 0, where a ≠ 0, a, b, and c are real numbers.

  • Roots/Zeros: The values of x that satisfy the quadratic equation (i.e., for which ax2 + bx + c = 0).

  • Standard Form: The form ax2 + bx + c = 0 (all terms on one side, equal to zero).

  • Identifying Quadratic Equations: You'll need to recognize the standard form (ax2 + bx + c = 0) and understand that a ≠ 0 (a cannot be zero).

  • Representing Situations as Quadratic Equations: Word problems might describe scenarios that can be modelled by a quadratic equation. You'll need to translate the given information into an equation with x2, x, and a constant term.

  • Coefficients: The numerical factors multiplied by the variables (a, b, and c in ax2 + bx + c).

  • Constant Term: The term that doesn't include any variable (c in ax2 + bx + c).

  • Discriminant: b2 - 4ac (used in the quadratic formula to determine the nature of roots).

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NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1 - Quadratic Equations
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Quadratic Equations in One Shot (Full Chapter) | CBSE 10 Math Chap 4 | Board 2021-22 | NCERT Vedantu
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Access PDF for Maths NCERT Chapter 4 Quadratic Equations Exercise 4.1 Class 10

Class 10 Maths Chapter 4 Exercise 4.1

1. Check whether the following are quadratic equations:

(i) ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

Ans: We are given an equation: ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

We will simplify the given equation.

${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 2x + 1 = 2x - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 1 = - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 7 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.


(ii) ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x = - 6 + 2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 4x + 6 =0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.


(iii) $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

Ans: We are given an equation: $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

We will simplify the given equation.

$\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x + x - 2 = }{{\text{x}}^{\text{2}}}\text{ - x + 3x - 3}$

$\Rightarrow \text{- 2x + x - 2 = - x + 3x - 3}$

$\Rightarrow \text{- x - 2 = 2x - 3}$

$\Rightarrow \text{3x - 1 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.


(iv) $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

Ans: We are given an equation: $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

We will simplify the given equation.

$\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{- 6x + x - 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{- 10x - 3 = 0}\]

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.


(v) $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

Ans: We are given an equation: $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

We will simplify the given equation.

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - x - 6x + 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x - x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 7x + 3 = 4x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 11x + 8 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.


(vi) ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\text{x}}^{\text{2}}}\text{ - 4x + 4}$

$\Rightarrow \text{3x + 1 = - 4x + 4}$

$\Rightarrow \text{7x - 3 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.


(vii) ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

Ans: We are given an equation: ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

We will simplify the given equation.

${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ + 6}{{\text{x}}^{\text{2}}}\text{ + 12x + 8 = 2}{{\text{x}}^{\text{3}}}\text{ - 2x}$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}\text{ + 14x + 8 = }{{\text{x}}^{\text{3}}}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ - 14x - 8 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.


(viii) \[{{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}\]

Ans: We are given an equation: ${{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}$

We will simplify the given equation.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{- 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - 13x + 9 =  0}$

This equation obtained is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, 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 $\text{528 }{{\text{m}}^{\text{2}}}$. 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.

Ans: It is given that the area of rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$.

Let us assume the breadth of the plot be $\text{b = x m}$.

It is given that the length is one more than twice the breadth.

So, the length will be $\text{l = }\left( \text{2x + 1} \right)\text{ m}$.

The area of the rectangle will be  $\text{A = l  }\!\!\times\!\!\text{  b}$

$\therefore \text{l  }\!\!\times\!\!\text{  b = 528}$

$\Rightarrow \text{x}\left( \text{2x + 1} \right)\text{ = 528}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$

Therefore, the required quadratic equation is $\text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$.


(ii) The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans: It is given that the product of two consecutive positive integers is $\text{306}$.

If we assume the first integer to be $\text{x}$, then the next consecutive integer will be $\left( \text{x+1} \right)$.

So, we can write $\text{x}\left( \text{x + 1} \right)\text{ = 306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$.


(iii) Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans: Let us assume Rohan’s age be $\text{x}$.

His mother is $\text{26}$ years older than him.

So, his mother’s age will be $\left( x+26 \right)$

After $\text{3}$ years, Rohan’s age $\text{= x + 3}$

His mother’s age $\text{= x + 26 + 3 = x + 29}$

It is given that after $\text{3}$ years, the product of their ages is $\text{360}$.

$\therefore \left( \text{x + 3} \right)\left( \text{x + 29} \right)\text{ = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x + 87 = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$.


(iv) A train travels a distance of $\text{480}$ $\text{km}$ at a uniform speed. If the speed had been $\text{8}$ $\text{km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans: Let us assume the speed of the train be $\text{x km/h}$.

The train travels a distance of $\text{480 km}$ at a uniform speed.

So, the time taken to travel the given distance $\text{= }\frac{\text{480}}{\text{x}}\text{ hrs}$.

It is given that the train would take $\text{3}$ hours more if the

speed had been $\text{8 km/h}$ less.

Now, the new speed of the train $\text{= }\left( \text{x - 8} \right)\text{ km/h}$.

The new time taken to travel the same distance $\text{= }\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ hr}$.

We know that $\text{speed  }\!\!\times\!\!\text{  time = distance}$

$\therefore \left( \text{x - 8} \right)\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ = 480}$

$\Rightarrow \text{480 + 3x - }\frac{\text{3840}}{\text{x}}\text{ - 24 = 480}$

$\Rightarrow \text{3x - }\frac{\text{3840}}{\text{x}}\text{ = 24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{ - 24x - 3840 = 0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$.


Conclusion

In Class 10 maths Exercise 4.1, students delve into the basics of quadratic equations. This exercise helps you understand how to identify quadratic equations and distinguish them from other polynomial equations. The key takeaway is recognizing the standard form of a quadratic equation, 𝑎𝑥2+𝑏𝑥+𝑐=0, and understanding the significance of each coefficient. Focus on identifying Quadratic Equations and ensure you can correctly identify equations of the form 𝑎𝑥2+𝑏𝑥+𝑐=0. Mastering these fundamentals is crucial for solving more complex quadratic equations in class 10 Maths ch 4 ex 4.1. Regular practice with Class 10th Exercise 4.1 Solutions will build a strong foundation for further mathematical concepts.


Class 10 Maths Chapter 4: Exercises Breakdown

Chapter 4 Quadratic Equations All Exercises in PDF Format

Exercise 4.2

6 Questions and Solutions

Exercise 4.3

5 Questions and Solutions



CBSE Class 10 Maths Chapter 4 Other Study Materials



Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these class 10th maths chapter 4 exercise 4.1 solutions to be thoroughly familiar with the concepts.



NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

FAQs on NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1 - Quadratic Equations

1.The Altitude of a Right Triangle is 7 cm Less than its Base. If the Hypotenuse is 13 cm, Find the Other Two Sides.

You can find the following in class 10 Maths ex 4.1 solutions:

Let x be the base, the altitude will be x - 7

Now as per Pythagoras theorem,

x2 + (x − 7)2 = 132

= x2 + x2 - 14x + 49 = 169

= 2x2 -14x = 120

= x2 - 7x = 60

= x2 - 12x + 5x - 60 = 0

= x(x - 12) + 5(x - 12) = 0

= (x - 12) (x + 5) = 0

x = 12 or -5

Since x cannot be negative x = 12.

So base is 12 cm and altitude is (12 - 7) = 5 cm.

2. If Zeba were Younger by 5 years than what she really is then the Square of her age Would have been 11 more than five times her actual age. What is her age now?

The solution according to exercise 4.1 class 10 Maths NCERT solutions is as follows:

Let age of Zeba be x years
As per the question:

(x − 5)2 = 11 + 5x(x − 5)2 = 11 + 5x
=> x2 + 25 − 10x = 11 + 5x(x)2 + 25 − 10x = 11 + 5x
=> x2 − 15x + 14 = 0 x 2 − 15x + 14 = 0 x 2 − 14x − x + 14 = 0 x 2 − 14x − x + 14 = 0
=> (x - 1)(x - 14)

Therefore, age of Zeba is 14 years.

3. What is a Quadratic Equation, and what is its general form?

When a quadratic order polynomial is of the type ax2 + bx + c = 0 (a ≠  0), we call it a Quadratic Equation. The general version of the Quadratic Equation is now ax2 + bx + c = 0 (a ≠ 0), where ‘a', ‘b' and ‘c’ are constants i.e. numbers. The constant ‘b’ and ‘c’ can also be equal to zero but ‘a’ cannot be zero.

4. How many questions are there in Chapter 4 Maths Class 10?

There are a total of four exercises. In the first exercise, there are two questions. In the second, there are six; in the third, there are 11; and in the fourth exercise, there are five. So, all in all, there are 23 questions and some of these questions have some sub-parts too. In theNCERT Solutions for Class 4, Maths Chapter 4 there are solved examples as well that will help in preparation for the board exams.

5. How to find the nature of roots?

The term b2 – 4ac is called the discriminant of a quadratic equation since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not. So, a Quadratic Equation ax2 + bx + c = 0 has 

(i) two distinct real roots, if b2 – 4ac > 0, 

(ii) two equal real roots, if b2 – 4ac = 0, 

(iii) no real roots, if b2 – 4ac < 0.

6. How can I top in Class 10 Maths?

In order to score the best marks in Maths, regular practice is required. You must practice and revise all the solved examples and questions given in the NCERT book. You must all be thorough with all the concepts, theorems and formulae of all the chapters. You can visit the page NCERT Solutions for Class 10 Maths, on the official website of Vedantu for a complete guide. 

7. Where can I get the NCERT Solution for Class 10 Chapter 4 Maths?

You can easily access the NCERT Solutions for Class 10 Maths, on the Vedantu official website (vedantu.com) and on the Vedantu app absolutely free of cost. You can also get revision notes and other study material free of cost on the Vedantu website (vedantu.com). The answers are provided comprehensively and are very easy to understand; all the formulae used in the questions are also provided along with the answers to facilitate a quick revision.