Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

ffImage
Last updated date: 29th Mar 2024
Total views: 812.4k
Views today: 19.12k
MVSAT 2024

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations - Free PDF

Maths is a subject that deals with lots of formulas and tricks. It requires a good understanding and lots of practice. Quadratic Equation is a chapter in class 10, which is quite tricky and difficult. But, it has good weightage from the board examination’s point of view. This is the reason why Vedantu has prepared NCERT Class 10 Maths Chapter 4 Solutions. NCERT Solutions of Class 10 Maths Chapter 4 available at Vedantu’s website and app is very simple and easy to understand. Students acquire a complete knowledge about the basic concepts in detail step by step first and then go through the solved examples. NCERT Solutions for Class 10 Science is also available on Vedantu.

 

Also, students can download Quadratic Equations Class 10 Solutions in PDF form to use offline. Those who want to secure the highest marks in their board examinations should not miss this Maths Chapter 4 Class 10. It has been recorded that every year, at least two or three questions are asked from this chapter. Therefore, it is advised to every student to start practice for Class 10 Maths Ch 4 with NCERT Solutions Class 10 Maths Ch 4 offered by Vedantu.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 4 - Quadratic Equations

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



All Topics of NCERT Class 10 Maths Chapter 4 - Quadratic Equations

The topics covered under Chapter 4 Maths Class 10 are given below.

S.No

Topic Name

4.1

Introduction

4.2

Quadratic Equations

4.3

Solution of a Quadratic Equation by Factorisation

4.4

Solution of a Quadratic Equation by Completing the Square

4.5

Nature of Roots

4.6

Summary


Important Points

  • A quadratic equation can be represented as:

ax2 + bx + c = 0

Where x is the variable of the equation and a, b and c are the real numbers. Also, a≠0.

  • The nature of roots of a quadratic equation ax2 + bx + c = 0 can be find as: 

Condition

Nature of Roots

b2 – 4ac >0

Two distinct real roots

b2 – 4ac = 0

Two equal roots

b2 – 4ac <0

No real roots

  • A real number α be root of quadratic equations ax2 + bx + c = 0 if and only if 

2 + bα + c = 0.


Quadratic equations are very important in real-life situations. Learn all the concepts deeply and understand each topic conceptually. And, now let us solve questions related to quadratic equations.

Competitive Exams after 12th Science
Watch videos on
NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations
icon
Quadratic Equations in One Shot (Full Chapter) | CBSE 10 Math Chap 4 | Board 2021-22 | NCERT Vedantu
Vedantu 9&10
Subscribe
iconShare
7.3K likes
148.2K Views
2 years ago
Download Notes
yt video
Quadratic Equations in One-Shot | CBSE Class 10 Maths NCERT Solutions | Vedantu Class 9 and 10
Vedantu 9&10
5.1K likes
157.7K Views
3 years ago
Download Notes
yt video
Quadratic Equations In one Shot | CBSE Class 10 Maths Chapter 4 | NCERT @VedantuClass910
Vedantu 9&10
5.4K likes
173.3K Views
4 years ago
More Free Study Material for Quadratic Equations
icons
Revision notes
739.2k views 12k downloads
icons
Important questions
648.9k views 15k downloads
icons
Ncert books
665.4k views 14k downloads

Exercises under NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Chapter 4 of NCERT Solutions for Class 10 Maths is titled "Quadratic Equations". In this chapter, students will learn about the standard form of a quadratic equation, methods for solving quadratic equations, and the nature of the roots of quadratic equations.


The chapter includes four exercises, each of which covers different aspects of quadratic equations.


Exercise 4.1:

This exercise covers the introduction to quadratic equations and the standard form of a quadratic equation. It also includes methods for solving quadratic equations by factorisation. In this exercise, students will learn how to identify a quadratic equation, how to convert a quadratic equation into standard form, and how to factorise quadratic equations using different methods. The exercise includes a set of questions that range from easy to difficult, allowing students to gradually build their understanding of the concepts.


Exercise 4.2:

This exercise covers more advanced methods for solving quadratic equations, such as completing the square and using the quadratic formula. It includes the derivation of the quadratic formula and shows how to apply it to solve quadratic equations. In this exercise, students will learn how to complete the square of a quadratic equation to convert it into standard form, and how to use the quadratic formula to solve quadratic equations. The exercise includes questions that require students to use both methods to solve quadratic equations.


Exercise 4.3:

This exercise focuses on the nature of the roots of quadratic equations and the discriminant of a quadratic equation. Students will learn how to determine the nature of the roots of a quadratic equation based on the value of its discriminant. The exercise covers the relationship between the coefficients and roots of a quadratic equation, and how to find the sum and product of the roots. The exercise includes a set of questions that require students to apply their knowledge of discriminant and the nature of roots to solve quadratic equations.


Exercise 4.4:

This exercise covers real-life applications of quadratic equations and includes word problems that require students to apply their knowledge of quadratic equations to solve practical problems. The exercise includes problems related to the trajectory of a projectile, finding the distance between two ships, and the dimensions of a garden. Students will learn how to formulate and solve quadratic equations to solve real-life problems. The exercise includes a set of word problems that gradually increase in difficulty, allowing students to develop their problem-solving skills.


Access NCERT Solutions for Class - 10 Maths Chapter 4 – Quadratic Equations

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

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

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

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

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

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

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

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

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

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

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

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

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

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

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

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.


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

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

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

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

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

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

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

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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: ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$ 

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

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

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

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

Ans: Let the breath of the plot be $\text{x m}$.

Thus, length would be-

$\text{Length=}\left( \text{2x+1} \right)\text{m}$

Hence, Area of rectangle $=$$\text{Length }\!\!\times\!\!\text{ breadth}$

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

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


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

Ans: Let the consecutive integers be $\text{x}$ and $\text{x+1}$.

Thus, according to question-

$\text{x}\left( \text{x+1} \right)\text{=306}$

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


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 Rohan’s age be $\text{x}$.

Hence, his mother’s age is $\text{x+26}$ .

Now, after $\text{3 years}$.

Rohan’s age will be $\text{x+3}$.

His mother’s age will be $\text{x+29}$ .

So, according to question-

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

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+29x+87=360}$

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


iv. A train travels a distance of $\text{480 km}$ at a uniform speed. If the speed had been $\text{8km/h}$ 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 the speed of train be $\text{x km/h}$.

Thus, time taken to travel $\text{482 km}$ is $\dfrac{\text{480}}{\text{x}}\text{hrs}$.

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

Therefore, time taken to travel $\text{480 km}$ is $\left( \dfrac{\text{480}}{\text{x}}+3 \right)\text{hrs}$.

Hence, $\text{speed }\!\!\times\!\!\text{ time=distance}$

i.e $\left( \text{x-8} \right)\left( \dfrac{\text{480}}{\text{x}}\text{+3} \right)\text{=480}$

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

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

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

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


Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans: ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$


ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

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

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

Therefore, roots of this equation are –

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$


iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans: $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

Therefore, roots of this equation are –

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$


iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow \dfrac{\text{1}}{\text{8}} {{\left( \text{4x-1} \right)}^{2}}$

Therefore, roots of this equation are –

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$


v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans: $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

Therefore, roots of this equation are –

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$


2. i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans: Let the number of john’s marbles be $\text{x}$.

Thus, number of Jivanti’s marble be $\text{45-x}$.

According to question i.e, 

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And number of Jivanti’s marble be $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

So now,

Case 1- If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles be $\text{36}$.

Thus, number of Jivanti’s marble be $\text{9}$.

Case 2- If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles be $9$.

Thus, number of Jivanti’s marble be $36$.


ii. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy be $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1- If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys be $25$.

Case 2- If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys be $30$.


3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$.

Ans: Let the first number be $\text{x}$ ,

Thus, the second number be $\text{27-x}$.

Therefore,

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1- If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number be $\text{14}$.

Case 2- If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number be $\text{14}$.

Thus, the second number be$13$.


4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1- If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because number is positive.

Case 2- If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $\text{13}$.

Thus, the second number be $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.


5. The altitude of a right triangle is $\text{7 cm}$ less than its base. If the hypotenuse is $\text{13 cm}$, find the other two sides.

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude be $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagores theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1- If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2- If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because side is always positive.

Hence, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$.


6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans:  Let the number of articles produced be $\text{x}$.

Therefore, cost of production of each article be $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1- If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because number of articles is always positive.

Case 2- If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced be $6$.

Therefore, cost of production of each article be $\text{Rs15}$.


Exercise 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-}\dfrac{\text{7}}{\text{2}}\text{x=-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x=-}\dfrac{\text{3}}{\text{2}}$

On adding ${{\left( \dfrac{7}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x+}{{\left( \dfrac{7}{4} \right)}^{2}}\text{=-}\dfrac{\text{3}}{\text{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49}}{\text{16}}\text{-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{25}}{\text{16}}$

$\Rightarrow \left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)\text{=}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}+\dfrac{5}{4}$ or $x\text{=}\dfrac{\text{7}}{\text{4}}-\dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{12}{\text{4}}$ or $x\text{=}\dfrac{2}{\text{4}}$

$\Rightarrow x\text{=}3$ or $x\text{=}\dfrac{\text{1}}{\text{2}}$


ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On dividing both sides of the equation by $\text{2}$.

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

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{1}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{33}}{\text{16}}$

\[\Rightarrow \left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)\text{=}\pm \dfrac{\sqrt{\text{33}}}{4}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}-1}{\text{4}}\]

\[\Rightarrow \text{x=}\dfrac{\sqrt{\text{33}}-1}{\text{4}}\] or \[\Rightarrow \text{x=}\dfrac{\sqrt{\text{-33}}-1}{\text{4}}\]


iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans: $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

$\Rightarrow {{\left( 2\text{x} \right)}^{\text{2}}}\text{+2}\left( 2\sqrt{3} \right)\text{x+}{{\left( \sqrt{\text{3}} \right)}^{2}}\text{=0}$

$\Rightarrow {{\left( \text{2x+}\sqrt{\text{3}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$ and $\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$ and $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$


iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{x+2=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x=-2}$

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=-2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{1}}{\text{16}}\text{-2}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=-}\dfrac{\text{31}}{\text{16}}$

Since, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.


2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-7}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ }\sqrt{\text{49-24}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ 5}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7+5}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{7-5}}{4}$

$\Rightarrow \text{x=}\dfrac{12}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{2}}{4}$

$\therefore \text{x=3 or }\dfrac{\text{1}}{\text{2}}$.


ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=-4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{33}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{-1-}\sqrt{33}}{4}$

$\therefore \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}\text{ or }\dfrac{\text{-1-}\sqrt{33}}{4}$.


iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans: $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=4}$, $\text{b=4}\sqrt{3}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}\text{ }\!\!\pm\!\!\text{ }\sqrt{\text{48-48}}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$  or $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$

$\therefore \text{x=}\dfrac{\text{-}\sqrt{3}}{2}\text{ or }\dfrac{\text{-}\sqrt{3}}{2}$.


iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

\[\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}\]

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{-31}}}{4}$

Since, there can not be any negative number inside square root for any real root to exist.

Therefore, there is no real root for the given equation.


3. Find the roots of the following equations:

i. $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

Ans: $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

$\Rightarrow\text{1}{{\text{x}}^{\text{2}}}\text{-3x-1=}0$.

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-3}$, $\text{c=-1}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{9+4}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{13}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}$  or $\Rightarrow \text{x=}\dfrac{\text{3-}\sqrt{\text{13}}}{2}$

$\therefore \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}\text{ or }\dfrac{\text{3-}\sqrt{\text{13}}}{2}$.


ii. $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

Ans: $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

$\Rightarrow \dfrac{\text{x-7-x-4}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \dfrac{\text{-11}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \left( \text{x+4} \right)\left( \text{x-7} \right)=-30$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x-28=-3}0$

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

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

$\Rightarrow \text{x}\left( \text{x-2} \right)\text{-1}\left( \text{x-2} \right)\text{=0}$

$\Rightarrow \left( \text{x-2} \right)\left( \text{x-1} \right)\text{=0}$

$\Rightarrow x\text{=1 or 2}$


4. The sum of the reciprocals of Rehman’s ages, (in years) $\text{3}$ years ago 

and $\text{5}$ years from now is \[\dfrac{\text{1}}{\text{3}}\]. Find his present age.

Ans: Let the present age of Rehman be $\text{x}$ years.

Three years ago, his age was $\left( \text{x-3} \right)\text{years}$.

Five years hence, his age will be $\left( \text{x+5} \right)\text{years}$.

Therefore,$\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x+5}}\text{=}\dfrac{\text{1}}{\text{3}}$

\[\Rightarrow \dfrac{\text{x+5+x-3}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

\[\Rightarrow \dfrac{\text{2x+2}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

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

$\Rightarrow \text{6x+6=}{{\text{x}}^{\text{2}}}\text{+2x-15}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x-21=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x+3x-21=0}$

$\Rightarrow \text{x}\left( \text{x-7} \right)\text{+3}\left( \text{x-7} \right)\text{=0}$

$\Rightarrow \left( \text{x-7} \right)\left( \text{x+3} \right)\text{=0}$

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

Therefore, Rehman’s age is $\text{7 years}$.


5. In a class test, the sum of Shefali’s marks in Mathematics and English is $\text{30}$. Had she got $\text{2}$ marks more in Mathematics and $\text{3}$ marks less in English, the product of their marks would have been $\text{210}$. Find her marks in the two subjects.

Ans: Let the marks in maths be $\text{x}$.

Thus, marks in English will be $\text{30-x}$.

Hence, according to question –

$\left( \text{x+2} \right)\left( \text{30-x-3} \right)\text{=210}$

$\left( \text{x+2} \right)\left( \text{27-x} \right)\text{=210}$

$\Rightarrow \text{-}{{\text{x}}^{\text{2}}}\text{+25x+54=210}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x+156=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-12x-13x+156=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)\text{-13}\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \text{x=12,13}$

Case 1- If the marks in mathematics are $\text{12}$ , then marks in English will be $18$.

Case 2- If the marks in mathematics are $\text{13}$ , then marks in English will be $17$.


6. The diagonal of a rectangular field is $\text{60}$ metres more than the shorter side. If the longer side is $\text{30}$ metres more than the shorter side, find the sides of the field.

Ans: Let the shorter side of the rectangle be $\text{x m}$.

Thus, Larger side of the rectangle will be $\left( \text{x+30} \right)\text{m}$.

Diagonal of the rectangle be $\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}$

Hence, according to question-

$\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}\text{=x+60}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}\text{=}{{\left( \text{x+60} \right)}^{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{+900+60x=}{{\text{x}}^{\text{2}}}\text{+3600+120x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-60x-2700=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-90x+30x-2700=0}$

$\Rightarrow \text{x}\left( \text{x-90} \right)+30\left( \text{x-90} \right)\text{=0}$

$\Rightarrow \left( \text{x-90} \right)\left( \text{x+30} \right)\text{=0}$

$\Rightarrow \text{x=90,-30}$

Since, side cannot be negative. 

Therefore, the length of the shorter side of rectangle is $\text{90 m}$.

Hence, length of the larger side of the rectangle be $\text{120 m}$.


7. The difference of squares of two numbers is $\text{180}$. The square of the smaller number is $\text{8}$ times the larger number. Find the two numbers.

Ans: Let the larger number be $\text{x}$ and smaller number be $\text{y}$.

According to question-

${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=180}$ and ${{\text{y}}^{\text{2}}}\text{=8x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x=180}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-180=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-18x+10x-180=0}\]

$\Rightarrow \text{x}\left( \text{x-18} \right)+10\left( \text{x-18} \right)\text{=0}$

$\Rightarrow \left( \text{x-18} \right)\left( \text{x+10} \right)\text{=0}$

$\Rightarrow \text{x=18,-10}$

Since, larger cannot be negative as $8$ times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be $18$.

$\therefore {{\text{y}}^{\text{2}}}\text{=8}\left( \text{18} \right)$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{=144}$

$\Rightarrow \text{y= }\!\!\pm\!\!\text{ 12}$

Hence, smaller number be $\pm 12$.

Therefore, the numbers are $18$ and $12$ or $18$ and $-12$ .


8. A train travels $\text{360 km}$km at a uniform speed. If the speed had been $\text{5km/h}$ more, it would have taken $\text{1}$hour less for the same journey. Find the speed of the train.

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

Time taken to cover $\text{360 km/h}$ be $\dfrac{\text{360}}{\text{x}}$.

According to question-

$\left( \text{x+5} \right)\left( \dfrac{\text{360}}{\text{x}}\text{-1} \right)\text{=360}$

$\Rightarrow \text{360-x+}\dfrac{\text{1800}}{\text{x}}\text{-5=360}$ 

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+5x-1800=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+45x-40x-1800=0}\]

$\Rightarrow \text{x}\left( \text{x+45} \right)-40\left( \text{x+45} \right)\text{=0}$

$\Rightarrow \left( \text{x+45} \right)\left( \text{x-40} \right)\text{=0}$

$\Rightarrow \text{x=40,-45}$

Since, the speed cannot be negative.

Therefore, the speed of the train is $\text{40 km/h}$.


9. Two water taps together can fill a tank in $\text{9}\dfrac{\text{3}}{\text{8}}$ hours. The tap of larger diameter takes $\text{10}$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans: Let the time taken by the smaller pipe to fill the tank be $\text{x hr}$.

So, time taken by larger pipe be $\left( \text{x-10} \right)\text{hr}$.

Part of the tank filled by smaller pipe in $1$ hour is $\dfrac{\text{1}}{\text{x}}$.

Part of the tank filled by larger pipe in $1$ hour is $\dfrac{\text{1}}{\text{x-10}}$.

So, according to the question-

$\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=9}\dfrac{\text{3}}{\text{8}}$

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=}\dfrac{\text{75}}{\text{8}}$

\[\Rightarrow \dfrac{\text{x-10+x}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

\[\Rightarrow \dfrac{\text{2x-10}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

$\Rightarrow \text{75}\left( \text{2x-10} \right)\text{=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{150x-750=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-230x+750=0}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-200x-30x+750=0}$

$\Rightarrow \text{8x}\left( \text{x-25} \right)\text{-30}\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{8x-30} \right)\text{=0}$

$\Rightarrow x\text{=25 or }\dfrac{\text{30}}{\text{8}}$

Case 1- If time taken by smaller pipe be $\dfrac{\text{30}}{\text{8}}$ i.e $\text{3}\text{.75 hours}$.

So, Time taken by larger pipe will be negative which is not possible.

Hence, this case is rejected.

Case 2- If the time taken by smaller pipe be $\text{25}$.Then, time taken by larger pipe will be $\text{15 hours}$.

Therefore, time taken by smaller pipe be $\text{25 hours}$ and time taken by larger pipe will be $\text{15 hours}$.


10. An express train takes $\text{1}$ hour less than a passenger train to travel $\text{132 km}$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is $\text{11 km/h}$ more than that of the passenger train, find the average speed of the two trains.

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

So, Average speed of express train be $\left( \text{x+11} \right)\text{km/h}$.

Thus, according to question.

$\therefore \dfrac{\text{132}}{\text{x}}\text{-}\dfrac{\text{132}}{\text{x+11}}\text{=1}$

 $\Rightarrow \text{132}\left[ \dfrac{\text{x+11-x}}{\text{x}\left( \text{x+11} \right)} \right]\text{=1}$

$\Rightarrow \dfrac{\text{132 }\!\!\times\!\!\text{ 11}}{\text{x}\left( \text{x+11} \right)}\text{=1}$

$\Rightarrow \text{132 }\!\!\times\!\!\text{ 11=x}\left( \text{x+11} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+11x-1452=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+44x-33x-1452=0}$

$\Rightarrow \text{x}\left( \text{x+44} \right)\text{-33}\left( \text{x+44} \right)\text{=0}$

$\Rightarrow \left( \text{x+44} \right)\left( \text{x-33} \right)\text{=0}$

$\Rightarrow x\text{=-44 or 33}$

Since, speed cannot be negative.

Therefore, the speed of the passenger train will be $\text{33 km/h}$ and thus, the speed of the express train will be $\text{44 km/h}$ .


11. Sum of the areas of two squares is $\text{468 }{{\text{m}}^{\text{2}}}$. If the difference of their perimeters are $\text{24 m}$, find the sides of the two squares.

Ans: Let the sides of the two squares be $\text{x m}$ and $\text{y m}$.

Thus, their perimeters will be $\text{4x}$ and $\text{4y}$ and areas will be ${{\text{x}}^{2}}$ and ${{\text{y}}^{2}}$.

Hence, according to question –

$\text{4x-4y=24}$

$\Rightarrow \text{x-y=6}$

$\Rightarrow \text{x=y+6}$

And ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

Substituting value of x-

${{\left( \text{y+6} \right)}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{36+}{{\text{y}}^{\text{2}}}\text{+12y+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+12y-432=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+6y-216=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+18y-12y-216=0}$

$\Rightarrow \text{y}\left( \text{y+18} \right)\text{-12}\left( \text{y+18} \right)\text{=0}$

$\Rightarrow \left( \text{y+18} \right)\left( \text{y-12} \right)\text{=0}$

$\Rightarrow \text{y=-18 or 12}$

Since, side cannot be negative.

Therefore, the sides of the square are $\text{12 m}$ and $\left( \text{12+6} \right)\text{m}$ i.e $\text{18 m}$.


Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. 

If the real roots exist, find them-

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-3}$, $\text{c=5}$.

Discriminant $\text{=}{{\left( \text{-3} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{5} \right)$

$\text{=9-40}$

$\text{=-31}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$.

Therefore, there is no real root for the given equation.


ii. $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac > 0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$ then there will be no real roots.

Thus, for $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=3}$, $\text{b=-4}\sqrt{\text{3}}$, $\text{c=4}$.

Discriminant $\text{=}{{\left( \text{-4}\sqrt{\text{3}} \right)}^{\text{2}}}\text{-4}\left( \text{3} \right)\left( \text{4} \right)$

$\text{=48-48}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real root for the given equation and the roots are-

$\dfrac{\text{-b}}{\text{2a}}$ and $\dfrac{\text{-b}}{\text{2a}}$.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( \text{-4}\sqrt{\text{3}} \right)}{\text{6}}$

$\text{=}\dfrac{\text{4}\sqrt{\text{3}}}{\text{6}}$

\[\text{=}\dfrac{\text{2}\sqrt{\text{3}}}{3}\]

Therefore, roots are \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\] and \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\].


iii. $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-6}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=36-24}$

$\text{=12}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exists for the given equation and the roots are-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

Hence, roots are-

$\text{x=}\dfrac{\text{-}\left( \text{-6} \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{36-24}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{12}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ 2}\sqrt{\text{3}}}{\text{4}}$

$\text{=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}}{\text{2}}$

Therefore, roots are $\dfrac{\text{3+}\sqrt{\text{3}}}{\text{2}}$ and $\dfrac{\text{3-}\sqrt{\text{3}}}{\text{2}}$.


2. Find the values of $\text{k}$ for each of the following quadratic equations, so 

that they have two equal roots.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$.

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=k}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{k} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=}{{\text{k}}^{2}}-24$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore {{\text{k}}^{\text{2}}}\text{-24=0}$

$\Rightarrow {{\text{k}}^{\text{2}}}\text{=24}$

$\Rightarrow \text{k=}\sqrt{\text{24}}$

$\Rightarrow \text{k=}\pm \text{2}\sqrt{\text{6}}$


ii. $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

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

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=k}$, $\text{b=-2k}$, $\text{c=6}$.

Discriminant $\text{=}{{\left( \text{-2k} \right)}^{\text{2}}}\text{-4}\left( \text{k} \right)\left( \text{6} \right)$

$\text{=4}{{\text{k}}^{\text{2}}}\text{-24k}$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore \text{4}{{\text{k}}^{\text{2}}}\text{-24k=0}$

$\Rightarrow \text{4k}\left( \text{k-6} \right)\text{=0}$

$\Rightarrow \text{k=0 or k=6}$

But $\text{k}$ cannot be zero. Thus, this equation has two equal roots when $\text{k}$ should be $\text{6}$ .


3. Is it possible to design a rectangular mango grove whose length is 

twice its breadth, and the area is $\text{800}{{\text{m}}^{\text{2}}}$ ? If so, find its length and breadth.

Ans: Let the breadth of mango grove be $\text{x}$.

So, length of mango grove will be $\text{2x}$.

Hence, Area of mango grove is $=\left( \text{2x} \right)\text{x}$

$\text{=2}{{\text{x}}^{\text{2}}}$.

So, $\text{2}{{\text{x}}^{\text{2}}}\text{=800}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-400=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=0}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{-400} \right)$

$\text{=1600}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exist for the given equation and the roots are-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

Hence, roots are-

$\text{x=}\dfrac{\text{-}\left( 0 \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( 0 \right)}^{\text{2}}}\text{-4}\left( 1 \right)\left( -400 \right)}}{2}$

$\text{=}\dfrac{\pm \sqrt{\text{1600}}}{2}$

$\text{=}\dfrac{\text{ }\!\!\pm\!\!\text{ 40}}{2}$

$\text{=}\pm \text{20}$

Since, length cannot be negative.

Therefore, breadth of the mango grove is $\text{20m}$.

And length of the mango grove be $\text{2}\left( \text{20} \right)\text{m}$ i.e., $\text{40m}$.


4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $\text{20}$ years. Four years ago, the product of their ages in years was $\text{48}$.

Ans: Let the age of one friend be $\text{x years}$.

So, age of the other friend will be $\left( \text{20-x} \right)\text{years}$.

Thus, four years ago, the age of one friend be $\left( \text{x-4} \right)\text{years}$.

And age of the other friend will be $\left( \text{16-x} \right)\text{years}$.

Hence, according to question-

$\left( \text{x-4} \right)\left( \text{16-x} \right)\text{=48}$

$\Rightarrow \text{16x-64-}{{\text{x}}^{\text{2}}}\text{+4x=48}$

$\Rightarrow 20\text{x-112-}{{\text{x}}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-20x+112-=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-20}$, $\text{c=112}$.

Discriminant $\text{=}{{\left( \text{-20} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{112} \right)$

$\text{=400-448}$

$\text{=-48}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac <0}$.

Therefore, there is no real root for the given equation and hence, this situation is not possible.


5. Is it possible to design a rectangular park of perimeter $\text{80 m}$ and area $\text{400}{{\text{m}}^{\text{2}}}$? If so find its length and breadth.

Ans: Let the length of the park be $\text{x m}$ and breadth of the park be $\text{x m}$.

Thus, $\text{Perimeter=2}\left( \text{x+y} \right)$.

Hence, according to question-

$\text{2}\left( \text{x+y} \right)\text{=80}$

$\Rightarrow \text{x+y=40}$

$\Rightarrow \text{y=40-x}$.

Now, $\text{Area=x }\!\!\times\!\!\text{ y}$.

Substituting value of y.

$\text{Area=x}\left( \text{40-x} \right)$

So, according to question-

$\text{x}\left( \text{40-x} \right)\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-40x+400=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-40}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{-40} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( 400 \right)$

$\text{=1600-1600}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real roots for the given equation and hence, this situation is possible.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( -40 \right)}{2}$

$\text{=}\dfrac{\text{40}}{2}$

\[\text{=20}\]

Therefore, length of park is $\text{x=20m}$ .

And breadth of park be $\text{y=}\left( \text{40-20} \right)\text{m}$ i.e., $\text{y=20m}$.


Overview of the Exercises Covered in NCERT Solutions for Class 10 Chapter 4 Quadratic Equations

Ex 4.1: There are 2 sums with a total of 12 sub-parts in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.1. Students will have to verify if the given equations are quadratic equations or not, for the first few sums. For the sums covered in the latter part of the exercise, students will have to form quadratic equations from the given word problems. These sums will familiarize them with the standard quadratic equation formula ax2+bx+c=0.  


Ex 4.2: There are a total of 6 sums covered in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.2. These sums are based on the application of the concept of factorisation in quadratic equations. Students will have to form the quadratic equation from the given word problems and find the roots of the quadratic equations by the middle term factor method.


Ex 4.3: There are a total of 11 sums in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.3. Students need to have a good understanding of the concepts of speed, time, and distance, time and work, average speed, perimeter and area, etc. to solve the sums covered in this exercise. The first few sums of this exercise require students to apply the method of completing the perfect square terms in quadratic equations. Also, the some of the sums covered in this exercise have the application of the quadratic formula, 

\[ x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \]

Ex.4.4: There are a total of 5 sums in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.4. The sums covered in this exercise will familiarize themselves with the nature of roots i.e., real, imaginary, equal, and unequal roots. The sums will help them to identify and deduce the discriminants of quadratic equations and solve the sums to find the nature of roots.


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

You can opt for Chapter 4 - Quadratic Equation NCERT Solutions for Class 10 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.


Quadratic Equations Class 10 NCERT Solutions have composed in such a way that every student can understand all concepts easily. The expert team of Vedantu has put all their possible efforts in Class 10 CBSE Maths Chapter 4 Solutions to make it exciting and fun-loving. No doubt that Maths is a subject which is not easy to learn. The formulas are not easy to learn, and students don't catch tricks easily where they should apply.

Chapter 4 Maths Class 10 contains quadratic equations to find the value of x. Apart from this, there are approximately three methods given in Class 10 Maths Chapter 4 for this. But, not all three methods are easy to understand to an individual. But in Chapter 4 Maths Class 10 NCERT Solutions, the experts of Vedantu have explained all three methods in very interesting ways that any student can learn quickly. All concepts of quadratic equations and formulas have been explained well. From where the formula comes, how it is discovered, by whom it is discovered, and many other things are mentioned first for the basic knowledge.

After that, many solved examples have been given in NCERT Solution Class 10 Maths Chapter 4 to teach students how to apply formulas and solve numerical problems. All NCERT exercises have been solved by the experts of Vedantu for the students. Some unsolved questions are also included in Class 10 Maths Chapter 4 NCERT Solutions for the practice of the students. Along with this, previous year's questions have also been stated in between the NCERT Solutions. For the ease of the students, it is also mentioned which question came in which year. Some of the mock test papers are also available at the end of pdf for the students' better preparation.  Aspirants who don't want to miss any single question in the board exam should practice regularly with Class 10 NCERT Maths Chapter 4 Solutions.

We can say with full confidence that Vedantu’s NCERT Maths Solution Class 10 Chapter 4 is the single key to get success in board examinations. Students can download the free PDF by a single click on the link available in the website of Vedantu.


Key Features for NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Class 10 Maths NCERT Solutions for Chapter 4 on Vedantu are highly recommended by the CBSE 10th Board teachers. These solutions are among the best guides for self-study purposes for the students of Class 10. The key features of these solutions for Quadratic Equations given below explain how these solutions are a must-have for CBSE Term 2 Class 10 Maths exam.

  • These NCERT Solutions PDFs for Class 10 Quadratic Equations are easy to access online and free to download for offline practice purposes.

  • Since these NCERT Solution PDFs are available on the Vedantu mobile app, students can download them on their mobile phones and study at any time.

  • Each sum covered in the four exercises of Class 10 Ch-4 Quadratic Equations is solved in a step-by-step manner for a thorough understanding of all students. 

  • Students can refer to these NCERT Solutions for their exam preparation as our highly experienced Maths teachers have prepared these solutions as per the latest CBSE guidelines for Class 10.

  • Students can refer to these NCERT Solutions whenever they are stuck with any sum. Thus, they can address their doubts in Quadratic Equations during the last-minute revisions without wasting any time or having to wait to consult any friend.


Chapter 4 - Quadratic Equations Exercises in PDF Format

Exercise 4.1

2 Questions & Solutions (1 Short Answer, 1 Long Answer)

Exercise 4.2

6 Questions & Solutions (6 Short Answers)

Exercise 4.3

11 Questions & Solutions (8 Short Answers, 3 Long Answers)

Exercise 4.4

5 Questions & Solutions (2 Short Answers, 3 Long Answer)


Along with this, students can also download additional study materials provided by Vedantu, for Chapter 4 of CBSE Class 10 Maths Solutions –

What is the Advantage of Class 10 Maths NCERT Solutions Chapter 4 Provided by Vedantu?

Maths NCERT Solutions Class 10 Chapter 4 includes all solutions for the numerical problem given in the NCERT book. It follows the latest instructions and guidelines announced by the NCERT. Vedantu doesn’t leave any question or concept that is important for the exams. All numerical problems are framed well and solved accurately in NCERT Class 10 Maths Chapter 4 solution. It would not be wrong if we say that the study materials provided by Vedantu is the bible for all those students who want to score good marks.


The subject experts at Vedantu have prepared the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations for the Term 2 exam preparation and revision purposes for students. To solve the sums covered in this chapter, students should have a sound understanding of other concepts of Maths, like perimeter and area, time and work, linear equations, etc. Therefore, downloading and referring to the Maths NCERT Solutions Class 10 Ch-4 will help students to prepare very well for their exams.

FAQs on NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

1. What is the Benefit of Taking Online Classes at Vedantu?

The subject’s experts take online classes available at Vedantu. They have lots of experience in their respective subjects. Along with this, they are working in their field with consistency. So, they have been faced with all kinds of problems and learned the tricks on how to come out from it.


In online classes, teachers share all their experiences with the students and teach some essential tricks. These tricks will be useful at the time of solving questions in the examination hall. Apart from the teaching concepts and formulas, they also motivate students and remove the fear of examinations from the student’s mind that is built up somewhere due to some kind of society’s pressure and other things.

2. Instead of Class 10th Maths Chapter 4, What is the Maths Syllabus for Class 10 for 2023-24?

Here is the complete syllabus for Class 10 Mathematics Revised Syllabus 2023-24:-

Unit- I: Number Systems

  • Real Numbers

Unit II: Algebra

  • Polynomials

  • Pair of Linear Equations in Two Variables

  • Quadratic Equations

  •  Arithmetic Progressions

Unit III: Coordinate Geometry

  • Lines (In two-dimensions)

 Unit IV: Geomtry

  • Triangles

  • Circles

  • Constructions

Unit V: Trigonometry

  • Introduction to Trigonometry

  • Trigonometric Identities

  •  Heights and Distances: Angle of elevation, Angle of Depression

Unit VI: Mensuration

  • Areas Related to Circles

  • Surface Areas and Volumes

Unit VII: Statistics and Probability

  •  Statistics

  • Probability

3. What is the Weightage for Class 10 Mathematics Unit-Wise?

Students who don’t know the weightage then they should go through the table given below. Here, we have mentioned the entire Class 10 Mathematics Unit-Wise Weightage for the knowledge of the students.


Units

Unit Name

Marks

I

NUMBER SYSTEMS

06

II

ALGEBRA

20

III

COORDINATE GEOMETRY

06

IV

GEOMETRY

15

V

TRIGONOMETRY

12

VI

MENSURATION

10

VII

STATISTICS & PROBABILITY

11

 

Total

80

4. Mention the important concepts that you learn in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

If you want to score 100 per cent marks in Class 10 Maths, you have to practice daily. Chapter 4 Quadratic Equations is an important chapter. Students can find NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations on Vedantu. There are five exercises in Chapter 4. Students can download the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to learn the important concepts that will help them understand the topic.

5. How to download Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Students can easily download Class 10 Maths Quadratic Equations NCERT textbooks PDF online. NCERT Solutions for Class 10 Maths Quadratic Equations are explained in an easy and simple language. Follow the given steps:

  • Click NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

  • Click on “Download PDF”.

  • Download and save it.

Students can use the PDF document without having an internet connection and can study Maths Quadratic Equations anytime. The NCERT Solutions give a clear understanding to them. 

6. What is the Quadratic formula Class 10th?

When a quadratic polynomial is equated to a constant, it forms a quadratic equation. An equation such as Ax = D, where Ax is a polynomial of degree two and D is a constant, forms a quadratic equation. The standard quadratic equation is ax2+bx+c=0 where a, b, and c are not equal to zero. You can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to understand more about the topic. You can also download Vedantu’s app. All the resources are available free of cost. 

7. What are the important topics covered in NCERT Solutions Class 10 Maths Chapter 4?

Chapter 4 Class 10 Maths is based on Quadratic Equations. The chapter includes important concepts about quadratic equations. This chapter includes five exercises that explain the different concepts of quadratic equations. The following topics are covered:

  • Exercise 4.1- Introduction

  • Exercise 4.2- Quadratic Equations

  • Exercise 4.3- Solution of a Quadratic Equation by Factorisation

  • Exercise 4.4- Solution of a Quadratic Equation by Completing the Square

  • Exercise 4.5- Nature of Roots

8. How do you solve Quadratic Equations in Class 10?

If you want to learn how to solve quadratic equations in Class 10, you can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. All the solutions are prepared by experts in an easy language. Students can understand the equations clearly. Students have to find the roots by using the quadratic formula. They can find the sum and product of both the roots. The method is simple and explained properly for easier understanding.