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# NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4

## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.4) Exercise 4.4

Last updated date: 04th Feb 2023
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Chapter 4 of the class 10 maths syllabus is on Quadratic Equations. It is an important chapter that is covered in class 10 and is divided into 6 major sections that we recommend that students go through very carefully to be able to learn, understand, and retain all the information provided in our solutions. These solutions are created by our expert teachers, keeping the students’ best interests in mind, and we are certain that covering these topics will aid students in scoring well in their exams.

The following is a table containing the 6 important topics covered in the chapter on quadratic equations.

 Sl. No. Important Topics 1. Introduction 2. Quadratic Equations 3. Solution to a Quadratic Equation using Factorisation 4. Solution to a Quadratic Equation using Completion of the Square 5. The Nature of Roots 6. Summary

NCERT solutions are the best guide for the students of class 10. It covers every detail about the chapter. Since class 10 is a very crucial year for the students, they need to have effective learning. A lot of practice is required to get good marks. Class 10 Maths chapter 4 deals with the Quadratic equation and exercise 4.4 of chapter 4 includes problems of finding the nature of the roots and some related questions. The stepwise solutions to the problems are of great help for the students in better practicing and scoring well. They also help in analyzing their weak points. You can easily download free PDF of NCERT Solution of Maths Class 10 Chapter 4 Exercise 4.4 to improve your exam preparation. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 10 Science, Maths solutions and solutions of other subjects.

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

Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;

I. $2{x^2} - 3x + 5 = 0$

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

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

To find: nature of the roots.

We know that for a quadratic equation $a{x^2} + bx + c = 0$

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

(A) If ${b^2} - 4ac > 0 \to$ two distinct real roots

(B) If ${b^2} - 4ac = 0 \to$ two equal real roots

(C) If ${b^2} - 4ac < 0 \to$ no real roots

(I) $\quad 2{{\text{x}}^2} - 3{\text{x}} + 5 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} = - 3,{\text{c}} = 5$

Discriminant : ${b^2} - 4ac$ $= {( - 3)^2} - 4(2)(5) = 9 - 40$

$= - 31$

As ${b^2} - 4ac < 0$

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

(II) $3{{\text{x}}^2} - 4\sqrt {3{\text{x}}} + 4 = 0$

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain $a = 3,b = - 4\sqrt 3 ,c = 4$

Discriminant $:{b^2} - 4ac = {( - 4\sqrt 3 )^2} - 4(3)(4)$

$= 48 - 48 = 0$

As ${b^2} - 4ac = 0$

Therefore, real roots exist for the given equation and they are equal to each other. And the roots will be $\dfrac{{ - b}}{{2a}}$ and $\dfrac{{ - b}}{{2a}}$.

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

Therefore, the roots are $\dfrac{2}{{\sqrt 3 }}$ and $\dfrac{2}{{\sqrt 3 }}$

(III) $2{{\text{x}}^2} - 6{\text{x}} + 3 = 0$

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} = - 6,{\text{c}} = 3$

Discriminant $= {b^2} - 4ac = {( - 6)^2} - 4(2)(3)$

$= 36 - 24 = 12$

As ${b^2} - 4ac > 0$

Therefore, distinct real roots exist for the given equation as follows. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$= \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)(3)} }}{{2(2)}}$

$= \dfrac{{6 \pm \sqrt {12} }}{4} = \dfrac{{6 \pm 2\sqrt 3 }}{4}$

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

Therefore, the root are $\dfrac{{3 + \sqrt 3 }}{2}$ or $\dfrac{{3 - \sqrt 3 }}{2}$

2. Find the values of ${\text{k}}$ for each of the following quadratic equations, so that they have two equal roots.

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

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

To find: k if the roots are equal

We know that if an equation $a{x^2} + bx + c = 0$ has two equal roots, its Discriminant $\left( {{b^2} - 4ac} \right)$ will be 0 .

(I) $2{{\text{x}}^2} + {\text{kx}} + 3 = 0$

Comparing equation with $a{x^2} + bx + c = 0$, we obtain

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

Discriminant $= {b^2} - 4ac = {(k)^2} - 4(2)(3)$

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

For equal roots, Discriminant $= 0$

${k^2} - 24 = 0$

${k^2} = 24$

$k = \pm \sqrt {24} = \pm 2\sqrt 6$

(II) ${\text{kx}}({\text{x}} - 2) + 6 = 0$

or $k{x^2} - 2kx + 6 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain $a = k,b = - 2k,c = 6$

Discriminant $= {b^2} - 4ac = {( - 2k)^2} - 4(k)(6)$

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

For equal roots, ${{\text{b}}^2} - 4{\text{ac}} = 0$

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

$4k(k - 6) = 0$

Either $4{\text{k}} = 0$ or ${\text{k}} = 6 = 0$

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

However, if ${\text{k}} = 0$, then the equation will not have the terms ' ${{\text{x}}^2}$ ' and ' ${\text{x}}$' .

Therefore, if the given equation has two equal roots, k should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\;{{\text{m}}^2}$ ? If so, find its length and breadth.

Ans: Given: A rectangular mango grove.

To find: its dimensions

Let us consider the breadth of mango grove be $l$

Length of mango grove will be 2l . Area of mango grove $= (2l)(l) = 2{l^2}$ $= 800$

${l^2} = \dfrac{{800}}{2} = 400$

${l^2} - 400 = 0$

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain $a = 1 ,b = 0,c = 400$

Discriminant $= {b^2} - 4ac = {(0)^2} - 4 \times (1) \times ( - 400) = 1600$

Here, ${b^2} - 4ac > 0$

Therefore, the equation will have real roots.

$l = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$l = \dfrac{{ - 0 \pm \sqrt {1600}}}{{2*1}}$

And Therefore, the desired rectangular mango grove can be designed. $l = \pm 20$

However, length cannot be negative. Therefore, breadth of mango grove $= 20\;{\text{m}}$

Length of mango grove $= 2 \times 20 = 40\;{\text{m}}$

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

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not?

Let us consider the age of one friend be $x$ years.

Age of the other friend will be $(20 - {\text{x}})$ years.

4 years ago, age of 1 st friend $= ({\text{x}} - 4)$ years

And, age of ${2^{{\text{nd }}}}$ friend $= (20 - {\text{x}} - 4)$ $= (16 - {\text{x}})$ years

Given that, $(x - 4)(16 - x) = 48$

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

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

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

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 1,\;{\text{b}} = - 20,{\text{c}} = 112$

Discriminant $= {b^2} - 4ac = {( - 20)^2} - 4(1)(112)$

$= 400 - 448 = - 48$

As ${b^2} - 4ac < 0$

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

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

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not?

Let us consider the length and breadth of the park be l and ${\text{b}}$

Perimeter $= 2(l + b) = 80$ $l + b = 40$

Or, ${\text{b}} = 40 - l$

Area $= l \times {\text{b}} = l(40 - l) = 40l - {l^2}$

$40l - {l^2} = 400$

${l^2} - 40l + 400 = 0$

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain

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

Discriminate $= {b^2} - 4ac = {( - 40)^2} - 4(1)(400)$

$= 1600 - 1600 = 0$

As ${b^2} - 4ac = 0$

Therefore, the given equation has equal real roots. And Therefore, the given situation

is possible. Root of the given equation, $l = - \dfrac{b}{{2a}}$

$l = - \dfrac{{( - 40)}}{{2(1)}} = \dfrac{{40}}{2} = 20$

Therefore, length of park, $l = 20\;{\text{m}}$

And breadth of park, $b = 40 - l = 40 - 20 = 20\;{\text{m}}$

## Class 10 Maths Chapter 4

In class 10th maths chapter 4 consists of Quadratic equation which includes topics like checking a quadratic equation, problem statements solving using equations, finding the roots of an equation using factorization and Sridhar Acharya formula, finding the roots (if they exist), using the method of completing the square and in exercise 4.4, the nature of roots is calculated using the value of the determinant.

### Importance of NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

NCERT solutions for class 10 maths chapter 4 exercise 4.4 helps in the following ways:

• It saves students from last-minute revision along with step by step solutions.

• It helps students in better practicing in a short period.

• It creates an optimum learning environment.

• By practicing, questions on the nature of roots and related topics, students will understand essential points in more depth.

### Solved Examples

1. From the following quadratic equations, find the nature of the roots. If the real roots exist, find them;

(i) 2x2 – 3x + 5 = 0

(ii) 3x2 – 4√3x + 4 = 0

(iii) 2x2 – 6x + 3 = 0

Solutions:

(i) Given,

2x2 – 3x + 5 = 0

Comparing it to the equation ax2 + bx + c = 0, we get

a, b and c are therefore 2, -3 and 5, respectively.

We know, Discriminant = b2 – 4ac

Putting the values, we get

= – 31

As we can see, b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0.

(ii) 3x2 – 4√3x + 4 = 0

Comparing it to the equation ax2 + bx + c = 0, we get

a, b and c are therefore, 3, -4√3 and 4, respectively.

We know, Discriminant = b2 – 4ac

Putting the values, we get 0.

As b2 – 4ac = 0,

For the given equation, real roots exist and they are equal to each other.

Hence the roots will be –b/2a and –b/2a.

–b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2x2 – 6x + 3 = 0

Comparing it to the equation ax2 + bx + c = 0, we get

a, b and c are therefore, 2, -6 and 3, respectively.

As we know, Discriminant = b2 – 4ac

Putting the values, we get

b2 – 4ac > 0,

Therefore, distinct real roots exist for this equation, 2x2 – 6x + 3 = 0.

Now, According to Sridhar Acharya’s formula,

= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)

= (6±2√3 )/4

= (3±√3)/2

For the given equation, the roots are (3+√3)/2 and (3-√3)/2.

2. For each of the following quadratic equations, find the value of k, so that they have two equal roots.

(i) 2x2 + kx + 3 = 0

(ii) kx2 – 2kx + 6 = 0

Solutions:

(i) 2x2 + kx + 3 = 0

Comparing it to the equation ax2 + bx + c = 0, we get,

a, b and c are therefore, 2, k and 3, respectively.

As we know, Discriminant = b2 – 4ac

= (k)2 – 4(2) (3)

= k2 – 24

For equal roots, we know,

Discriminant = 0

k2 – 24 = 0

k2 = 24

k = ±√24 = ±2√6

(ii) kx2 – 2kx + 6 = 0

Comparing it to the equation ax2 + bx + c = 0, we get

a, b and c are therefore, k, 2k and 6, respectively.

We know, Discriminant = b2 – 4ac

= ( – 2k)2 – 4 (k) (6)

= 4k2 – 24k

For equal roots, we know,

b2 – 4ac = 0

4k2 – 24k = 0

4k (k – 6) = 0

Therefore,

Either 4k=0

or k-6=0

k = 0 or k = 6

If k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘, so it is not possible.

Therefore, k should be 6, if this equation has two equal roots.

### NCERT Solutions for Class 10 Maths Chapter 4 Exercises

Chapter 4 Quadratic Equations All Exercises in PDF Format

Exercise 4.1

2 Questions and Solutions

Exercise 4.2

6 Questions and Solutions

Exercise 4.3

11 Questions and Solutions

## FAQs on NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.4

1. Which Topic is Included in Class 10 Maths Chapter 4 Exercise 4.4?

In class 10th maths chapter 4 consists of Quadratic equation which includes topics like checking whether a given equation is quadratic or not, real-life problem statements which are to be solved with equations, finding the roots of an equation using factorization, finding the roots of the equation (if they exist), using the method of completing the square, finding roots of a quadratic equation using SridharAcharya formula.

Exercise 4.4 consists of finding the roots of an equation, discussing real roots, equal roots, and imaginary roots cases, finding the discriminant and then depending on the value finding the roots of the equation.

2. What is the Advantage of NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4?

NCERT solutions for class 10 maths chapter 4 exercise 4.4 is advantageous to the students in the following ways:

• The solutions contain accurate answers, step by step procedures, that facilitates clarity in concepts of the students.

• NCERT solutions are precise yet comprehensive. They are most important for effective learning.

• In the crucial year of class 10, the students would need a lot of practice to score good marks. The NCERT solutions would help them in doing so.

• The students would also be able in analysing their weak points in the course of practice. This would help them in concentrating on the weaker parts.

3. How many questions are there in Exercise 4.4 of Class 10 Maths?

There are a total of five questions in Exercise 4.4 of Class 10 Maths. Chapter 4 in Class 10 Maths deals with Quadratic Equations. This chapter includes topics such as quadratic equations, finding the roots of quadratic equations, nature of roots etc. In Exercise 4.4, the questions involve finding the nature of roots, finding the unknown values and few word problems that make the students learn the application of quadratic equations. The NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.4 are available on the official website of Vedantu.

4. How do we find Discriminants?

For any equation in the form ax²+bx+c, where a, b, and c are constant terms and a≠0, Discriminant is given by: D = b² - 4ac. The discriminant of an equation is denoted by D. The discriminant helps in finding the nature of roots of quadratic equations.

• If D = 0, the roots are real and equal.

• If D < 0, the roots are not real.

• If D > 0, the roots are real and unequal.

5. Where can I get NCERT Solutions for Exercise 4.4 of 10th Maths?

Students may find NCERT Solutions For Class 10  Maths Exercise 4.4 on the official website of Vedantu. It is available for free download. The NCERT Solutions are presented in an easy and detailed manner. They have thorough explanations of topics. Students will get answers to all NCERT questions here. Subject-Matter Experts have made sure to curate the best answers for students. You can go through these solutions to clear your queries or doubts. You can access the study material on Vedantu App as well. All the material is available free of cost.

6. How to find the nature of roots of a quadratic equation?

To establish the nature of quadratic equation roots (of the form ax2 + bx + c = 0), we must compute the discriminant, which is b2 - 4ac. When the discriminant exceeds 0, the roots are unequal and real. When the discriminant equals zero, the roots are both equal and real. The roots are imaginary when the discriminant is less than zero.

7. How can I score good marks in Class 10 Maths?

Class 10 is an extremely important class for every student. Mathematics plays a vital role in building the bases of almost every subject. Among all the subjects, most of the students find Maths as a difficult subject. To solve this problem, Vedantu brings to you a student-friendly platform that helps them clear all the important concepts in Maths with notes and NCERT Solutions. With the help of this and lots of determination, you are ready to excel in Class 10. SHARE TWEET SHARE SUBSCRIBE