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

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

Edited By Ramraj Saini | Updated on Nov 16, 2023 11:56 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.4

NCERT Solutions for Exercise 4.4 Class 10 Maths Chapter 4 Quadratic Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 4.4 class 10 deals with the various types of questions that covers the whole topic of the nature of roots with the discriminant that is referred as “D”. The nature of the root depends on what the value of the discriminant has occurred. Discriminant of quadratic equation is (b2 - 4ac ). In exercise 4.4 Class 10 Maths, there are three main methods to solve the quadratic equation i.e. Factorizations, Completing the square and Quadratic Formula.

NCERT solutions for Class 10 Maths exercise 4.4, focused on the concepts of how we can find out the nature of roots of the equation ax2+ bx + c = 0 with the help of the discriminant of the quadratic equation. These class 10 maths ex 4.4 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Quadratic Equations class 10 chapter 4 Exercise: 4.4

Q1 (i) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

2x^2 - 3x +5 = 0

Answer:

For a quadratic equation, ax^2+bx+c = 0 the value of discriminant determines the nature of roots and is equal to:

D = b^2-4ac

If D>0 then roots are distinct and real.

If D<0 then no real roots.

If D= 0 then there exists two equal real roots.

Given the quadratic equation, 2x^2 - 3x +5 = 0 .

Comparing with general to get the values of a,b,c.

a = 2, b =-3,\ c= 5

Finding the discriminant:

D= (-3)^2 - 4(2)(5) = 9-40 = -31

\because D<0

Here D is negative hence there are no real roots possible for the given equation.

Q1 (ii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 3x^2 - 4\sqrt3x + 4 = 0

Answer:

b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0

Here the value of discriminant =0, which implies that roots exist and the roots are equal.

The roots are given by the formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}

So the roots are

\frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}

Q1 (iii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

2x^2 - 6x + 3 = 0

Answer:

The value of the discriminant

b^2-4ac=(-6)^2-4\times2\times3=12

The discriminant > 0. Therefore the given quadratic equation has two distinct real root

roots are

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}

So the roots are

\frac{3}{2}+\frac{\sqrt{3}}{2}, \frac{3}{2}-\frac{\sqrt{3}}{2}

Q2 (i) Find the values of k for each of the following quadratic equations so that they have two equal roots.

2x^2 + kx + 3 = 0

Answer:

For two equal roots for the quadratic equation: ax^2+bx+c =0

The value of the discriminant D= 0 .

Given equation: 2x^2 + kx + 3 = 0

Comparing and getting the values of a,b, and, c.

a = 2, \ b = k,\ c = 3

The value of D = b^2-4ac = (k)^2 - 4(2)(3)

\Rightarrow (k)^2 = 24

Or, \Rightarrow k=\pm \sqrt{24} = \pm 2\sqrt{6}

Q2 (ii) Find the values of k for each of the following quadratic equations so that they have two equal roots

kx(x-2) + 6 = 0

Answer:

For two equal roots for the quadratic equation: ax^2+bx+c =0

The value of the discriminant D= 0 .

Given equation: kx(x-2) + 6 = 0

Can be written as: kx^2-2kx+6 = 0

Comparing and getting the values of a,b, and, c.

a = k, \ b = -2k,\ c = 6

The value of D = b^2-4ac = (-2k)^2 - 4(k)(6) = 0

\Rightarrow 4k^2 - 24k = 0

\Rightarrow 4k(k-6) = 0

\Rightarrow k= 0\ or\ 6

But k= 0 is NOT possible because it will not satisfy the given equation.

Hence the only value of k is 6 to get two equal roots.

Q3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m 2 ? If so, find its length and breadth.

Answer:

Let the breadth of mango grove be 'b' .

Then the length of mango grove will be '2b' .

And the area will be:

Area = (2b)(b) = 2b^2

Which will be equal to 800m^2 according to question.

\Rightarrow 2b^2 = 800m^2

\Rightarrow b^2 - 400 = 0

Comparing to get the values of a,b,c .

a=1, \ b= 0 , \ c = -400

Finding the discriminant value:

D = b^2-4ac

\Rightarrow 0^2-4(1)(-400) = 1600

Here, D>0

Therefore, the equation will have real roots.

And hence finding the dimensions:

\Rightarrow b^2 - 400 = 0

\Rightarrow b = \pm 20

As negative value is not possible, hence the value of breadth of mango grove will be 20m.

And the length of mango grove will be: = 2\times10 = 40m

Q4 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.

Answer:

Let the age of one friend be x\ years.

and the age of another friend will be: (20-x)\ years.

4 years ago, their ages were, x-4\ years and 20-x-4 \ years .

According to the question, the product of their ages in years was 48.

\therefore (x-4)(20-x-4) = 48

\Rightarrow 16x-64-x^2+4x= 48

\Rightarrow -x^2+20x-112 = 0 or \Rightarrow x^2-20x+112 = 0

Now, comparing to get the values of a,\ b,\ c .

a = 1,\ b= -20,\ c =112

Discriminant value D = b^2-4ac = (-20)^2 -4(1)(112) = 400-448 = -48

As D<0 .

Therefore, there are no real roots possible for this given equation and hence,

This situation is NOT possible.

Q5 Is it possible to design a rectangular park of perimeter 80 m and area 400 m 2 ? If so, find its length and breadth.

Answer:

Let us assume the length and breadth of the park be 'l'\ and\ 'b' respectively.

Then, the perimeter will be P = 2(l+b) = 80

\Rightarrow l+b = 40\ or\ b = 40 - l

The area of the park is:

Area = l\times b = l(40-l) = 40l - l^2

Given : 40l - l^2 = 400

l^2 - 40l +400 = 0

Comparing to get the values of a, b and c.

The value of the discriminant D = b^2-4ac

\Rightarrow = b^2-4ac = (-40)^2 - 4(1)(400) = 1600 -1600 = 0

As D = 0 .

Therefore, this equation will have two equal roots.

And hence the roots will be:

l =\frac{-b}{2a}

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

Therefore, the length of the park, l =20\ m

and breadth of the park b = 40-l = 40 -20 = 20\ m .



More About NCERT Solutions for Class 10 Maths Exercise 4.4

Class 10th Maths chapter 4 exercise 4: The questions in exercise 4.4 Class 10 Maths, broadly consist of three types of questions. Firstly we have a direct question in which we are given a quadratic equation in which we have to find the nature of the root .then we have given the nature of the root and we have to find the missing variable in the quadratic equation along with this we have word problems. NCERT syllabus Class 10 Maths exercise 4.4 mainly focus on the application based situation question on nature of roots of the quadratic equation. Exercise 4.4 Class 10 Maths covers all types of questions that can be formed on the nature of roots. Also students can get access of Quadratic Equations Class 10 Notes to revise all the concepts quickly.

Benefits of NCERT Solutions for Class 10 Maths Exercise 4.4

  • Class 10 Maths chapter 4 exercise 4.4 broadly covers all kinds of questions that can be formed on the nature of roots of a quadratic equation


  • NCERT Class 10 Maths chapter 4 exercise 4.4, will be helpful in JEE Main (Joint Entrance Exam) as quadratic equations are a major part of some chapters.


  • Exercise 4.4 Class 10 Maths, is based on the nature of the roots of the quadratic equation and its application, which is an important concept of the chapter.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What does the nature of root mean according to NCERT solutions for Class 10 Maths 1 exercise 4.4 ?

Nature of root tell us that whether the roots are equal or distinct and they also tell us roots are real or non real

2. What is discriminant according to NCERT solutions for Class 10 Maths 1 exercise 4.4 ?

Discriminant of quadratic equation is (b- 4ac )  and we substitute the value of b, c, a from ax2+ bx + c = 0

3. What are non real roots according to NCERT solutions for Class 10 Maths 1 exercise 4.4 ?

There are three ways to categorise discriminant: 9I) two distinct real roots, if (b- 4ac)> 0; (ii) two equal real roots, if b- 4ac)= 0; (iii) no real roots, if (b- 4ac)=0

4. When roots are equal what will be the general value of roots?

Non-real roots are imaginary roots because discriminant is negative that is why they are unsolvable under root.

5. What is the number of solved examples before the Exercise 4.4 Class 10 Maths which are based on the nature of root?

There are mainly 3 questions that are solved before the Class 10 Maths chapter 4 exercise 4.4 which are based on the nature of the root 

6. How many questions are there in the Exercise 4.4 Class 10 Maths ?

There are five questions in exercise 4.4 Class 10 Maths question one has three subparts and question two have two subparts.

7. How many types of questions are there in the Exercise 4.4 Class 10 Maths and explain each type?

There are three types of questions in exercise 4.4 Class 10 Maths question one has direct subparts to find the nature of the root of the quadratic equation then in question two we have given nature of root and we have to find the missing variable in the quadratic equation .question three, four and five are word problem which is based on the real-world application.

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

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According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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0.34\; J

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1.00\; J

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0.67\; J

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