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# NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.3

Last updated date: 10th Aug 2024
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## NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 Quadratic Equations - FREE PDF Download

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3, "Quadratic Equations," focuses on understanding and solving quadratic equations. Exercise 4.3 specifically deals with solving these equations using various methods like factorization, completing the square, and the quadratic formula. This exercise is crucial for building a strong foundation in algebra and enhancing problem-solving skills.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 Quadratic Equations - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 4 Exercise 4.3 Class 10 | Vedantu
3. Formulas Used in Class 10 Chapter 4 Exercise 4.3
4. Access NCERT Solutions for Maths Class 10 Chapter 4 - Quadratic Equations Exercise 4.3
5. Class 10 Maths Chapter 4: Exercises Breakdown
6. CBSE Class 10 Maths Chapter 4 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

Students should focus on mastering the quadratic formula and understanding how to apply it to different problems. The comprehensive solutions provided by Vedantu ensure that students grasp the concepts thoroughly, making it easier to tackle similar problems in exams.

## Glance on NCERT Solutions Maths Chapter 4 Exercise 4.3 Class 10 | Vedantu

• NCERT for Class 10 Maths Chapter 4 Exercise 4.3, deals with using the discriminant, and how to determine the nature of roots (solutions) of quadratic equations.

• Discriminant value indicates how many and what kind of roots a quadratic equation has.

• Nature of Roots based on Discriminant:

1. D > 0: Two distinct real roots (solutions can be calculated using the quadratic formula).

2. D = 0: Two equal real roots (solutions will be identical).

3. D < 0: No real root (solutions will be complex numbers).

• Completing the Square Method transforms a standard quadratic equation $ax^2+bx+c=0$ into a perfect square trinomial, allowing for straightforward solutions.

• Class 10 Math Chapter 4 Exercise 4.3 Solutions NCERT Solutions has overall 5 fully solved questions.

## Formulas Used in Class 10 Chapter 4 Exercise 4.3

### Cross Multiplication Method:

• For solving the pair of linear equations:

$a_{1} x + b_{1} y +c_{1} = 0$

$a_{2} x + b_{2} y +c_{2} = 0$

• The cross-multiplication formulas are:

$x = \dfrac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

$y = \dfrac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

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## Access NCERT Solutions for Maths Class 10 Chapter 4 - Quadratic Equations Exercise 4.3

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, $b = 40 - l = 40 - 20 = 20\;{\text{m}}$

## Conclusion

Class 10 Maths Exercise 4.3 Chapter 4 covers "Quadratic Equations." This exercise focuses on solving quadratic equations by completing the square, a key algebraic technique. Using NCERT Solutions by Vedantu enhances understanding of Quadratic Equations and the discriminant concept. These equations are vital in physics, engineering, economics, and more. Regular practice with Vedantu's NCERT solutions boosts comprehension and problem-solving skills. Focus on step-by-step solutions to grasp principles and ensure concept clarity.

## Class 10 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Exercise 4.1 2 Questions and Solutions Exercise 4.2 6 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

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

## FAQs on NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.3

1. Which Topic is Included in Class 10 Maths Chapter 4 Exercise 4.3?

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 4.3 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 4.3?

NCERT solutions for class 10 maths chapter 4 exercise 4.3 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 4.3 of Class 10 Maths?

There are a total of five questions in Exercise 4.3 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 4.3, 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 4.3 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 4.3 of 10th Maths?

Students may find NCERT Solutions For Class 10  Maths Exercise 4.3 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.

8. What is covered in ex 4.3 Class 10 Maths Chapter 4?

Ex 4.3 Class 10 Maths in Chapter 4 focuses on solving quadratic equations using the method of completing the square. This exercise is designed to help students understand and apply this method to find the roots of quadratic equations.

9. Why is completing the square an important method to learn in class 10 ex 4.3?

In Class 10 ex 4.3, completing the square is a crucial algebraic method that not only helps in solving quadratic equations but also lays the foundation for understanding more advanced mathematical concepts, such as the derivation of the quadratic formula and the analysis of quadratic functions.

10. Are there any common mistakes to avoid while solving class 10 maths ex 4.3 problems?

Yes, some common mistakes included in class 10 maths ex 4.3 are:

• Forgetting to divide the entire equation by the coefficient of x^2  when it is not 1.

• Incorrectly adding and subtracting the square of half the coefficient of x.

• Arithmetic errors while simplifying the equation.

• Not considering both the positive and negative roots when taking the square root.

11. How important is Class 10th Exercise 4.3 for the board exams?

Class 10th Exercise 4.3 is very important for the board exams as it strengthens your problem-solving skills and understanding of quadratic equations, which are frequently tested topics. Mastering this exercise can significantly boost your confidence and performance in the exams. By practicing these solutions, students can ensure they are well-prepared for their exams and have a solid understanding of the concepts involved in solving quadratic equations using the method of completing the square.