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

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

The full form of NCERT stands for National Council of Educational Research and Training, which was founded in 1961 and offers students a number of services and advantages. The Ministry of Education, a division of the Indian Government, established the NCERT as an educational organisation and independent intelligence service.

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There are a few relevant topics that are covered in Class 10 Maths Chapter 4 Exercise 4.4 which are given below:

- Nature of roots of the Quadratic Equations
- The value of Discriminant

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

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### Class 10 Maths Chapter 4

Chapter 4 of the subject Mathematics in Class 10 is Quadratic Equations. Quadratic Equations are the Mathematical equations that consist of any formula that may be transformed into standard form, including ax2 + bx + c = 0.

Equations in Mathematics that have a squared variable are known as Quadratic Equations. The three different types of Quadratic Equations are Cubic, Quadratic, and Linear. The Quadratic Formula is used to solve the Quadratic Equation, which is the most challenging of the three.

The Mathematical equation known as the Quadratic Formula is used to resolve Quadratic Equations.

The Quadratic Formula is:

x = −b ± √b2 − 4ac2

Where:

x is the solution to the equation

b is the coefficient of the squared variable

a is the coefficient of the variable

c is the constant

Quadratic Equations have a variety of real-world applications. One such application is in the field of Engineering. Engineers often use Quadratic equations to model the motion of objects. For example, they might use a Quadratic equation to model the motion of a car as it speeds up and slows down. Quadratic equations can also be used to duplicate the motion of objects in planetary orbits.

Another real-world application of Quadratic equations is in the field of Economics. Economists often use Quadratic equations to model the behaviour of financial markets. For example, they might use a Quadratic equation to replicate the behaviour of stock prices.

Contrary to that, Quadratic equations can be used in the Physical Sciences. Physicists frequently use Quadratic equations to copy the actions of gases. For example, they may use a Quadratic equation to model the behaviour of air pressure.

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

NCERT Solutions are collections of questions and solutions that may be found on the Extramarks website. These NCERT Solutions are made to aid students in their examinations and provide them with a deeper comprehension of the subjects covered in the syllabus.The NCERT Solutions are helpful to many students since they not only help them understand the topics better but also provide answers to their queries.

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The NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 can be helpful in order to better understand the subject content. Moreover, because the NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 are straightforward to memorise and follow, students can successfully study for their exams with their assistance.Further, the solutions could help students enhance their Mathematical aptitudes. Extramarks’ NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 should be fully utilised by students looking for answers to NCERT Textbook questions.Solved Examples

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

There are in total 4 Exercises in Chapter 4, Quadratic Equations of Class 10. Each exercise holds different concepts and importance. In Exercise 4.1 students can discover questions related to the Quadratic Equation, how to solve them if certain scenarios occur, and the present age of a person. Exercise 4.2 is all about solving Quadratic Equations and obtaining their roots with the help of Factorisation.

On the other hand, in Exercise 4.3 students can explore the questions regarding solving the Quadratic Equations by Completing the square and Quadratic formula. Finally, Class 10 Maths Ch 4 Ex 4.4 of Quadratic Equations focuses on how to solve these equations using the Nature of Roots and determining the discriminants of the equations.. If students want to solve the questions mentioned in Class 10 Chapter 4 Exercise 4.4, they can check the NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 on the Extramarks website. These solutions are approved by the experts and, hence, are exact and trustworthy.

### NCERT Solutions for Class 10 Maths PDFs (Chapter-wise)

The NCERT Solutions, for example, NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 guides students in solving queries without making any Mathematical errors or mistakes. The solutions suggested by Extramarks are reviewed by subject matter experts or specialists. This advantage helps students to rely on them without any confusion. The Chapters that are there in the Mathematics NCERT Textbook of Class 10 are as follows:

- Real Numbers
- Polynomials
- Pair of Linear Equations in Two Variables
- Quadratic Equations
- Arithmetic Progressions
- Triangles
- Coordinate Geometry
- Introduction to Trigonometry
- Some applications of Trigonometry
- Circles
- Constructions
- Areas Related to Circles
- Surface Areas and Volumes
- Statistics
- Probability

The solutions for the chapters introduced in the Mathematics textbook of Class 10 can be found on the website of Extramarks. For the solutions for Class 10 Mathematical questions, students can download the mobile application of Extramarks and take benefit of its facilities.

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**Q.1 **

$\begin{array}{l}\text{Find the nature of the roots of the following quadratic}\\ \text{equations. If the real roots exist, find them.}\\ \text{(i)}2{\mathrm{x}}^{2}-3\mathrm{x}+5=0\text{(ii)\hspace{0.17em}\hspace{0.17em}}3{\mathrm{x}}^{2}-4\sqrt{3}\mathrm{x}+4=0\\ \text{(iii)}2{\mathrm{x}}^{2}-6\mathrm{x}+3=0\end{array}$

**Ans.**

$\begin{array}{l}\text{(i)}2{\mathrm{x}}^{2}-3\mathrm{x}+5=0\text{}\\ \text{Here,}\mathrm{a}=2,\text{}\mathrm{b}=-3,\text{}\mathrm{c}=5\\ \text{Therefore, discriminant}{\mathrm{b}}^{2}-4\mathrm{ac}={\left(-3\right)}^{2}-4\times 2\times 5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}9-40\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-31<0\\ \text{So, the given equation has no real roots.}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}}3{\mathrm{x}}^{2}-4\sqrt{3}\mathrm{x}+4=0\\ \text{Here,}\mathrm{a}=3,\text{}\mathrm{b}=-4\sqrt{3},\text{}\mathrm{c}=4\\ \text{Therefore, discriminant}{\mathrm{b}}^{2}-4\mathrm{ac}={\left(-4\sqrt{3}\right)}^{2}-4\times 3\times 4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}48-48\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \text{So, the given equation has two equal real roots.}\\ \text{The roots are}\frac{-\mathrm{b}}{2\mathrm{a}},\text{\hspace{0.17em}\hspace{0.17em}}\frac{-\mathrm{b}}{2\mathrm{a}},\text{\hspace{0.17em}\hspace{0.17em}i.e.,}\frac{4\sqrt{3}}{6},\text{\hspace{0.17em}\hspace{0.17em}}\frac{4\sqrt{3}}{6},\text{i.e.,}\frac{2}{\sqrt{3}},\text{}\frac{2}{\sqrt{3}}.\\ \text{(iii)}2{\mathrm{x}}^{2}-6\mathrm{x}+3=0\\ \text{Here,}\mathrm{a}=2,\text{}\mathrm{b}=-6,\text{}\mathrm{c}=3\\ \text{Therefore, discriminant}{\mathrm{b}}^{2}-4\mathrm{ac}={\left(-6\right)}^{2}-4\times 2\times 3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36-24\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12>0\\ \text{So, the given equation has two distinct real roots.}\\ \text{The roots are given by}\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^{2}-4\mathrm{ac}}}{2\mathrm{a}}\text{and \hspace{0.17em}}\frac{-\mathrm{b}\u2013\sqrt{{\mathrm{b}}^{2}-4\mathrm{ac}}}{2\mathrm{a}}.\\ \text{}\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^{2}-4\mathrm{ac}}}{2\mathrm{a}}=\frac{6\pm \sqrt{12}}{4}=\frac{3\pm \sqrt{3}}{2}\\ \text{Hence, the roots are}\frac{3+\sqrt{3}}{2}\text{and}\frac{3-\sqrt{3}}{2}.\end{array}$

**Q.2 ** Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x^{2} + kx +3 = 0

(ii) kx(x – 2) + 6 = 0

**Ans.**

$\begin{array}{l}\text{(i)}2{\mathrm{x}}^{2}+\mathrm{kx}+3=0\\ \text{Discriminant must be zero for two roots of the given}\\ \text{equation to be equal}.\\ \text{Therefore,}\\ \text{}{\mathrm{b}}^{2}-4\mathrm{ac}=0\\ \Rightarrow \text{\hspace{0.17em}}{\mathrm{k}}^{2}-4\times 2\times 3=0\\ \Rightarrow {\mathrm{k}}^{2}-24=0\\ \Rightarrow {\mathrm{k}}^{2}=24\\ \Rightarrow \mathrm{k}=\sqrt{24}=\pm 2\sqrt{6}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{kx}(\mathrm{x}-2)+6=0\\ \Rightarrow {\mathrm{kx}}^{2}-2\mathrm{kx}+6=0\\ \text{Discriminant must be zero for two roots of the given}\\ \text{equation to be equal}.\\ \text{Therefore,}\\ \text{}{\mathrm{b}}^{2}-4\mathrm{ac}=0\\ \Rightarrow \text{\hspace{0.17em}}4{\mathrm{k}}^{2}-4\times \mathrm{k}\times 6=0\\ \Rightarrow {\mathrm{k}}^{2}-6\mathrm{k}=0\\ \Rightarrow \mathrm{k}(\mathrm{k}-6)=0\\ \Rightarrow \mathrm{k}=0\text{or}\mathrm{k}=6\\ \text{But}\mathrm{k}=0\text{does not satisfy the given equation.}\\ \text{Thus, the given equation has equal roots if the value of}\mathrm{k}\text{}\\ \text{is}6\text{only.}\end{array}$

**Q.3 ** 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.

**Ans.**

$\begin{array}{l}\text{Let breadth of the rectangular mango grove is}\mathrm{x}\text{.}\\ \text{According to question,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\times 2\mathrm{x}=\text{800}\\ \Rightarrow {\text{x}}^{2}=400\\ \Rightarrow {\text{x}}^{2}-400=0\\ \text{Discriminant}={\mathrm{b}}^{2}-4\mathrm{ac}=0-4\times 1\times (-400)=1600>0\\ \text{Therefore, the above equation has two distinct real roots.}\\ \text{Now,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{2}-400=0\\ \Rightarrow {\text{x}}^{2}=400\\ \Rightarrow \text{x}=\pm 20\\ \text{But breadth can not be negative.}\\ \text{So,}\mathrm{x}=20\text{and 2}\mathrm{x}=40\\ \text{Hence, length}=40\text{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}breadth}=20\text{m}\end{array}$

**Q.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.**

$\begin{array}{l}\text{Let age of one friend be}\mathrm{x}\text{years.}\\ \text{Age of the other friend}=(20-\mathrm{x})\text{years}\\ \text{According to question,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(\mathrm{x}-4)(20-\mathrm{x}-4)=4\text{8}\\ \Rightarrow (\mathrm{x}-4)(16-\mathrm{x})=4\text{8}\\ \Rightarrow -{\text{x}}^{2}+16\mathrm{x}-64+4\mathrm{x}-48=0\\ \Rightarrow {\mathrm{x}}^{2}-20\mathrm{x}+112=0\\ \text{Discriminant}={\mathrm{b}}^{2}-4\mathrm{ac}={(-20)}^{2}-4\times 112=400-448=-48<0\\ \text{Therefore, no real root is possible and so the given situation}\\ \text{is not possible.}\end{array}$

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

**Ans.**

$\begin{array}{l}\text{Let length and breadth of the rectangular park be}\mathrm{x}\text{.}\\ \text{and}\mathrm{y}\text{respectively.}\\ \text{According to question, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2(\mathrm{x}+\mathrm{y})=\text{80}\\ \Rightarrow \mathrm{x}+\mathrm{y}=40\\ \Rightarrow \mathrm{y}=40-\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{Also, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{xy}=400\\ \Rightarrow \mathrm{x}(40-\mathrm{x})=400\\ \Rightarrow 40\mathrm{x}-{\mathrm{x}}^{2}=400\\ \Rightarrow {\text{x}}^{2}-40\mathrm{x}+400=0\\ \text{Discriminant}={\mathrm{b}}^{2}-4\mathrm{ac}={(-40)}^{2}-4\times 400=0\\ \text{So, the above equation has two equal real roots.}\\ \text{Now,}\\ \text{}{\mathrm{x}}^{2}-40\mathrm{x}+400=0\\ \Rightarrow {(\mathrm{x}-20)}^{2}=0\\ \Rightarrow \mathrm{x}=20\text{or}\mathrm{x}=20\\ \text{From equation (1), we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=40-\mathrm{x}=40-20=20\\ \text{Hence, length and breadth of the rectangular park are}\\ \text{20 m and 20 m respectively.}\end{array}$

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