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

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

Ans: Given: Quadratic equations

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

Ans: Given: A quadratic equation

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.

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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

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

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability