NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4
Obtain Complimentary NCERT Solutions for Chapter 4 Ex 4.4 Quadratic Equations in Class 10 Maths. These Class 10 Maths NCERT Solutions are invaluable aids for completing homework assignments. The Exercise 4.4 Class 10 Maths NCERT Solutions were meticulously crafted by seasoned educators from School Notes Wallah. You can find comprehensive solutions to all the questions in Chapter 4 of the Class 10 Maths NCERT Textbook, specifically focusing on Quadratic Equations, in Exercise 4.4.
Topics and Sub Topics in Class 10 Maths Chapter 4 Quadratic Equations:
| Section Name | Topic Name |
| 4 | Quadratic Equations |
| 4.1 | Introduction |
| 4.2 | Quadratic Equations |
| 4.3 | Solution of a Quadratic Equation by Factorisation |
| 4.4 | Solution of a Quadratic Equation by Completing the Square |
| 4.5 | Nature of Roots |
| 4.6 | Summary |
You can also download the free PDF of Chapter 4 Ex 4.4 Quadratic Equations NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Quadratic Equations |
| Exercise | Ex 4.4 |
| Number of Questions Solved | 5 |
| Category | NCERT Solutions |
NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4.
EXERCISE 4.4
Q.1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0 (ii) 
(iii) 2x2 – 6x + 3 – 0
Sol. (i) 2x2 – 3x + 5 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 2
b= – 3
c = 5
∴ The discriminant = b2 – 4ac
= ( – 3)2 – 4(2)(5)
= – 9 – 40
= – 31 < 0
Since b2 – 4ac is negative.
∴ The given quadratic equation has no real roots.
(i) 
Comparing the given quadratic equation with ax2 + bx + c= 0, we get
Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:
(iii) 2x2 – 6x + 3 = 0
Comparing it with the general quadratic equation, we have:
a = 2
b = –6
c = 3
∴ The given quadratic equation has two real and distinct roots, which are given by
Q.2. Find the values of k for each of the fallowing quadratic equations, so that they have two equal roots:
(i) 2x2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Sol. (i) 2x2 + kx + 3 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = 2
b = k
c = 3
b2 – 4ac = ( – k)2 – 4 (2) (3)
= k2 – 24
∵ For a quadratic equation to have equal roots,
b2 – 4ac = 0
(ii) kx(x – 2) + 6 = 0
Comparing kx (x – 2) + 6 = 0 i.e., kx2 – 2kx + 6 = 0 with ax2 + bx + c = 0, we get
a = k
b = – a
c = 6
∴ b2 – 4ac = ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
Since, the roots are real and equal,
∴ b2 – 4ac = 0
⇒ 4k2 – 24k = 0
⇒ 4k(K – 6) = 0
⇒ 4k = 0 or k – 6=0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.
Q.3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m? If so, find its length and breadth.
Sol. Let the breadth be x metres.
∴ Length = 2x metres
Now, Area = Length × Breadth
= 2x × x metre2
= 2x2 sq. metre.
According to the given condition,
2x2 = 800
Therefore, x = 20 and x = – 20
But x = – 20 is possible (∵ breadth cannot be negative).
∴ x = 20
⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m
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.
Sol. Let the age of one friend = x years
∴ The age of the other friend = (20 – x – 4) years [∵ Sum of their ages is 20 years]
Four years ago
Age of one friend = (x – 4) years
Age of the other friend = (20 – x – 4) years
= (16 – x) years
According to the condition,
(x – 4) × (16 – x) = 48
⇒ 16x – 64 – x2 – 4x = 48
⇒ – x2 – 20x – 64 – 48 = 0
⇒ – x2 – 20x – 112 = 0
⇒ x2 + 20x + 112 – 0 (1)
Here, a = 1, b = 20 and c = 112
∴ b2 – 4ac = (20)2 – 4 (1) (112)
= 400 – 448
= – 48 < 0
Since b2 – 4ac is Is than O.
∴ The quadratic equation (1) has no real roots.
Thus, the given equation is not possible.
Q.5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Sol. Let the breadth of the rectangle be x metres.
Since, the parimetee of the rectangle = 80 m
∴ 2 [Length + Breadth] = 80
2 [Length + x] 80
⇒ Length = (40 – x) metres
∴ Area of the rectangle = Length × breadth
= (40 – x) × x sq. m
= 40x – x2
Now, according to the given condition,
Area of the rectangle = 400 m2
∴ 40x – x2 = 400
⇒ – x2 + 40x – 400 = 0
⇒ x2 – 40x + 400 = 0
Comparing (1) with axe + bx + c = 0, we get
a = 1
b = –40
c = 400
∴ b2 – 4ac = ( – 40)2 – 4 (1) (400)
= 1600 – 1600 – 0
Thus, the equation (1) has two equal and real roots.
∴ Breadth, x = 20 m
∴ Length = (40 – x) = (40 – 20) m = 20 m.
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