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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4




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

In this page we have NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations for EXERCISE 4.4 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Nature of roots of Quadratic equation
S.no
Condition
Nature of roots
1
b2 -4ac > 0
Two distinct real roots
 
2
b2-4ac =0
One real root
3
b2-4ac < 0
No real roots
Question 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) 3x2 - 4√3x + 4 = 0
(iii) 2x2 - 6x + 3 = 0

Answer

(i) Consider the equation
x2 - 3x + 5 = 0
Comparing it with ax2 + bx c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
( - 3)2 - 4 (2) (5) = 9 - 40
= - 31
As b2 - 4ac < 0,
Therefore, no real root is possible for the given equation.

(ii) 3x2 - 4√3x + 4 = 0
Comparing it with ax2 + bx c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 - 4ac
= (-4√3)- 4(3)(4)
= 48 - 48 = 0
As b2 - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be -b/2a and -b/2a.-b/2= -(-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 this equation with ax2 + bx c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)2 - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation:
$x  =\frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
 Substituting the values of a, b, c,
x = (3+√3)/2 or (3-√3)/2
 

Question 2
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x - 2) + 6 = 0

Answer

(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 - 4ac
= (k)2 - 4(2) (3)
k2 - 24
For equal roots,
Discriminant = 0
k2 - 24 = 0
k2 = 24
k = ±√24 = ±2√6

(ii) kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx c = 0, we get
a = kb = - 2k and c = 6
Discriminant = b2 - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
b2 - 4ac = 0
4k2 - 24k = 0
4k (k - 6) = 0
Either 4k = 0
or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x2' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.
 
Question 3
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Answer

Let the breadth of mango grove be x.
Length of mango grove will be 2x.
Area of mango grove = (2x) (x)= 2x2
2x= 800
x= 800/2 = 400
x- 400 =0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = 0, c = 400
Discriminant = b2 - 4ac
= 1600
Here, b2 - 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
x = ±20
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m

Question 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.

Answer

Let the age of one friend be x years.
then the age of the other friend will be (20 - x) years.
4 years ago,
Age of 1st friend = (x - 4) years
Age of 2nd friend = (20 - x - 4) = (16 - x) years
A/q we get that,
(x - 4) (16 - x) = 48
16x - x2 - 64 + 4x = 48
 - x2 + 20x - 112 = 0
x2 - 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 - 4ac
= (-20)2 - 4 × 112
= 400 - 448 = -48
b2 - 4ac < 0
As discriminant is negative, there will be no real solution possible for the equations. Such type of condition doesn't exist.

Question 5
Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Answer

Let the length and breadth of the park be x and y.
Perimeter = 2 (x + y) = 80
x + y = 40
Or, y = 40 – x   ---(A)
Area = x×y
Substituting value of y from equation A
= x (40 - x) = 40x – x2 = 400
x2 -  40x + 400 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 - 4ac
(-40)2 - 4 × 400
= 1600 - 1600 = 0
b2 - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,x= -b/2a
x = (40)/2(1) = 40/2 = 20
Therefore, length of park, x = 20 m
And breadth of park, y = 40 - a = 40 - 20 = 20 m.

Download NCERT solution Quadratic Equation Exercise 4.4 as pdf
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Also Read



Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

  1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
  2. Mathematics for Class 10 by R D Sharma
  3. Pearson IIT Foundation Maths Class 10
  4. Secondary School Mathematics for Class 10
  5. Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.


Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20






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