NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4
NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations Exercise 4.4
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Nature of roots of Quadratic equation
S.no
Condition
Nature of roots
1
b^{2} -4ac > 0
Two distinct real roots
2
b^{2}-4ac =0
One real root
3
b^{2}-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) 2x^{2} - 3x + 5 = 0
(ii) 3x^{2} - 4√3x + 4 = 0
(iii) 2x^{2} - 6x + 3 = 0
Answer
(i) Consider the equation x^{2} - 3x + 5 = 0
Comparing it with ax^{2} + bx + c = 0, we get a = 2, b = -3 and 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) 3x^{2} - 4√3x + 4 = 0
Comparing it with ax^{2} + bx + c = 0, we get a = 3, b = -4√3 and c = 4
Discriminant = b^{2} - 4ac
= (-4√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 -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) 2x^{2} - 6x + 3 = 0
Comparing this equation with ax^{2} + bx + c = 0, we get a = 2, b = -6, 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 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) 2x^{2} + kx + 3 = 0
(ii) kx (x - 2) + 6 = 0
Answer
(i) 2x^{2} + kx + 3 = 0
Comparing equation with ax^{2} + bx + c = 0, we get a = 2, b = k and c = 3
Discriminant = b^{2} - 4ac
= (k)^{2} - 4(2) (3)
= k^{2} - 24
For equal roots,
Discriminant = 0 k^{2} - 24 = 0 k^{2} = 24
k = ±√24 = ±2√6
(ii) kx(x - 2) + 6 = 0
or kx^{2} - 2kx + 6 = 0
Comparing this equation with ax^{2} + bx + c = 0, we get a = k, b = - 2k and c = 6
Discriminant = b^{2} - 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 'x^{2}' 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 m^{2}? 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)= 2x^{2}
2x^{2 }= 800 x^{2 }= 800/2 = 400 x^{2 }- 400 =0
Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = 0, c = 400
Discriminant = b^{2} - 4ac
= 1600
Here, b^{2} - 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 - x^{2} - 64 + 4x = 48
- x^{2} + 20x - 112 = 0
x^{2} - 20x + 112 = 0
Comparing this equation with ax^{2} + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b^{2} - 4ac
= (-20)^{2} - 4 × 112
= 400 - 448 = -48
b^{2} - 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 m^{2}? 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 – x^{2} = 400
x^{2} - 40x + 400 = 0
Comparing this equation with ax^{2} + bx + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b^{2} - 4ac
(-40)^{2} - 4 × 400
= 1600 - 1600 = 0
b^{2} - 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.
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