In this post, you will find the NCERT solutions for class 10 maths ex 4.3. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 4.3
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
Solution: (i) 2x2 − 3x + 5 = 0
Comparing the equation 2x2 − 3x + 5 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 2, b = −3 and c = 5
Discriminant (D) = b2 − 4ac = (−3)2 – 4 × 2 × 5
= 9 – 40 = −31
Here, D < 0, which means that the equation has no real roots.
(ii) 3x2 − 4√3x + 4 = 0
Comparing the equation 3x2 − 4√3x + 4 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 3, b = −4√3 and c = 4
Discriminant (D) = b2 − 4ac = (−4√3)2 − 4 × 3 × 4
= 48 – 48 = 0
Here, D = 0, which means that the equation has two equal real roots.
Using the quadratic formula, we have
⇒ 
⇒ 
Since the equation has two equal roots, it means
(iii) 2x2 − 6x + 3 = 0
Comparing the equation 2x2 − 6x + 3 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 2, b = −6 and c = 3
Discriminant (D) = b2 − 4ac = (−6)2 − 4 × 2 × 3
= 36 – 24 = 12
Here, D > 0, which means that the equation has two distinct real roots.
Using the quadratic formula, we have
⇒
⇒ 
⇒ 
⇒ 
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
Solution: (i) 2x2 + kx + 3 = 0
We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.
Comparing the equation 2x2 + kx + 3 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant (D) = b2 − 4ac = k2 – 4 × 2 × 3 = k2 − 24
Since the equation has two equal roots, we have
D = 0
k2 − 24 = 0
k2 = 24
⇒ k = ±√24 = ±2√6
⇒ k = 2√6, −2√6
(ii) kx(x − 2) + 6 = 0
⇒ kx2 – 2kx + 6 = 0
Comparing the equation kx2 – 2kx + 6 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = k, b = −2k and c = 6
Discriminant (D) = b2 − 4ac = (−2k)2 − 4 × k × 6 = 4k2 − 24k
We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.
Since the equation has two equal roots, we have
D = 0
4k2 − 24k = 0
⇒ 4k(k − 6) = 0
⇒ k = 0, 6
According to the definition of a quadratic equation, we have
A quadratic equation is the equation of the form ax2 + bx + c = 0, where a ≠ 0.
Therefore, in equation kx2 − 2kx + 6 = 0, we cannot have k = 0.
Therefore, we discard k = 0.
Hence, the answer is k = 6.
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.
Solution: Let the breadth of the rectangular mango grove be x m.
Then, the length of the rectangular mango grove = 2x m
Area of the rectangular mango grove = length × breadth= x × 2x = 2x2 m2
According to the question, we have
2x2 = 800
⇒ 2x2 − 800 = 0
⇒ x2 − 400 = 0
Comparing the equation x2 − 400 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 1, b = 0 and c = −400
Discriminant (D) = b2 − 4ac = (0)2 − 4 × 1 × (−400) = 1600
Since D > 0, it means that the equation has two distinct real roots.
Therefore, it is possible to design a rectangular mango grove.
Using the quadratic formula, we have
⇒
Putting the values of a, b and c, we get
⇒ x = 20, −20
We discard the negative value of x because the breadth of a rectangle cannot be negative.
Therefore, x = breadth of the rectangle = 20 m
And, length of the rectangle = 2x = 2 × 20 = 40 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.
Solution: Let the age of first friend be x years.
Then, the age of second friend = (20 − x) years.
Four years ago, the age of first friend = (x − 4) years
Four years ago, the age of second friend = (20 − x) − 4 = (16 − x) years
According to the question, we have
(x − 4)(16 − x) = 48
⇒ 16x − x2 – 64 + 4x = 48
⇒ 20x − x2 − 112 = 0
⇒ x2 − 20x + 112 = 0
Comparing the equation x2 − 20x + 112 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 1, b = −20 and c = 112
Discriminant (D) = b2 − 4ac = (−20)2 – 4 × 1 × 112 = 400 – 448 = −48
Here D < 0, which means we have no real roots for this equation.
Therefore, the give situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution: Let the length of the park be x m.
It is given that the area of the rectangular park = 400 m2
Length × breadth = 400 m2
Therefore, breadth of park = 400/x m
Perimeter of the rectangular park = 2(length + breath) = 2(x + 400/x) m
It is given that the perimeter of the rectangular park = 80 m
According to the question, we have
⇒ 
⇒ 2x2 + 800 = 80x
⇒ 2x2 − 80x + 800 = 0
⇒ x2 − 40x + 400 = 0
Comparing the equation x2 − 40x + 400 = 0 with the general form of a quadratic equation ax2 + bx + c = 0, we get
a = 1, b = −40 and c = 400
Discriminant (D) = b2 − 4ac = (−40)2 − 4 × 1 × 400 = 1600 – 1600 = 0
Here D = 0.
Therefore, the two roots of the equation are real and equal, which means that it is possible to design a rectangular park of the perimeter 80 m and area 400 m2.
Using the quadratic formula, we have
⇒
Putting the values of a, b and c, we get
⇒
Here, both the roots are equal to 20.
Therefore, the length of the rectangular park = 20 m
And, the breadth of the rectangular park = 400/x = 400/20 = 20 m