NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 Quadratic Equations PDF

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 Quadratic Equations: Exercise 4.3 introduces students to solving quadratic equations using the quadratic formula, one of the most important tools in algebra.

This method helps determine roots accurately by using the coefficients and the discriminant. The exercise strengthens conceptual clarity and prepares students for exams and higher-level mathematics. These explanations support learners looking for quadratic equation class 10 exercise 4.3.

Quadratic Equations Class 10 Exercise 4.3 Solutions 

These NCERT solutions  help students understand how to apply the quadratic formula to find the roots of quadratic equations accurately. Exercise 4.3 strengthens problem-solving skills and builds confidence for board exam preparation.

Solve the following Questions

1. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:

(i) 2x2 – 3x + 5 = 0

(ii) 3x 2 -4√3x+4 = 0

(iii) 2x2– 6x + 3 = 0

Answer:

We know that the quadratic equation ax 2 +bx+c = 0 has
(a) Two distinct real roots, if b 2 -4ac > 0,
(b) Two equal real roots, if b 2 -4ac = 0,
(c) No real roots, if b 2 -4ac < 0.

(i) 2x 2 -3x+5 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0,

we get a = 2, b = -3 and c = 5
Then, b 2 -4ac = (-3) 2 -4×2×5 = 9-40 = -31 < 0
Hence, the given quadratic equation has no real roots.

(ii) 3x 2 -4√3x+4 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0

we get a = 3, b = -4√3 and c = 4
Then, b 2 -4ac = (-4√3) 2 -4×3×4 = 48-48 = 0
Therefore, real roots exist for the given equation,

 and they are equal to each other.

And the roots will be

Therefore, the roots are

(iii) 2x 2 -6x+3 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0,
we get a = 2, b = -6 and c = 3
then, b 2 -4ac = (-6) 2 -4×2×3 = 36-24 = 12 > 0
Hence, the given quadratic equation has two distinct real roots.

Applying the quadratic formulato find roots,

2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x 2 +kx+3 = 0

Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0, we get
a = 2, b = k and c = 3
Now the given quadratic equation have two equal roots if
b 2 -4ac = 0

(k)2-4×2×3 = 0
k 2 -24 = 0
k 2 = 24
k = ±√24
k = ± 2√6

Therefore, the required value of k is ± 2√6

(ii)  kx(x-2)+6 = 0

Answer:
The given quadratic equation can be written as
kx 2 -2kx+6 = 0….(1)
Comparing the quadratic equation (1) with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = k, b = -2k and c = 6

Now the given quadratic equation have two equal roots if
b 2 -4ac = 0
(-2k) 2 -4×k×6 = 0
4k 2 -24k = 0
4k(k-6) = 0
k(k-6) = 0
k = 0
Or, k-6 = 0
k = 6

If k = 0, then equation will not have x 2 and x, which is not possible because the given equation is quadratic.
Therefore, the required value of k is 6.

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 the mango grove be x m and the length is 2x m.
Area = length × breadth
= x × 2x
= 2x2 m 2
Then by the given condition,
2x 2 = 800
x 2 = 400
x = ±√400
x = ±20
Since length cannot be negative, then x ≠ -20
Hence, it is possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m 2, and its breadth is 20 m, and length is 20×2 = 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.

Answer:
Let the age of 1st friend is x.
Then the age of 2nd friend is (20-x)
Four years ago their age was (x-4) and (20-x-4)
Then, using the given condition, we have
(x-4)(16-x) = 48
16x – 64 – x 2 +4x = 48
20x – x 2 – 64 – 48 = 0
x 2 – 20x + 112 = 0……(1)

Now, comparing the above quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = 1, b = -20 and c = 112
then, b 2 -4ac = (-20) 2 -4×1×112 = 400-448 = -48 > 0

Therefore, no real root is possible for this equation, and hence, this situation is not possible.


5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m 2? If so, find its length and breadth.

Answer:
Let the length and breadth of the park be “l” and “b”
Area of rectangle = 2(l + b)
2(l + b) = 80
l + b = 40
b = 40 – l
Then, Area = l(40 – l)
Then by the given condition,
l(40 – l) = 400
40l  – l 2 -400 = 0
l 2 – 40l + 400 = 0….(1)

Now, comparing the above quadratic equation with the general form of quadratic equation ax 2 +bx+c = 0 we get
a = 1, b = -40 and c = 400
Then, b 2 -4ac = (-40) 2 -4×1×400 = 1600-1600 = 0
As the quadratic equation has two equal roots, the given situation is possible.

Therefore, length of park, l = 20 m

And breadth of park, b = 40 - l = 40 - 20 = 20 m.

Class 10 Maths Quadratic Equation Exercise 4.3 PDF

Exercise 4.3 focuses on mastering the quadratic formula through structured practice.

The provided PDF contains detailed, easy-to-follow solutions that guide students through each step, improving confidence and accuracy.

This resource is perfect for revision and exam preparation, particularly for learners needing help with class 10 maths quadratic equation exercise 4.3.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 PDF