NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 helps students extend their problem-solving expertise in quadratic equations with two new methods. The exercise goes beyond the traditional factorisation method for solving the quadratic equations with the method of completing the square & the quadratic formula. These methods are mostly used when the quadratic equations don’t have a common term to form two binomials. Here, you can find NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.3 PDF, along with some important key concepts of the exercise in brief.
Let us first recall what solutions of a quadratic equation are before proceeding to the steps of solving quadratic equations. The solutions of any given quadratic equation are the variable values (mainly x), which set the equation to zero. They are determined when the quadratic equations have been expressed in their standard form. That is:
In this expression, a and b are coefficients of the variable x, and c is the constant term. Solutions of quadratic equations are also referred to as the roots or the zeroes of the equations.
Exercise 4.3 will look forward to the new methods of solving quadratic equations, understanding which will equip students with extra tools to find the zeroes of equations. Let’s explore important key concepts of these methods:
Completing the square method requires a proper approach to solving the quadratic equation, which includes the following:
The expression derived from the method of completing the square forms a generalised formula for the quadratic equation . Which can be expressed as:
In the formula, b2 – 4ac can never be negative if the roots of a quadratic equation are real values. This generalised formula is also known as the quadratic formula for any equation. The formula is a useful tool in solving complex quadratic equations, which require long-form solutions of the equations.
To thrive in solving quadratic equations and build up your knowledge, start practising with the help of NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 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
Here a = 2, b = -3 and c = 5
Therefore, discriminant D = b² - 4ac
= (-3)² - 4 × 2 × 5
= 9 - 40 = -31
As D < 0
Hence, no real root.
(ii) 3x² - 4√3x + 4 = 0
Here a = 3, b = -4√3 and c = 4
Therefore, discriminant D = b² - 4ac
= (-4√3)² - 4(3)(4) = 48 - 48 = 0
As D = 0
Hence, two equal real roots.
Now, the roots are
= (-b ± √D) / 2a = (4√3 ± 0) / (2 × 3) = 2/√3
Hence, the roots are 2/√3 and 2/√3.
(iii) 2x² - 6x + 3 = 0
a = 2, b = -6 and c = 3
Discriminant D = b² - 4ac
= (-6)² - 4(2)(3)
= 36 - 24 = 12
=> D > 0
Hence, roots are distinct and real.
The roots are
x = (-b ± √D) / 2a
= (6 ± √12) / (2 × 2) = (6 ± 2√3) / 4 = (3 ± √3) / 2
Therefore, the roots are (3 + √3) / 2 and (3 - √3) / 2.
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
Here a = 2, b = k and c = 3
Therefore, D = b² - 4ac
= k² - 4 × 2 × 3 = k² - 24
Two roots will be equal if D = 0
=> k² - 24 = 0
=> k = ±√24
=> k = ±2√6
(ii) kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Here, a = k, b = -2k and c = 6
Therefore, D = b² - 4ac
= (-2k)² - 4(k)(6) = 4k² - 24k
Two roots will be equal if D = 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'. Hence, k = 6.
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and area is 800 m²? If so, find its length and breadth.
Solution:
Let x be the breadth and 2x be the length of the rectangle.
x × 2x = 800
=> 2x² = 800
=> x² = 400 = (20)²
=> x = 20
Hence, the rectangle is possible and it has breadth = 20 m and length = 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 one friend be x years.
The age of the other friend will be (20 - x) years.
4 years ago, age of first friend = (x - 4) years and age of second friend = (20 - x - 4) = (16 - x) years
Given that,
(x - 4)(16 - x) = 48
16x - 64 - x² + 4x = 48
-x² + 20x - 112 = 0
x² - 20x + 112 = 0
Here, a = 1, b = -20 and c = 112
Therefore, D = b² - 4ac
= (-20)² - 4(1)(112)
= 400 - 448 = -48
As D < 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 m²? If so, find its length and breadth.
Solution:
Perimeter of the rectangular park = 80 m
=> Length + Breadth of the park = 80/2 m = 40 m
Let the breadth be x metres, then length = (40 - x) m.
Here, x < 40.
x × (40 - x) = 400 [Area of the park]
=> -x² + 40x - 400 = 0
=> x² - 40x + 400 = 0
=> (x - 20)² = 0
=> x = 20
Thus, we have length = breadth = 20 m.
Therefore, the park is in the shape of a square having 20 m side.
(Session 2025 - 26)