NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Class 10 Maths Exercise 4.1

1. Checking whether the following are quadratic equations:
(i). (x + 1)² = 2(x – 3)
See Solution(i) (x + 1)² = 2(x – 3)
Simplifying the given equation, we get
(x + 1)² = 2(x – 3)
=> x² + 1² + 2x * 1 = 2x – 6
=> x² + 1 + 2x = 2x – 6
=> x² + 2x – 2x + 1 + 6 = 0
or x² + 0x + 7 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

(ii). x² – 2x = (-2)(3 – x)
See Solution(ii) x² – 2x = (-2)(3 – x)
Simplifying the given equation, we get
x² – 2x = (-2)(3 – x)
=> x² – 2x = -6 + 2x
=> x² – 2x – 2x + 6 = 0
or x² – 4x + 6 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

(iii). (x – 2)(x + 1) = (x – 1)(x + 3)
See Solution(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
Simplifying the given equation, we get
(x – 2)(x + 1) = (x – 1)(x + 3)
=> x² + x – 2x – 2 = x² + 3x – x – 3
=> x² – x – 2 = x² + 2x – 3
=> x² – x – x² – 2x – 2 + 3 = 0
or -3x + 1 = 0
This is not an equation of type ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(iv). (x – 3)(2x + 1) = x(x + 5)
See Solution(iv) (x – 3)(2x + 1) = x(x + 5)
Simplifying the given equation, we get
(x – 3)(2x + 1) = x(x + 5)
=> 2x² + x – 6x – 3 = x² + 5x
=> 2x² – 5x – 3 = x² + 5x
=> 2x² – x² – 5x – 5x – 3 = 0
or x² – 10x – 3 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

(v). (2x – 1)(x – 3) = (x + 5)(x – 1)
See Solution(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Simplifying the given equation, we get
(2x – 1)(x – 3) = (x + 5)(x – 1)
=> 2x² – 6x – x + 3 = x² – x + 5x – 5
=> 2x² – 7x + 3 = x² + 4x – 5
=> 2x² – x² – 7x – 4x + 3 + 5 = 0
or x² – 11x + 8 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

(vi). x² + 3x + 1 = (x – 2)²
See Solution(vi) x² + 3x + 1 = (x – 2)²
Simplifying the given equation, we get
x² + 3x + 1 = (x – 2)²
=> x² + 3x + 1 = x² – 4x + 4
=> x² – x² + 3x + 4x + 1 – 4 = 0
or 7x – 3 = 0
This is not an equation of type ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(vii). (x + 2)³ = 2x(x² – 1)
See Solution(vii) (x + 2)³ = 2x(x² – 1)
Simplifying the given equation, we get
(x + 2)³ = 2x(x² – 1)
=> x³ + 2³ + 3 * x * 2 * (x + 2) = 2x³ – 2x
=> x³ + 8 + 6x(x + 2) = 2x³ – 2x
=> x³ + 8 + 6x² + 12x = 2x³ – 2x
=> x³ – 2x³ + 6x² + 12x + 2x + 8 = 0
or -x³ + 6x² + 14x + 8 = 0
This is not an equation of type ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(viii). x³ – 4x² – x + 1 = (x – 2)³
See Solution(viii) x³ – 4x² – x + 1 = (x – 2)³
Simplifying the given equation, we get
x³ – 4x² – x + 1 = (x – 2)³
=> x³ – 4x² – x + 1 = x³ – 2³ – 3 * x * 2 * (x – 2)
=> x³ – 4x² – x + 1 = x³ – 8 – 6x(x – 2)
=> x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x
=> x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0
or 2x² – 13x + 9 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
See Solution(i) Let, the breadth of plot = x m
Therefore, the length of plot = 2x + 1 m
Hence, area = x(2x + 1) m²
According to questions, x(2x + 1) = 528
=> 2x² + x = 528
=> 2x² + x – 528 = 0
Hence, the length and breadth of plot satisfies the equation 2x² + x – 528 = 0.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.
See Solution(ii) Let the first integer = x
Therefore, the second integer = x + 1
Hence, the product = x(x + 1)
According to questions, x(x + 1) = 306
=> x² + x = 306
=> x² + x – 306 = 0
Hence, the two consecutive integers satisfies the quadratic equation x² + x – 306 = 0.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
See Solution(iii) Let, Rohan’s age = x years
Therefore, mother’s age = x + 26 years
After three years,
Rohan’s age = x + 3 years
And mother’s age = x + 29 years
Hence, the product of ages = (x + 3)(x + 29) years
According to questions, (x + 3)(x + 29) = 360
=> x² + 3x + 29x + 87 = 360
=> x² + 32x – 273 = 0
Hence, the age of Rohan satisfies the quadratic equation x² + x – 306 = 0.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
See Solution(iv) Let the speed of train = x km/h
Total distance = 480 km
Therefore, time taken = 480/x hours
If the speed had been 8 km/h less, then time taken = 480/(x-8) hours
According to questions,
480/(x – 8) – 480/x = 3
=> [480x – 480(x – 8)] / [x(x – 8)] = 3
=> 480x – 480x + 3640 = 3(x – 8)x
=> 3640 = 3x² – 24x
=> 3x² – 24x – 3640 = 0
Hence, the speed of train satisfies the quadratic equation 3x² – 24x – 3640 = 0.

Chapter 4 Board Questions

Class 10 Maths Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:
(i). x² – 3x – 10 = 0
See Solution(i) x² – 3x – 10 = 0
Solving the quadratic equation, we get, x² – 3x – 10 = 0
=> x² – 5x + 2x – 10 = 0
=> x(x – 5) + 2(x – 5) = 0
=> (x – 5)(x + 2) = 0
=> (x – 5) = 0 or (x + 2) = 0
Either x = 5 or x = -2
Hence, the roots of the given quadratic equation are 5 and -2.

(ii). 2x² + x – 6 = 0
See Solution(ii) 2x² + x – 6 = 0
Solving the quadratic equation, we get, 2x² + x – 6 = 0 => 2x² + 4x – 3x – 6 = 0 [Note: The image might have a typo here, it should likely be 2x² + 4x – 3x – 6 or 2x² – 3x + 4x – 6 to factor correctly. Transcribing as shown, but the factorization shown proceeds from 2x² + 4x – 3x – 6]
=> 2x(x + 2) + 3(x – 2) = 0 [Note: Step shown in image doesn’t follow directly from previous line. Correct factorization based on 2x² + 4x – 3x – 6 = 0 would be 2x(x+2)-3(x+2)=0. The factorization shown in the image is 2x(x-2)+3(x-2)=0 which implies the middle terms were -4x and +3x]
=> (x – 2)(2x + 3) = 0
=> (x – 2) = 0 or (2x + 3) = 0
Either x = 2 or x = -3/2
Hence, the roots of the given quadratic equation are 2 and -3/2.

(iii). √2x² + 7x + 5√2 = 0
See Solution(iii) √2x² + 7x + 5√2 = 0
Solving the quadratic equation, we get, √2x² + 7x + 5√2 = 0
=> √2x² + 5x + 2x + 5√2 = 0
=> x(√2x + 5) + √2(√2x + 5) = 0 [Note: √2 * √2x = 2x, which matches the line above]
=> (√2x + 5)(x + √2) = 0
=> (√2x + 5) = 0 or (x + √2) = 0
Either x = -5/√2 or x = -√2
Hence, the roots of the given quadratic equation are -5/√2 and -√2.

(iv). 2x² – x + 1/8 = 0
See Solution(iv) 2x² – x + 1/8 = 0
Solving the quadratic equation, we get, 16x² – 8x + 1 = 0
=> 16x² – 4x – 4x + 1 = 0
=> 4x(4x – 1) – 1(4x – 1) = 0
=> (4x – 1)(4x – 1) = 0
=> (4x – 1) = 0 or (4x – 1) = 0
Either x = 1/4 or x = 1/4
Hence, the roots of the given quadratic equation are 1/4 and 1/4.

(v). 100x² – 20x + 1 = 0
See Solution(v) 100x² – 20x + 1 = 0
Solving the quadratic equation, we get,
100x² – 20x + 1 = 0
=> 100x² – 10x – 10x + 1 = 0
=> 10x(10x – 1) – 1(10x – 1) = 0 [Note: Image shows -10(10x – 1), which is a typo. Corrected to -1(10x – 1)]
=> (10x – 1)(10x – 1) = 0
=> (10x – 1) = 0 or (10x – 1) = 0
Either x = 1/10 or x = 1/10
Hence, the roots of the given quadratic equation are 1/10 and 1/10.

2. Solve the problems given in Example 1. [The problems given in the example 1 are x² – 45x + 324 = 0 and x² – 55x + 750 = 0.]
See Solutionx² – 45x + 324 = 0
Solving the quadratic equation, we get, x² – 45x + 324 = 0
=> x² – 36x – 9x + 324 = 0
=> x(x – 36) – 9(x – 36) = 0
=> (x – 36)(x – 9) = 0
=> (x – 36) = 0 or (x – 9) = 0
Either x = 36 or x = 9
Hence, John and Jivanti have 36 and 9 marbles respectively in the beginning.
x² – 55x + 750 = 0
Solving the quadratic equation, we get
x² – 55x + 750 = 0
=> x² – 30x – 25x + 750 = 0
=> x(x – 30) – 25(x – 30) = 0
=> (x – 30)(x – 25) = 0
=> (x – 30) = 0 or (x – 25) = 0
Either x = 30 or x = 25
Hence, the number of toys on that day was 30 or 25.

3. Find two numbers whose sum is 27 and product is 182.
See SolutionLet the first number = x
Therefore, the second number = 27 – x
According to question,
Product = x(27 – x) = 182
=> 27x – x² = 182
=> x² – 27x + 182 = 0
=> x² – 13x – 14x + 182 = 0
=> x(x – 13) – 14(x – 13) = 0
=> (x – 13)(x – 14) = 0
=> (x – 13) = 0 or (x – 14) = 0
Either x = 13 or x = 14
Hence, the two required numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.
See SolutionLet the first number = x, Therefore, the second number = x + 1
According to questions, x² + (x + 1)² = 365
=> x² + x² + 2x + 1 = 365
=> 2x² + 2x – 364 = 0
=> x² + x – 182 = 0
=> x² – 13x + 14x – 182 = 0
=> x(x – 13) + 14(x – 13) = 0
=> (x – 13)(x + 14) = 0
=> (x – 13) = 0 or (x + 14) = 0
=> Either x = 13 or x = -14
Hence, the two consecutive positive integers are 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
See SolutionLet the base = x cm
Therefore, the height = x – 7 cm
Given that: Hypotenuse = 13 cm
Using Pythagoras theorem, x² + (x – 7)² = 13²
=> x² + x² – 14x + 49 = 169
=> 2x² – 14x – 120 = 0
=> x² – 7x – 60 = 0
=> x² – 12x + 5x – 60 = 0
=> x(x – 12) + 5(x – 12) = 0
=> (x – 12)(x + 5) = 0
=> (x – 12) = 0 or (x + 5) = 0
Either x = 12 or x = -5
But x ≠ -5, as x is side of triangle.
Therefore, x = 12 and the second side = x – 7 = 12 – 7 = 5
Hence, the other two sides are 5 cm and 12 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
See SolutionLet, the number of article = x
Therefore, the cost of one article = 2x + 3
According to question, the total cost = x(2x + 3) = 90
=> 2x² + 3x = 90
=> 2x² + 3x – 90 = 0
=> 2x² + 15x – 12x – 90 = 0
=> x(2x + 15) – 6(2x + 15) = 0
=> (2x + 15)(x – 6) = 0
=> (2x + 15) = 0 or (x – 6) = 0
Either x = -15/2 or x = 6
But x ≠ -15/2, as x is number of articles.
Therefore, x = 6 and the cost of each article = 2x + 3 = 2 * 6 + 3 = ₹15
Hence, the number of articles = 6 and the cost of each article is ₹15.

Class 10 Maths Chapter 4 MCQ

Class 10 Maths 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
See Solution(i) 2x² – 3x + 5 = 0
The given equations is of the form ax² + bx + c = 0, in which a = 2, b = -3, c = 5.
Therefore, D = b² – 4ac = (-3)² – 4 × 2 × 5 = 9 – 40 = -31 < 0
Hence, there is no real roots for this quadratic equation.

(ii). 3x² – 4√3x + 4 = 0
See Solution(ii) 3x² – 4√3x + 4 = 0
The given equations is of the form ax² + bx + c = 0 in which a = 3, b = -4√3, c = 4. Therefore, D = b² – 4ac = (-4√3)² – 4 × 3 × 4 = 48 – 48 = 0
So, the roots of quadratic equation are real and equal.
Hence, x = (4√3 ± √0) / 6 = 4√3 / 6 = 2√3 / 3 [As x = (-b ± √(b² – 4ac)) / 2a]
Hence, the roots of the quadratic equation are 2√3 / 3 and 2√3 / 3.

(iii). 2x² – 6x + 3 = 0
See Solution(iii) 2x² – 6x + 3 = 0
The given equations is of the form ax² + bx + c = 0 in which a = 2, b = -6, c = 3. Therefore, D = b² – 4ac = (-6)² – 4 × 2 × 3 = 36 – 24 = 12 > 0
So, the roots of quadratic equation are real and unequal.
Hence, x = (6 ± √12) / 4 = (6 ± 2√3) / 4 = (3 ± √3) / 2 [As x = (-b ± √(b² – 4ac)) / 2a]
Either x = (3 + √3) / 2 or x = (3 – √3) / 2
Hence, the roots of the quadratic equation 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
See Solution(i) 2x² + kx + 3 = 0
For the quadratic equation 2x² + kx + 3 = 0 we have a = 2, b = k, c = 3.
Therefore,
b² – 4ac = (k)² – 4 × 2 × 3 = k² – 24
For two equal roots, we have k² – 24 = 0
=> k² = 24 => k = ±√24 => k = ±2√6

(ii) kx(x – 2) + 6 = 0
See Solution(ii) kx(x – 2) + 6 = 0
On simplification, we get kx² – 2kx + 6 = 0
For the quadratic equation kx² – 2kx + 6 = 0 we have a = k, b = -2k, c = 6.
Therefore, b² – 4ac = (-2k)² – 4 × k × 6 = 4k² – 24k
For equal roots, we have 4k² – 24k = 0 => 4k(k – 6) = 0
=> 4k = 0 or (k – 6) = 0 => k = 0 or k = 6
But k ≠ 0, as it doesn’t satisfies the equation kx(x – 2) + 6 = 0.
Hence, 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.
See SolutionLet, the breadth of mango grove = x m
Therefore, the length = 2x m
So, the area = x × 2x = 2x²
According to question, 2x² = 800
=> x² = 400
=> x = ±20
Since, the breadth of the mango grove can’t be negative, so the breadth = 20 m
Hence, the length of the mango grove = 2 x 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.
See SolutionLet, the age of the first friend = x years
So, the age of the other friend = 20 – x years
Four years ago:
Age of the first friend = x – 4 years
Age of the second friend = 20 – x – 4 = 16 – x years
According to question,
(x – 4)(16 – x) = 48
=> 16x – x² – 64 + 4x = 48
=> -x² + 20x – 112 = 0
=> x² – 20x + 112 = 0
For the quadratic equation x² – 20x + 112 = 0, we have a = 1, b = -20, c = 112.
Therefore, D = b² – 4ac = (-20)² – 4 * 1 * 112 = 400 – 448 = -48 < 0
So, there is no real roots of this quadratic equation.
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.
See SolutionLet the length of the park = x
Perimeter = 80 m
Therefore, the breadth = 40 – x m [As Perimeter = 2(Length + Breadth)]
According to question,
Area = x(40 – x) = 400
=> 40x – x² = 400
=> x² – 40x + 400 = 0
=> x² – 20x – 20x + 400 = 0
=> x(x – 20) – 20(x – 20) = 0
=> (x – 20)(x – 20) = 0
=> (x – 20)² = 0 [Note: Corrected from image which incorrectly shows (x-12)²=0 or (x-13)=0]
=> (x – 20) = 0
=> x = 20
Hence the length = 20 m and the breadth of the park = 40 – 20 = 20 m.

Class 10 Maths Chapter 4 Important Questions

Class 10 Maths Board Questions

1. The sides of a right triangle are such that the longest side is 4m more than the shortest side and the third side is 2m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle. [CBSE 2025]
See SolutionLet’s denote the sides as follows:
Let a = shortest side
Let c = longest side (hypotenuse)
Let b = third side
Given:
c = a + 4
b = c – 2 = a + 4 – 2 = a + 2
Since this is a right triangle, we can use the Pythagorean theorem:
a² + b² = c²
Substituting the expressions:
a² + (a + 2)² = (a + 4)²
Expanding: a² + a² + 4a + 4 = a² + 8a + 16
2a² + 4a + 4 = a² + 8a + 16
a² – 4a – 12 = 0
Using the quadratic formula:
a = (-(-4) ± √(16 + 48))/2 = (4 ± √64)/2 = (4 ± 8)/2
So a = 6 or a = -2 Since a represents length, a = 6 is our solution.
Therefore:
a = 6 (shortest side)
b = a + 2 = 8 (third side)
c = a + 4 = 10 (longest side/hypotenuse)
Area of the triangle = (1/2) × a × b = (1/2) × 6 × 8 = 24 square meters
Perimeter of the triangle = a + b + c = 6 + 8 + 10 = 24 meters
Difference between area and perimeter = |24 – 24| = 0.

2. Express the equation (x-2)/(x-3) + (x-4)/(x-5) = 10; (x ≠ 3, 5) as a quadratic equation in standard form. Hence, find the roots of the equation so formed. [CBSE 2025]
See SolutionLet’s start by finding the LCM of the denominators:
(x-2)/(x-3) + (x-4)/(x-5) = 10
Multiply all terms by (x-3)(x-5):
(x-2)(x-5) + (x-4)(x-3) = 10(x-3)(x-5)
Expanding:
(x-2)(x-5) = x² – 5x – 2x + 10 = x² – 7x + 10
(x-4)(x-3) = x² – 3x – 4x + 12 = x² – 7x + 12
10(x-3)(x-5) = 10(x² – 5x – 3x + 15) = 10x² – 80x + 150
Now our equation becomes:
x² – 7x + 10 + x² – 7x + 12 = 10x² – 80x + 150
2x² – 14x + 22 = 10x² – 80x + 150
-8x² + 66x – 128 = 0
8x² – 66x + 128 = 0
Dividing by 2:
4x² – 33x + 64 = 0
Using the quadratic formula: x = (33 ± √(33² – 4×4×64))/8
x = (33 ± √(1089 – 1024))/8
x = (33 ± √65)/8
So the roots are x = (33 + √65)/8 and x = (33 – √65)/8.

3. The sum of two numbers is 15. If the sum of their reciprocals is 3/10. Find the two numbers. [CBSE 2023]
See SolutionLet x and y be two numbers.
According to the question,
x + y = 15 …(i)
and
1⁄x + 1⁄y = 3⁄10
⇒ (x + y)⁄(xy) = 3⁄10
⇒ 15⁄(xy) = 3⁄10 [using Eq. (i)]
⇒ xy = 50 ⇒ x = 50⁄y
Substituting x = 50⁄y in Eq. (i), we get:
50⁄y + y = 15
⇒ 50 + y² = 15y
⇒ y² − 15y + 50 = 0
⇒ y² − 10y − 5y + 50 = 0
⇒ y(y − 10) − 5(y − 10) = 0
⇒ (y − 10)(y − 5) = 0
⇒ y = 5 or y = 10
When y = 5, then x = 15 − 5 = 10
When y = 10, then x = 15 − 10 = 5
Thus, the two numbers are 5 and 10.

4. Three consecutive odd numbers are such that the  sum of the squares of the first two numbers is greater than the square of the third by 65. Find  the numbers.  [CBSE 2023 Compartment]
See SolutionLet the number be 2n − 1, 2n + 1 and 2n + 3.
Then, (2n − 1)² + (2n + 1)² = (2n + 3)² + 65
⇒ 4n² + 1 − 4n + 4n² + 1 + 4n = 4n² + 9 + 12n + 65
∴ 4n² − 12n − 72 = 0
⇒ n² − 3n − 18 = 0
⇒ n² − 6n + 3n − 18 = 0
[by splitting the middle term]
⇒ n(n − 6) + 3(n − 6) = 0
⇒ (n + 3)(n − 6) = 0
⇒ n = −3, 6
When n = −3
2n − 1 = 2(−3) − 1 = −7
2n + 1 = 2(−3) + 1 = −5
2n + 3 = 2(−3) + 3 = −3
When n = 6
2n − 1 = 2(6) − 1 = 11
2n + 1 = 2(6) + 1 = 13
2n + 3 = 2(6) + 3 = 15
The three consecutive odd numbers are −7, −5 and −3  or 11, 13 and 15.

5. Solve for x, 2x² – 2√2x + 1 = 0 [CBSE 2022]
See SolutionGiven, quadratic equation is 2x² – 2√2x + 1 = 0.
On comparing with standard form of quadratic equation ax² + bx + c = 0, we get
a = 2, b = -2√2 and c = 1
On substituting the values of a, b and c in the quadratic formula,
x = [-b ± √(b² – 4ac)]/2a
x = [-(-2√2) ± √((-2√2)² – 4(2)(1))]/[2(2)]
x = [2√2 ± √(8 – 8)]/4
x = (2√2 ± 0) / 4
=> x = 2√2 / 4 => x = √2 / 2 => x = 1/√2
Hence, 1/√2 is the required solution of the given equation.

6. Find the values of k for which the quadratic equation x² + 5kx + 16 = 0 has real and equal roots. [CBSE 2022]
See SolutionGiven, quadratic equation is x² + 5kx + 16 = 0
On comparing with standard form of quadratic equation ax² + bx + c = 0, we get
a = 1, b = 5k and c = 16
Since, the roots are real and equal,
∴ D = b² – 4ac = 0
=> (5k)² – 4(1)(16) = 0
=> 25k² – 64 = 0
=> 25k² = 64
=> k² = 64 / 25 => k = ± 8/5
Hence, the value of k is 8/5 and -8/5.

7. Find the value of k for which the quadratic equation kx² – 5x + k = 0 has real and equal roots. [CBSE 2022]
See SolutionGiven, quadratic equation is kx² – 5x + k = 0
On comparing with standard form of quadratic equation ax² + bx + c = 0, we get
a = k, b = -5 and c = k
Since, the roots are real and equal,
∴ D = b² – 4ac = 0
=> (-5)² – 4(k)(k) = 0
=> 25 – 4k² = 0 => 4k² = 25
=> k² = 25 / 4 => k = ± 5/2
Hence, the value k is 5/2 and -5/2.

8. If α and β are roots of the quadratic equation  x² − 7x + 10 = 0, find the quadratic equation  whose roots are α² and β².  [CBSE 2022]
See SolutionGiven, quadratic equation is
x² − 7x + 10 = 0
⇒ x² − 5x − 2x + 10 = 0
[by splitting the middle term]
⇒ x(x − 5) − 2(x − 5) = 0
⇒ (x − 5)(x − 2) = 0
∴ x = 5 and x = 2
So, the roots are α = 5 and β = 2
Now, α² = 5² = 25 and β² = 2² = 4
Then, α² + β² = 25 + 4 = 29
and α²β² = 25 × 4 = 100
So, the quadratic equation whose roots are
α² and β² will be
x² − (α² + β²)x + α²β² = 0
⇒ x² − 29x + 100 = 0

9. A 2-digit number is such that the product of its  digits is 24. If 18 is subtracted from the number, the digits interchange their places. Find the  number.  [CBSE 2022]
See SolutionLet the tens digit be x and the unit digit be y.
Then, xy = 24 …(i)
Now, 10x + y − 18 = 10y + x
⇒ 9x − 9y = 18 ⇒ x − y = 2 …(ii)
From Eqs. (i) and (ii), we get
x − 24/x = 2
⇒ x² − 24 = 2x
⇒ x² − 2x − 24 = 0
⇒ x² − 6x + 4x − 24 = 0 [by splitting the middle term]
⇒ (x − 6)(x + 4) = 0
⇒ x = 6, −4 (−4 not possible)
Thus, x = 6
y = 24/6 = 4
Hence, the required number is 10x + y = 10(6) + 4 = 64

10. The difference of the squares of two numbers is  180. The square of the smaller number is 8 times  the greater number. Find the two numbers.  [CBSE 2022]
See SolutionLet the larger number be x and the smaller number be y.
According to the question,
x² − y² = 180 …(i)
y² = 8x …(ii)
So, x² − 8x = 180 [from Eqs. (i) and (ii)]
⇒ x² − 8x − 180 = 0
⇒ x² + 10x − 18x − 180 = 0
⇒ x(x + 10) − 18(x + 10) = 0 [by splitting the middle term]
⇒ (x − 18)(x + 10) = 0
⇒ x = 18 or x = −10
When x = 18
y² = 8 × 18 = 144
y = ±12
When x = −10
y² = 8 × (−10) = −80 → y = ±√(−80) → not possible
Thus, the required numbers are 8, 12 and 8, −12

11. Solve for x:
1⁄(x + 4) − 1⁄(x + 7) = 11⁄30 , x ≠ −4, 7.  [CBSE 2022]
See SolutionGiven,
1/(x + 4) − 1/(x + 7) = 11/30, x ≠ −4, 7
⇒ [(x + 7) − (x + 4)] / [(x + 4)(x + 7)] = 11/30
⇒ 3 / [(x + 4)(x + 7)] = 11/30
⇒ 3 × 30 = 11 × (x + 4)(x + 7)
⇒ 11(x² + 11x + 28) = 90
⇒ 11x² + 121x + 308 − 90 = 0
⇒ 11x² + 121x + 218 = 0
Comparing with standard form:
ax² + bx + c = 0,
a = 11, b = 121, c = 218
Using quadratic formula:
x = (−b ± √(b² − 4ac)) / 2a
= (−121 ± √(121)² − 4 × 11 × 218) / (2 × 11)
= (−121 ± √14641 − 9592) / 22
= (−121 ± √5049) / 22
= (−121 ± 3√561) / 22
Hence, the solutions are:
x = (−121 + 3√561) / 22
and
x = (−121 − 3√561) / 22

12. A fast train takes 3 h less than a slow train for a  journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find  the speed of each train.  [CBSE 2022]
See SolutionLet the speed of the slow train be x km/h.
⇒ Speed of fast train = (x + 10) km/h
We know that, Time = Distance / Speed
⇒ Time taken by slow train to cover 600 km = 600 / x
Time taken by fast train to cover 600 km = 600 / (x + 10)
According to the question,
Time taken by slow train − Time taken by fast train = 3
⇒ 600 / x − 600 / (x + 10) = 3
⇒ 600[(x + 10 − x) / x(x + 10)] = 3
⇒ 600 × 10 / [x(x + 10)] = 3
⇒ 200 × 10 = x(x + 10)
⇒ x² + 10x − 2000 = 0
⇒ x² + (50 − 40)x − 2000 = 0
(by splitting the middle term)
⇒ x² + 50x − 40x − 2000 = 0
⇒ x(x + 50) − 40(x + 50) = 0
⇒ (x − 40)(x + 50) = 0
⇒ x = 40, −50
Since speed cannot be negative, we reject x = −50
⇒ x = 40
Hence, speed of slow train = 40 km/h
speed of fast train = 50 km/h

13. Solve for y, y² + (3√5/2)y – 5 = 0 [CBSE 2022]
See SolutionGiven, quadratic equation is
y² + (3√5/2)y – 5 = 0
On multiplying both sides of the given equations by 2, we get
2y² + 3√5y – 10 = 0
By splitting middle term,
2y² + 3√5y – 10 = 0
2y² + 4√5y – √5y – 10 = 0
2y(y + 2√5) – √5(y + 2√5) = 0
(y + 2√5)(2y – √5) = 0
if y + 2√5 = 0 => y = -2√5
if 2y – √5 = 0 => 2y = √5 => y = √5 / 2
Hence, √5 / 2 and -2√5 are the required solutions of the given equation.

14. Solve for x, 4√3x² + 5x – 2√3 = 0 [CBSE 2020 Compartment]
See SolutionGiven, quadratic equation is
4√3x² + 5x – 2√3 = 0
Factorise by splitting the middle term
=> 4√3x² + (8 – 3)x – 2√3 = 0
[∵ ac = 4√3 * (-2√3) = -24 and b = 5 = 8 – 3]
=> 4√3x² + 8x – 3x – 2√3 = 0
=> 4x(√3x + 2) – √3(√3x + 2) = 0
=> (4x – √3)(√3x + 2) = 0
=> (4x – √3) = 0 or (√3x + 2) = 0
=> 4x = √3 or √3x = -2
=> x = √3 / 4 or x = -2 / √3
=> x = √3 / 4 or x = (-2 / √3) * (√3 / √3) = -2√3 / 3
Hence, √3 / 4 and -2√3 / 3 are the required solutions of the given equation.

15. A train travels at a certain average speed for a distance of 360 km. It would have taken 48 min less to travel the same distance if its speed was 5 km/h more. Find the original speed of the train. [CBSE 2020 Compartment]
See SolutionLet original speed of train be x km/h. So, time taken to cover 360 km is t₁ = 360/x h and time taken to cover 360 km when the speed is 5 km/h more is t₂ = 360 / (x + 5) h.
According to the question,
t₁ = t₂ + 48/60 [∵ 48 min = 48/60 h]
=> t₁ – t₂ = 48/60
=> 360/x – 360/(x + 5) = 48/60
=> 360[(x + 5 – x) / (x(x + 5))] = 4/5
=> 360 * 5 * 5 = 4(x² + 5x) => 90 * 25 = x² + 5x
=> x² + 5x – 2250 = 0
=> x² + (50 – 45)x – 2250 = 0 [by splitting middle term]
=> x² + 50x – 45x – 2250 = 0
=> x(x + 50) – 45(x + 50) = 0
=> (x – 45)(x + 50) = 0
=> x = 45, -50
But speed can not be negative, so we reject the negative value of x. Therefore, we consider only positive value of x.
∴ x = 45 km/h
Hence, the original speed of train is 45 km/h.

16. Find the nature of roots of the quadratic equation 3x² – 4√3x + 4 = 0. If the roots are real, find them. [CBSE 2020 Compartment]
See SolutionGiven, quadratic equation is
3x² – 4√3x + 4 = 0
On comparing with standard quadratic equation ax² + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Now, discriminant = b² – 4ac = (-4√3)² – 4 × 3 × 4
= 48 – 48 = 0
Hence, the roots are real and equal.
By using quadratic formula,
x = [-b ± √D] / 2a = [-(-4√3) ± √0] / (2 × 3)
= 4√3 / (2 × 3) = 2√3 / 3
Hence, the roots of the given quadratic equation are 2√3 / 3 and 2√3 / 3.

17. Amisha bought a number of books for ₹ 1800. If she had bought 10 more books for the same amount, each book would have cost her ₹ 30 less. How many books did she buy originally?. [CBSE 2020 Compartment]
See SolutionLet the number of books originally bought by Amisha be x.
Then, the cost of each book is ₹ 1800 / x
and if she had bought 10 more books, then cost of each book is ₹ 1800 / (x + 10).
According to the given condition,
1800/x – 1800/(x + 10) = 30
=> 1800[ (x + 10 – x) / (x(x + 10)) ] = 30
=> 60 * 10 = x(x + 10)
=> x² + 10x – 600 = 0
=> x² + (30 – 20)x – 600 = 0 [by splitting the middle term]
=> x² + 30x – 20x – 600 = 0
=> x(x + 30) – 20(x + 30) = 0
=> (x – 20)(x + 30) = 0
=> x = 20, -30
But the number of books can not be negative, so consider only positive value.
∴ x = 20
Hence, originally the number of books bought was 20.

18. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0. [CBSE 2019]
See SolutionGiven, quadratic equation is
2x² – 4x + 3 = 0
On comparing with standard form of quadratic equation ax² + bx + c = 0, we get
a = 2, b = -4 and c = 3
Now, discriminant, D = b² – 4ac
= (-4)² – 4 * 2 * 3 = 16 – 24 = -8 < 0
Hence, the given equation has no real roots.

19. For what value of k, the roots of the equation x² + 4x + k = 0 are real? [CBSE 2019]
See SolutionGiven, quadratic equation is x² + 4x + k = 0
On comparing with standard form of quadratic equation ax² + bx + c = 0, we get
a = 1, b = 4 and c = k
The condition for real roots,
b² – 4ac ≥ 0
∴ (4)² – 4(1)(k) ≥ 0 => 16 – 4k ≥ 0
=> 16 ≥ 4k => k ≤ 4

20. If one root of the quadratic equation 2x² + 2x + k = 0 is -1/3, then find the value of k. [CBSE 2019 Compartment]
See SolutionGiven, -1/3 is a root of quadratic equation
2x² + 2x + k = 0.
Therefore, it satisfies the given equation.
Put x = -1/3 in given equation, we get
2(-1/3)² + 2(-1/3) + k = 0
=> 2/9 – 2/3 + k = 0 => (2 – 6)/9 + k = 0
=> -4/9 + k = 0
=> k = 4/9
Hence, the value of k is 4/9.

Class 10 Maths Chapter 4 Exercise 4.1
Class 10 Maths Chapter 4 Exercise 4.2
Class 10 Maths Chapter 4 Exercise 4.3

Class 10 Maths Chapter 4 Solutions for State Boards
With detailed explanations and illustrations, students can grasp the logic behind solving quadratic equations effectively. By referring to CBSE Class 10 Maths Chapter 4 NCERT Exercises Solutions PDF, students can practice and review concepts repeatedly, enhancing their problem-solving skills and building confidence for exams.
Class 10 Maths Exercise 4.1
Class 10 Maths Exercise 4.2
Class 10 Maths Exercise 4.3
Class 10 Maths Exercise 4.4

The Quadratic Equations NCERT Exercises Solutions Class 10 offer a perfect approach to understanding this chapter’s various exercises. CBSE Class 10 Mathematics Quadratic Equations NCERT Textbook Solutions describe real-life applications, helping students relate theoretical concepts to practical scenarios. NCERT 10th Math solutions simplify complex questions and ensure students gain clarity in solving equations systematically. Whether it’s learning to determine the roots of an equation or analyzing the nature of roots, the Detailed Solutions Chapter 4 Class 10 Maths are invaluable resources.
NCERT Mathematics Chapter 4 Exercise Solutions PDF provides a convenient way for students to study on-the-go, catering to the needs of tech-savvy learners. Quadratic Equations Math Chapter 4 Class 10 explains critical concepts with examples, ensuring students are well-prepared for board exams. Students can also use Grade 10 Maths Chapter 4 Complete Solutions for revision and quick recaps, saving time during exam preparations.

Key Points for Class 10 Maths Chapter 4 Quadratic Equations in Board Exams

Class 10 Maths Chapter 4, Quadratic Equations, focuses on solving equations using factorization, completing the square and the quadratic formula. Key topics include finding roots, analyzing the nature of roots (real and imaginary) and practical applications. Understanding formulas, step-by-step methods and real-world problem-solving is essential for board exam success.

For those looking to excel in board exams, the NCERT Textbook Solutions for Class 10 Mathematics Quadratic Equations are an indispensable study tool. These solutions break down the complexities of Quadratic Equations Class 10 Math Exercise Solutions into digestible parts, enabling students to develop a strong conceptual understanding. NCERT Detailed Solutions for Class 10 Maths Chapter 4 include worked-out examples, detailed explanations and practice questions to ensure thorough preparation. By using the CBSE Class 10 Mathematics Chapter 4 Solutions PDF, students can focus on solving problems efficiently and understanding the underlying principles.
The Quadratic Equations Solutions for Class 10 Math emphasize both theoretical and application-based learning, ensuring students are exam-ready. With regular practice and a clear understanding of Quadratic Equations solutions Class 10, students can significantly improve their problem-solving speed and accuracy, achieving excellent results in mathematics.

The Hindi Medium solutions for the academic session 2025-26 for class 10 Maths chapter 4 are given here. These solutions are helpful for UP Board students studying in Class 10. 10th Maths Chapter 4 solutions are online for study or download in PDF format.
Download Assignments for practice with Solutions
10th Maths Chapter 4 Assignment 1
10th Maths Chapter 4 Assignment 2
10th Maths Chapter 4 Assignment 3
10th Maths Chapter 4 Assignment 4

10th Maths Chapter 4 NCERT Solutions follows the current CBSE syllabus. Students of MP, UP, Gujarat board and CBSE can use it for Board exams. Class 10 Maths NCERT Solutions Offline Apps in Hindi or English for offline use. For any scholarly help, contact us. We will try to help you in the best possible ways.

Textbook Solutions for class 10 Maths Chapter 4 are given below in PDF format or view online. Solutions are in Hindi and English Medium. Uttar Pradesh students also can download UP Board Solutions for Class 10 Maths Chapter 4 here in Hindi Medium.
Learning quadratic equations is very important because they have many uses in various fields. Not only are they applied in other areas of mathematics and physics, but they are also useful in real-life situations. Understanding quadratic equations helps in solving many practical problems. You can download the NCERT books for the 2025-26 session, as well as revision materials and solutions, from the links provided here.

1. Two marks questions
Find the roots of the quadratic equation √2 x² + 7x + 5√2 = 0. [CBSE 2017] 2. Find the value of k for which the equation x²+k(2x + k – 1) + 2 = 0 has real and equal roots. [CBSE 2017]
2. Three marks questions
If the equation (1 + m² ) x² +2mcx + c² – a² = 0 has equal roots then show that c² = a² (1 + m²).
3. Four marks questions
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE 2017]

The word quadratic is derived from the Latin word “Quadratum” which means “A square figure”.
Brahmagupta (an ancient Indian Mathematician)(A.D. 598-665) gave an explicit formula to solve a quadratic equation. Later Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-khwarizni(about A.D. 800) also studied quadratic equations of different types. It is believed that Babylonians were the first to solve quadratic equations. Greek mathematician Euclid developed a geometrical approach for finding lengths, which are nothing but solutions of quadratic equations.

Preparing for Quadratic Equations in Class 10 requires a structured approach to grasp concepts thoroughly and practice systematically. Start with understanding the basics, such as the definition of a quadratic equation, its standard form, and its graphical representation. Dedicate the first day to studying these concepts from NCERT and solving simple examples. Next, move to solving quadratic equations by factorization. Practice problems from Exercise 4.2 to understand step-by-step methods and focus on avoiding calculation errors. This method builds a strong foundation, ensuring you are ready to tackle more advanced techniques in the upcoming days.

Mastering Advanced Techniques for Quadratic Equations
On subsequent days, focus on advanced techniques like the completing the square method and the quadratic formula. These methods are crucial for solving complex problems and analyzing the nature of roots. Practice Examples 6-10 and solve questions from Exercises 4.3 and 4.4, emphasizing accuracy and speed. Allocate the last day for reviewing all methods, solving past board exam questions and identifying weak areas. Consistent revision of formulas and key concepts ensures confidence during exams. This structured day-wise plan, combined with regular practice, guarantees a solid preparation for board exams.