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By Rohit RP
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Updated on 5 Aug 2026, 16:45 IST
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations provide step-by-step answers for Exercises 4.1, 4.2 and 4.3 from the current textbook. This chapter explains how to identify, form and solve quadratic equations using factorisation, the quadratic formula and the discriminant.
You will also learn how to translate word problems into equations, choose the correct solving method, check your answers and avoid common mistakes.
The exercise structure and solutions on this page follow the official NCERT Class 10 Mathematics Chapter 4 reprint for 2026–27. The current chapter contains Exercises 4.1, 4.2 and 4.3.
The current NCERT Chapter 4 contains three exercises: Exercise 4.1, Exercise 4.2 and Exercise 4.3.
The official chapter is marked “Reprint 2026–27.” Its main sections are Introduction, Quadratic Equations, Solution by Factorisation, Nature of Roots and Summary.
There are three exercises in the current chapter:
| Exercise | Main topic |
| Exercise 4.1 | Identifying and forming quadratic equations |
| Exercise 4.2 | Solving quadratic equations by factorisation |
| Exercise 4.3 | Discriminant, nature of roots and applications |
No. In the current textbook, 4.4 is the section titled “Nature of Roots”; it is not Exercise 4.4.
Some older books and websites still show an Exercise 4.4. Students should check the edition before using such material.
Completing the square is not listed as a prescribed Class 10 solving method in the current CBSE syllabus. The listed methods are factorisation and the quadratic formula for equations with real roots.
NCERT mentions completing the square in a short historical note about the development of the quadratic formula, but the current chapter does not contain a separate completing-the-square exercise.
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Also Check: CBSE Mathematics Class X Syllabus
| Feature | Older books or webpages | Current NCERT 2026–27 |
| Number of exercises | May show Exercises 4.1 to 4.4 | Exercises 4.1 to 4.3 |
| Section 4.4 | Sometimes treated as Exercise 4.4 | Nature of Roots |
| Completing the square | May appear as a separate method | Not a separate current exercise |
| Prescribed solving methods | May include older material | Factorisation and quadratic formula |
| Nature of roots | May be placed later | Covered in Exercise 4.3 |
A quadratic equation is a polynomial equation of degree two that can be written in the standard form ax² + bx + c = 0, where a is not zero.
The chapter teaches four main skills:
An editorial classification of the NCERT textbook exercises gives the following approximate distribution:
| Skill group | Number of tasks | Approximate share |
| Identifying or forming an equation | 12 | 38.7% |
| Solving by factorisation | 11 | 35.5% |
| Discriminant, roots and feasibility | 8 | 25.8% |
This count treats each individual subpart or situation as one task.
The three essential formulas are the standard form, the quadratic formula and the discriminant formula.
ax² + bx + c = 0
Here:
If a = 0, the x² term disappears.
The equation then becomes:
bx + c = 0
This is a linear equation, not a quadratic equation.
For the equation:
ax² + bx + c = 0
The roots are:
x = [−b ± √(b² − 4ac)] / (2a)
The entire expression −b ± √(b² − 4ac) is divided by 2a.
D = b² − 4ac
The value of D tells us the nature of the roots.
| Value of D | Nature of roots |
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots |
| D < 0 | No real roots |
When D = 0, both roots have the same value:
x = −b / (2a)
| Concept | Formula or condition |
| Standard form | ax² + bx + c = 0 |
| Quadratic formula | x = [−b ± √(b² − 4ac)] / (2a) |
| Discriminant | D = b² − 4ac |
| Distinct real roots | D > 0 |
| Equal real roots | D = 0 |
| No real roots | D < 0 |
An equation is quadratic when its simplified standard form has degree two and a non-zero coefficient of x².
If the final equation has the form ax² + bx + c = 0 with a ≠ 0, it is quadratic.
Check:
(x − 2)² + 1 = 2x − 3
Expand the square:
x² − 4x + 4 + 1 = 2x − 3
x² − 4x + 5 = 2x − 3
Move all terms to the left:
x² − 6x + 8 = 0
The highest power is 2, so it is a quadratic equation.
Check:
x(x + 1) + 8 = (x + 2)(x − 2)
Expand both sides:
x² + x + 8 = x² − 4
Cancel x²:
x + 12 = 0
The simplified equation has degree one, so it is not quadratic.
Check:
(x + 2)³ = x³ − 4
Expand:
x³ + 6x² + 12x + 8 = x³ − 4
Cancel x³:
6x² + 12x + 12 = 0
Divide by 6:
x² + 2x + 2 = 0
The simplified equation is quadratic.
| Quadratic polynomial | Quadratic equation |
| An expression such as ax² + bx + c | An equation such as ax² + bx + c = 0 |
| Does not contain an equality sign | Contains an equality sign |
| Has zeroes | Has roots or solutions |
| May be evaluated for different values of x | Is solved to find values of x |
The zeroes of ax² + bx + c are the same as the roots of ax² + bx + c = 0.
A word problem can be converted into a quadratic equation by defining one unknown, expressing the other quantities through it and applying the relationship stated in the problem.
| Wording | Algebraic expression |
| One more than x | x + 1 |
| Five less than x | x − 5 |
| Twice x | 2x |
| Three more than twice x | 2x + 3 |
| Two consecutive integers | x and x + 1 |
| Consecutive even integers | x and x + 2 |
| Three years from now | x + 3 |
| Four years ago | x − 4 |
| Product of two quantities | Multiply the expressions |
| Speed reduced by 8 km/h | x − 8 |
| Time taken | Distance / Speed |
Suppose the breadth of a rectangle is x metres.
Its length is one metre more than twice its breadth:
Length = 2x + 1
Area = Length × Breadth
If the area is 528 m²:
x(2x + 1) = 528
2x² + x = 528
2x² + x − 528 = 0
This is the required quadratic equation.
Let the first positive integer be x.
The next integer is x + 1.
If their product is 306:
x(x + 1) = 306
x² + x − 306 = 0
Let a student’s present age be x years.
Suppose the parent is 26 years older:
Parent’s present age = x + 26
After three years:
Student’s age = x + 3
Parent’s age = x + 29
If the product of their ages then is 360:
(x + 3)(x + 29) = 360
x² + 32x + 87 = 360
x² + 32x − 273 = 0
Let the original speed be x km/h.
Original time:
480 / x hours
Reduced speed:
x − 8 km/h
New time:
480 / (x − 8) hours
If the journey takes three extra hours:
480 / (x − 8) = 480 / x + 3
On simplifying:
x² − 8x − 1280 = 0
Students sometimes write:
Time = Speed / Distance
This is incorrect.
The correct formula is:
Time = Distance / Speed
After solving a word problem, ask:
A negative number may satisfy the equation but still be invalid as a length, age, speed or number of objects.
Use factorisation when the factors are easy to identify, the quadratic formula when they are not, and the discriminant when only the nature of the roots is required.
| Question type | Recommended method |
| Equation factorises easily | Factorisation |
| Factors are difficult to find | Quadratic formula |
| Question asks only for nature of roots | Discriminant |
| Question asks for k and equal roots | Set D = 0 |
| Question asks for distinct real roots | Use D > 0 |
| Question asks whether a situation is possible | Check D and practical restrictions |
| Word problem | Form equation, solve and validate roots |
Factorisation is usually efficient when:
Use the quadratic formula when:
Do not calculate the complete roots if the question asks only whether they are:
Calculate D = b² − 4ac and state the result.
Factorisation solves a quadratic equation by rewriting its quadratic polynomial as the product of two linear factors.
For:
ax² + bx + c
Find two numbers p and q such that:
p × q = ac
p + q = b
Then replace bx with px + qx and factor by grouping.
Here:
a = 1, b = −3 and c = −10
We need two numbers whose product is −10 and sum is −3.
The numbers are −5 and 2.
x² − 5x + 2x − 10 = 0
x(x − 5) + 2(x − 5) = 0
(x − 5)(x + 2) = 0
Therefore:
x − 5 = 0 or x + 2 = 0
x = 5 or x = −2
If the product of two numbers is zero, at least one of them must be zero.
Therefore, if:
A × B = 0
Then:
A = 0 or B = 0
This is called the zero-product property.
Solve:
6x² − x − 2 = 0
Here:
ac = 6 × (−2) = −12
We need numbers with product −12 and sum −1.
The numbers are 3 and −4.
6x² + 3x − 4x − 2 = 0
3x(2x + 1) − 2(2x + 1) = 0
(3x − 2)(2x + 1) = 0
Therefore:
3x − 2 = 0 or 2x + 1 = 0
x = 2/3 or x = −1/2
Solve:
100x² − 20x + 1 = 0
Recognise the perfect square:
100x² − 20x + 1 = (10x − 1)²
Therefore:
(10x − 1)² = 0
10x − 1 = 0
x = 1/10
The two roots are equal:
x = 1/10 and x = 1/10
| Product ac | Sum b | Signs of the two numbers |
| Positive | Positive | Both positive |
| Positive | Negative | Both negative |
| Negative | Positive | Positive number has greater magnitude |
| Negative | Negative | Negative number has greater magnitude |
The quadratic formula can solve any quadratic equation within the real-number scope when its discriminant is zero or positive.
For:
ax² + bx + c = 0
Solve:
2x² − 6x + 3 = 0
Identify the coefficients:
a = 2
b = −6
c = 3
Calculate the discriminant:
D = b² − 4ac
D = (−6)² − 4(2)(3)
D = 36 − 24
D = 12
Apply the formula:
x = [−b ± √D] / (2a)
x = [6 ± √12] / 4
Since:
√12 = 2√3
x = [6 ± 2√3] / 4
x = (3 ± √3) / 2
Therefore, the roots are:
x = (3 + √3) / 2
and
x = (3 − √3) / 2
Suppose b = −6.
Then:
b² = (−6)² = 36
Writing −6² without brackets may be read as:
−(6²) = −36
Always use brackets when substituting negative coefficients.
The discriminant D = b² − 4ac determines how many real roots a quadratic equation has.
If:
D > 0
The equation has two distinct real roots.
Example:
x² − 5x + 6 = 0
D = (−5)² − 4(1)(6)
D = 25 − 24
D = 1
Since D > 0, the equation has two distinct real roots.
If:
D = 0
The equation has two equal real roots.
Example:
x² − 6x + 9 = 0
D = (−6)² − 4(1)(9)
D = 36 − 36
D = 0
The equal root is:
x = −b / (2a)
x = 6/2
x = 3
The equation has one distinct numerical value, but that value occurs twice.
For example:
(x − 3)² = 0
The roots may be written as:
3 and 3
If:
D < 0
The equation has no real roots.
Example:
2x² − 3x + 5 = 0
D = (−3)² − 4(2)(5)
D = 9 − 40
D = −31
Since D < 0, the equation has no real roots.
When the coefficients are rational:
| Discriminant | Graph of the quadratic polynomial |
| D > 0 | The graph crosses the x-axis at two points |
| D = 0 | The graph touches the x-axis at one point |
| D < 0 | The graph does not meet the x-axis |
This graphical interpretation is useful enrichment. Detailed graphing of parabolas is not required to solve the current NCERT exercises.
Questions involving k are solved by translating the required nature of roots into a condition on the discriminant.
For equal roots:
D = 0
For distinct real roots:
D > 0
For no real roots:
D < 0
Consider:
2x² + kx + 3 = 0
Here:
a = 2
b = k
c = 3
For equal roots:
D = 0
k² − 4(2)(3) = 0
k² − 24 = 0
k² = 24
k = ±√24
k = ±2√6
Consider:
kx² − 2kx + 6 = 0
The coefficient of x² is k.
For the equation to remain quadratic:
k ≠ 0
For equal roots:
D = (−2k)² − 4(k)(6)
D = 4k² − 24k
D = 4k(k − 6)
Set D = 0:
4k(k − 6) = 0
k = 0 or k = 6
But k = 0 makes the equation non-quadratic. Therefore:
k = 6
Exercise 4.1 tests whether students can simplify equations, identify quadratic equations and form equations from everyday situations. The equations and exercise structure below follow the official current chapter.
Determine whether:
(x + 1)² = 2(x − 3)
Expand:
x² + 2x + 1 = 2x − 6
Move all terms to the left:
x² + 7 = 0
This is of the form ax² + bx + c = 0.
Answer: It is a quadratic equation.
Determine whether:
x² − 2x = −2(3 − x)
Expand the right side:
x² − 2x = −6 + 2x
Move all terms to the left:
x² − 4x + 6 = 0
Answer: It is a quadratic equation.
Determine whether:
(x − 2)(x + 1) = (x − 1)(x + 3)
Expand both sides:
x² − x − 2 = x² + 2x − 3
Cancel x²:
−x − 2 = 2x − 3
−3x + 1 = 0
The simplified equation is linear.
Answer: It is not a quadratic equation.
Determine whether:
(x − 3)(2x + 1) = x(x + 5)
Expand:
2x² − 5x − 3 = x² + 5x
Move all terms to the left:
x² − 10x − 3 = 0
Answer: It is a quadratic equation.
Determine whether:
(2x − 1)(x − 3) = (x + 5)(x − 1)
Expand:
2x² − 7x + 3 = x² + 4x − 5
Move all terms to the left:
x² − 11x + 8 = 0
Answer: It is a quadratic equation.
Determine whether:
x² + 3x + 1 = (x − 2)²
Expand the right side:
x² + 3x + 1 = x² − 4x + 4
Cancel x²:
7x − 3 = 0
The simplified equation is linear.
Answer: It is not a quadratic equation.
Determine whether:
(x + 2)³ = 2x(x² − 1)
Expand:
x³ + 6x² + 12x + 8 = 2x³ − 2x
Move all terms to one side:
x³ − 6x² − 14x − 8 = 0
The highest power is 3.
Answer: It is a cubic equation, not a quadratic equation.
Determine whether:
x³ − 4x² − x + 1 = (x − 2)³
Expand the right side:
x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8
Cancel x³ and simplify:
2x² − 13x + 9 = 0
Answer: It is a quadratic equation.
A rectangular plot has area 528 m². Its length is one metre more than twice its breadth.
Let the breadth be x metres.
Length = 2x + 1 metres
Area = Breadth × Length
x(2x + 1) = 528
2x² + x − 528 = 0
Answer: The required quadratic equation is 2x² + x − 528 = 0.
The product of two consecutive positive integers is 306.
Let the smaller integer be x.
The next integer is x + 1.
x(x + 1) = 306
x² + x − 306 = 0
Answer: The required quadratic equation is x² + x − 306 = 0.
A mother is 26 years older than her son. Three years from now, the product of their ages will be 360.
Let the son’s present age be x years.
Mother’s present age = x + 26
Their ages after three years are:
x + 3 and x + 29
Therefore:
(x + 3)(x + 29) = 360
x² + 32x + 87 = 360
x² + 32x − 273 = 0
Answer: The required quadratic equation is x² + 32x − 273 = 0.
A train covers 480 km. At a speed 8 km/h lower, it would take three additional hours.
Let the original speed be x km/h.
Original time = 480/x hours
Reduced speed = x − 8 km/h
New time = 480/(x − 8) hours
According to the condition:
480/(x − 8) = 480/x + 3
Multiply by x(x − 8):
480x = 480(x − 8) + 3x(x − 8)
Simplify:
480x = 480x − 3840 + 3x² − 24x
3x² − 24x − 3840 = 0
Divide by 3:
x² − 8x − 1280 = 0
Answer: The required quadratic equation is x² − 8x − 1280 = 0.
Exercise 4.2 focuses on finding roots by factorisation and applying quadratic equations to numbers, geometry and production problems.
Solve:
x² − 3x − 10 = 0
Split the middle term:
x² − 5x + 2x − 10 = 0
x(x − 5) + 2(x − 5) = 0
(x − 5)(x + 2) = 0
Therefore:
x = 5 or x = −2
Answer: The roots are 5 and −2.
Solve:
2x² + x − 6 = 0
Split x as 4x − 3x:
2x² + 4x − 3x − 6 = 0
2x(x + 2) − 3(x + 2) = 0
(2x − 3)(x + 2) = 0
Therefore:
2x − 3 = 0 or x + 2 = 0
x = 3/2 or x = −2
Answer: The roots are 3/2 and −2.
Solve:
√2x² + 7x + 5√2 = 0
Split 7x as 5x + 2x:
√2x² + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2(√2x + 5) = 0
(√2x + 5)(x + √2) = 0
Therefore:
√2x + 5 = 0
x = −5/√2 = −5√2/2
Or:
x + √2 = 0
x = −√2
Answer: The roots are −5√2/2 and −√2.
Solve:
2x² − x + 1/8 = 0
Multiply the equation by 8:
16x² − 8x + 1 = 0
Recognise a perfect square:
(4x − 1)² = 0
4x − 1 = 0
x = 1/4
Answer: The equal roots are 1/4 and 1/4.
Solve:
100x² − 20x + 1 = 0
Recognise a perfect square:
(10x − 1)² = 0
10x − 1 = 0
x = 1/10
Answer: The equal roots are 1/10 and 1/10.
Two people initially have a total of 45 marbles. After each loses five marbles, the product of the remaining numbers is 124.
Let the first person initially have x marbles.
The other person has:
45 − x
After losing five marbles, they have:
x − 5 and 40 − x
According to the condition:
(x − 5)(40 − x) = 124
Expand:
−x² + 45x − 200 = 124
x² − 45x + 324 = 0
Factorise:
(x − 9)(x − 36) = 0
x = 9 or x = 36
Answer: They initially had 9 marbles and 36 marbles.
A workshop produces x toys in a day. The cost per toy is ₹(55 − x), and the total daily production cost is ₹750.
x(55 − x) = 750
55x − x² = 750
x² − 55x + 750 = 0
Factorise:
(x − 25)(x − 30) = 0
x = 25 or x = 30
If 25 toys are produced:
Cost per toy = ₹30
If 30 toys are produced:
Cost per toy = ₹25
Answer: The possible combinations are 25 toys at ₹30 each or 30 toys at ₹25 each.
Find two numbers whose sum is 27 and product is 182.
Let one number be x.
The other number is 27 − x.
x(27 − x) = 182
27x − x² = 182
x² − 27x + 182 = 0
Factorise:
(x − 13)(x − 14) = 0
x = 13 or x = 14
Answer: The numbers are 13 and 14.
Find two consecutive positive integers whose squares have a sum of 365.
Let the smaller integer be x.
The next integer is x + 1.
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x − 364 = 0
Divide by 2:
x² + x − 182 = 0
Factorise:
(x − 13)(x + 14) = 0
x = 13 or x = −14
The integer must be positive, so x = 13.
Answer: The integers are 13 and 14.
The altitude of a right triangle is 7 cm shorter than its base. Its hypotenuse is 13 cm.
Let the altitude be x cm.
Base = x + 7 cm
Using Pythagoras’ theorem:
x² + (x + 7)² = 13²
x² + x² + 14x + 49 = 169
2x² + 14x − 120 = 0
Divide by 2:
x² + 7x − 60 = 0
Factorise:
(x + 12)(x − 5) = 0
x = −12 or x = 5
A length cannot be negative, so x = 5.
Base = 5 + 7 = 12 cm
Answer: The altitude is 5 cm and the base is 12 cm.
Pythagoras Theorem Questions
A pottery unit produces x articles in a day. The cost per article is ₹(2x + 3), and the total production cost is ₹90.
x(2x + 3) = 90
2x² + 3x − 90 = 0
Split the middle term:
2x² + 15x − 12x − 90 = 0
x(2x + 15) − 6(2x + 15) = 0
(2x + 15)(x − 6) = 0
x = −15/2 or x = 6
The number of articles must be positive.
Therefore:
x = 6
Cost per article:
2(6) + 3 = ₹15
Answer: Six articles were produced, and the cost of each article was ₹15.
Exercise 4.3 uses the discriminant to determine the nature of roots, find parameter values and test whether real-life situations are mathematically possible.
Find the nature of the roots of:
2x² − 3x + 5 = 0
Here:
a = 2
b = −3
c = 5
D = b² − 4ac
D = (−3)² − 4(2)(5)
D = 9 − 40
D = −31
Since D < 0, there are no real roots.
Answer: The equation has no real roots.
Find the nature of the roots of:
3x² − 4√3x + 4 = 0
Here:
a = 3
b = −4√3
c = 4
D = (−4√3)² − 4(3)(4)
D = 48 − 48
D = 0
Therefore, the roots are equal.
Equal root:
x = −b / (2a)
x = 4√3/6
x = 2√3/3
Answer: The equation has two equal real roots, each equal to 2√3/3.
Find the nature and values of the roots of:
2x² − 6x + 3 = 0
Here:
a = 2
b = −6
c = 3
D = (−6)² − 4(2)(3)
D = 36 − 24
D = 12
Since D > 0, the roots are real and distinct.
Using the formula:
x = [6 ± √12] / 4
x = [6 ± 2√3] / 4
x = (3 ± √3) / 2
Answer: The distinct real roots are (3 + √3)/2 and (3 − √3)/2.
Find k so that:
2x² + kx + 3 = 0
has equal roots.
For equal roots:
D = 0
k² − 4(2)(3) = 0
k² − 24 = 0
k² = 24
k = ±2√6
Answer: k = 2√6 or k = −2√6.
Find k so that:
kx(x − 2) + 6 = 0
has equal roots.
Expand:
kx² − 2kx + 6 = 0
Here:
a = k
b = −2k
c = 6
For equal roots:
D = 0
(−2k)² − 4(k)(6) = 0
4k² − 24k = 0
4k(k − 6) = 0
k = 0 or k = 6
For the equation to be quadratic:
k ≠ 0
Therefore:
k = 6
Answer: k = 6.
Determine whether a rectangular mango grove can have area 800 m² when its length is twice its breadth.
Let the breadth be x metres.
Length = 2x metres
Area:
x × 2x = 800
2x² = 800
x² = 400
x = ±20
A breadth cannot be negative, so:
Breadth = 20 m
Length = 40 m
Answer: Yes, the design is possible. Its dimensions are 20 m by 40 m.
Two friends have a total present age of 20 years. Four years ago, the product of their ages was 48.
Let one friend’s present age be x.
The other friend’s present age is:
20 − x
Four years ago, their ages were:
x − 4 and 16 − x
According to the condition:
(x − 4)(16 − x) = 48
Expand:
−x² + 20x − 64 = 48
x² − 20x + 112 = 0
Calculate the discriminant:
D = (−20)² − 4(1)(112)
D = 400 − 448
D = −48
Since D < 0, the equation has no real roots.
Answer: The stated situation is not possible.
Determine whether a rectangular park can have a perimeter of 80 m and area of 400 m².
Let the breadth be x metres.
Since:
2(Length + Breadth) = 80
Length + Breadth = 40
Length = 40 − x
Using the area:
x(40 − x) = 400
40x − x² = 400
x² − 40x + 400 = 0
Calculate the discriminant:
D = (−40)² − 4(1)(400)
D = 1600 − 1600
D = 0
Therefore, the equation has equal roots.
x = −b / (2a)
x = 40/2
x = 20
Length = 40 − 20 = 20
Answer: Yes, the park is possible. It must be a square measuring 20 m by 20 m.
A calculated root must satisfy both the quadratic equation and the practical conditions of the original problem.
A negative root is normally rejected when x represents:
Suppose a breadth equation gives:
x = 12 or x = −12.5
Both numbers may satisfy the algebraic equation.
However, −12.5 metres cannot be a physical breadth.
Therefore, only x = 12 is accepted.
Zero may also be unsuitable when x represents:
Most errors occur while simplifying the equation, handling signs, splitting the middle term or applying the quadratic formula.
| Common mistake | Why it is wrong | How to avoid it |
| Deciding the degree before simplifying | Terms may cancel | Expand and simplify first |
| Forgetting that b can be negative | The formula uses the signed coefficient | Write a, b and c separately |
| Finding factors of c instead of ac | Middle-term splitting requires product ac | Write “product = ac, sum = b” |
| Using the wrong signs in factorisation | The product or sum becomes incorrect | Check by multiplying the factors |
| Writing b² − 4a + c | The formula is b² − 4ac | Keep 4ac together |
| Dividing only √D by 2a | The whole numerator is divided | Use brackets around the numerator |
| Forgetting the ± sign | One root may be lost | Calculate both cases |
| Keeping an impossible negative answer | It may violate the context | Check the original situation |
| Accepting k = 0 | It may remove the x² term | Confirm a ≠ 0 |
| Converting surds to decimals too early | Accuracy may be lost | Keep exact values until the end |
Write a, b and c before using the discriminant or quadratic formula.
For example:
For 2x² − 7x − 4 = 0:
a = 2
b = −7
c = −4
This simple step prevents many sign errors.
Competency-based questions test whether students can model, interpret and check a situation instead of only repeating a memorised method.
CBSE publishes competency-focused Grade 10 practice material containing a dedicated Quadratic Equations section.
A rectangular board is 5 cm longer than its breadth and has an area of 84 cm².
Which equation represents the situation?
Answer: A
Explanation:
Breadth = x
Length = x + 5
x(x + 5) = 84
x² + 5x − 84 = 0
A student solves:
x² − 7x + 12 = 0
and writes:
(x − 6)(x − 2) = 0
The factorisation is incorrect because:
−6 + (−2) = −8, not −7
The correct factors are:
(x − 3)(x − 4) = 0
Therefore:
x = 3 or x = 4
A rectangle has perimeter 20 cm and area 30 cm².
Let one side be x.
The other side is:
10 − x
Area equation:
x(10 − x) = 30
x² − 10x + 30 = 0
D = (−10)² − 4(1)(30)
D = 100 − 120
D = −20
Since D < 0, such a rectangle is not possible using real dimensions.
A word problem produces the roots:
x = 8 and x = −11
If x is the number of notebooks purchased, the correct answer is:
x = 8
The value −11 is mathematically obtained but cannot represent a number of notebooks.
These questions revise identification, factorisation, the quadratic formula, the discriminant and word-problem modelling.
Check whether the following is quadratic:
(x + 3)² = x² + 5
Answer:
x² + 6x + 9 = x² + 5
6x + 4 = 0
It is linear, not quadratic.
Solve by factorisation:
3x² − 11x + 6 = 0
Answer:
3x² − 9x − 2x + 6 = 0
3x(x − 3) − 2(x − 3) = 0
(3x − 2)(x − 3) = 0
x = 2/3 or x = 3
Solve using the quadratic formula:
x² − 4x − 1 = 0
Answer:
a = 1, b = −4, c = −1
D = 16 + 4 = 20
x = [4 ± √20]/2
x = [4 ± 2√5]/2
x = 2 ± √5
Find the nature of roots:
5x² + 2x + 3 = 0
Answer:
D = 2² − 4(5)(3)
D = 4 − 60
D = −56
The equation has no real roots.
Find k if:
x² + kx + 16 = 0
has equal roots.
Answer:
k² − 64 = 0
k = ±8
Two consecutive positive integers have product 156.
Answer:
Let the integers be x and x + 1.
x(x + 1) = 156
x² + x − 156 = 0
(x − 12)(x + 13) = 0
x = 12 or x = −13
The positive integers are 12 and 13.
Use these internal links to help students revise connected chapters and question types.
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The standard form is ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0.
If a = 0, the x² term disappears and the equation becomes linear rather than quadratic.
The current NCERT 2026–27 chapter contains Exercises 4.1, 4.2 and 4.3. Section 4.4 is Nature of Roots, not Exercise 4.4.
No. Exercise 4.4 may appear in older editions or older online resources, but it is not part of the current exercise structure.
It is not listed as one of the prescribed current solving methods. Students are expected to solve real-root equations using factorisation and the quadratic formula.
The quadratic formula is:
x = [−b ± √(b² − 4ac)] / (2a)
The discriminant is D = b² − 4ac. Its sign tells us whether the equation has two distinct real roots, equal real roots or no real roots.
Yes. Calculate D = b² − 4ac and examine whether D is positive, zero or negative.
It means the equation has two equal real roots. There is one distinct root value repeated twice.
Write the equation in standard form, split the middle term, group the terms, form two linear factors and set each factor equal to zero.
No. Some equations do not have convenient rational factors. The quadratic formula is more efficient in such cases.
Use it when factorisation is difficult, when the roots contain surds or when a general solving method is required.
A negative root is rejected when it cannot represent the quantity in the question, such as a positive length, age, speed or number of objects.
Write the equation in standard form and set its discriminant equal to zero:
b² − 4ac = 0
The roots of ax² + bx + c = 0 are the same values as the zeroes of the polynomial ax² + bx + c.
NCERT exercises build the essential concepts, but students should also practise competency-based, case-based and previous-year questions.
Students can download the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations PDF for free from Infinity Learn. The platform provides chapter-wise, step-by-step NCERT solutions designed according to the current Class 10 Maths syllabus.