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Updated on 21 Jul 2026, 12:55 IST
These Quadratic Equations Class 10 Notes PDF explain the standard form, factorisation method, quadratic formula, discriminant, nature of roots, graphs, value-of-k questions, and real-life applications required for Class 10 Maths. The chapter also includes solved examples, a method-selection guide, important questions, common mistakes, revision exercises, and a downloadable formula sheet.
For the CBSE 2026–27 syllabus, students are expected to solve quadratic equations with real roots by factorisation and the quadratic formula, determine the nature of roots using the discriminant, and apply quadratic equations to daily-life situations.
Also Check:
A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a ≠ 0.
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ax² + bx + c = 0
Here:
Quadratic formula:
x = (−b ± √(b² − 4ac))/(2a)
Discriminant:
D = b² − 4ac
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Nature of roots:
| Condition | Nature of roots |
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots |
| D < 0 | No real roots |
Repeated root when D = 0:
x = −b/(2a)
| Situation | Best method |
| Factors are easy to identify | Factorisation |
| Factors are not obvious | Quadratic formula |
| Only the nature of roots is required | Discriminant |
| Equal roots are mentioned | Set D = 0 |
| Real roots are required | Use D ≥ 0 |
| A word problem is given | Form the equation first |
The current Class 10 syllabus covers standard form, real roots, factorisation, the quadratic formula, discriminant, and situational problems.
The prescribed areas are:
Completing the square is useful for understanding the quadratic formula, but it is not listed as a separate required solving method in the CBSE 2026–27 Class 10 syllabus. Complex-number solutions are also outside the stated chapter scope because the syllabus specifies real roots.
CBSE assigns 20 marks to the complete Algebra unit, but it does not publish a guaranteed marks allocation for Quadratic Equations alone.
The Algebra unit also contains Polynomials, Pair of Linear Equations in Two Variables, and Arithmetic Progressions. Claims such as “Quadratic Equations always carries six marks” should not be made unless they refer to a specific examination paper.
A quadratic equation is an equation whose highest power is 2 after all terms have been simplified and arranged on one side.
The standard form is:
ax² + bx + c = 0
where:
a, b, and c are real numbers, and a ≠ 0.
Quadratic expression: An expression of degree 2 without an equals sign.
Example:
2x² − 5x + 3
Quadratic polynomial: A polynomial whose highest power is 2.
Example:
x² + 4x − 7
Quadratic equation: A quadratic expression set equal to zero or another expression.
Example:
x² + 4x − 7 = 0
Root or solution: A value of x that makes the equation true.
Zero of a polynomial: A value of x that makes the related polynomial equal to zero.
For:
ax² + bx + c = 0
| Symbol | Meaning |
| a | Coefficient of x² |
| b | Coefficient of x |
| c | Constant term |
| a ≠ 0 | Ensures that the equation is quadratic |
Consider:
3x² − 7x + 2 = 0
The coefficients are:
The sign belongs to the coefficient. Therefore, b is −7, not 7.
An equation should be expanded, simplified, and rearranged before deciding whether it is quadratic.
An equation that appears quadratic may become linear after cancellation. An equation written with brackets may become quadratic only after expansion.
Determine whether:
(x − 3)² = 5x − 7
is quadratic.
Expand:
x² − 6x + 9 = 5x − 7
Move all terms to the left:
x² − 6x − 5x + 9 + 7 = 0
x² − 11x + 16 = 0
The highest power is 2.
Therefore, the equation is quadratic.
Determine whether:
(x + 2)(x − 1) = x² − 2x − 3
is quadratic.
Expand the left side:
x² + x − 2 = x² − 2x − 3
Cancel x² from both sides:
x − 2 = −2x − 3
3x + 1 = 0
The simplified equation is linear, not quadratic.
Determine whether:
x(x + 2)/3 = 4
is quadratic.
Multiply both sides by 3:
x(x + 2) = 12
Expand:
x² + 2x = 12
Write in standard form:
x² + 2x − 12 = 0
The equation is quadratic.
A quadratic expression has no equality sign, a quadratic polynomial is a degree-2 polynomial, and a quadratic equation states that two expressions are equal.
| Form | Example | Can it be solved for roots? |
| Quadratic expression | x² − 5x + 6 | Not until it is made into an equation |
| Quadratic polynomial | p(x) = x² − 5x + 6 | Its zeroes can be found |
| Quadratic equation | x² − 5x + 6 = 0 | Yes |
The zeroes of the polynomial:
p(x) = x² − 5x + 6
are the same numerical values as the roots of:
x² − 5x + 6 = 0
[INTERNAL LINK: Zeroes of polynomials and coefficient relationships → /class-10-polynomials-notes/]
A root of a quadratic equation is a value of x that makes the left side equal to zero.
For:
ax² + bx + c = 0
a number α is a root if:
aα² + bα + c = 0
Check whether x = 2 is a root of:
x² − 5x + 6 = 0
Substitute x = 2:
2² − 5(2) + 6
= 4 − 10 + 6
= 0
Therefore, x = 2 is a root.
A quadratic equation can have:
A quadratic equation cannot have more than two roots because the related polynomial has degree 2.
To verify a calculated root:
The two main Class 10 methods are factorisation and the quadratic formula.
The discriminant is used when the question asks for the nature of roots without requiring their exact values.
| Question type | Recommended approach |
| Simple integer factors | Factorisation |
| Difficult or irrational roots | Quadratic formula |
| Nature of roots only | Discriminant |
| Equal roots | Set D = 0 |
| Value-of-k range | Use D > 0, D = 0, D ≥ 0, or D < 0 |
| Word problem | Form the equation, then choose a method |
Factorisation solves a quadratic equation by expressing it as a product of two linear factors and setting each factor equal to zero.
For:
ax² + bx + c = 0
find two numbers whose:
Use these two numbers to split the middle term.
Solve:
2x² − 7x + 3 = 0
Here:
a = 2
b = −7
c = 3
Calculate:
ac = 2 × 3
ac = 6
We need two numbers whose product is 6 and whose sum is −7.
The numbers are:
−6 and −1
Split the middle term:
2x² − 6x − x + 3 = 0
Group the terms:
2x(x − 3) − 1(x − 3) = 0
Factorise:
(2x − 1)(x − 3) = 0
Use the zero-product property:
2x − 1 = 0
or:
x − 3 = 0
First root:
2x = 1
x = 1/2
Second root:
x = 3
Therefore:
x = 1/2 or x = 3
For x = 3:
2(3)² − 7(3) + 3
= 18 − 21 + 3
= 0
For x = 1/2:
2(1/2)² − 7(1/2) + 3
= 2(1/4) − 7/2 + 3
= 1/2 − 7/2 + 3
= −3 + 3
= 0
Both roots are correct.
For:
2x² − 7x + 3
students may search for two numbers whose product is only c = 3.
The required product is:
a × c = 2 × 3 = 6
When the coefficient of x² is 1, find two numbers whose product is c and whose sum is b.
Solve:
x² − 9x + 20 = 0
We need two numbers whose:
The numbers are −4 and −5.
Therefore:
x² − 4x − 5x + 20 = 0
x(x − 4) − 5(x − 4) = 0
(x − 4)(x − 5) = 0
Hence:
x = 4 or x = 5
The quadratic formula gives the roots of ax² + bx + c = 0 directly from its coefficients.
The formula is:
x = (−b ± √(b² − 4ac))/(2a)
The symbol ± means that two calculations must be completed:
x = (−b + √(b² − 4ac))/(2a)
and:
x = (−b − √(b² − 4ac))/(2a)
Solve:
3x² + x − 2 = 0
Here:
a = 3
b = 1
c = −2
Calculate the discriminant:
D = b² − 4ac
D = 1² − 4(3)(−2)
D = 1 + 24
D = 25
Apply the formula:
x = (−1 ± √25)/(2 × 3)
x = (−1 ± 5)/6
Using the plus sign:
x = (−1 + 5)/6
x = 4/6
x = 2/3
Using the minus sign:
x = (−1 − 5)/6
x = −6/6
x = −1
Therefore:
x = 2/3 or x = −1
Check that:
The quadratic formula can produce irrational roots when the discriminant is positive but not a perfect square.
Solve:
2x² + x − 7 = 0
Here:
a = 2
b = 1
c = −7
Calculate:
D = b² − 4ac
D = 1² − 4(2)(−7)
D = 1 + 56
D = 57
Use the formula:
x = (−1 ± √57)/(2 × 2)
Therefore:
x = (−1 ± √57)/4
The equation has two distinct irrational real roots.
The discriminant D = b² − 4ac determines whether a quadratic equation has two, one, or no real roots.
| Discriminant | Nature of roots | Graphical meaning |
| D > 0 | Two distinct real roots | Parabola crosses the x-axis twice |
| D = 0 | Two equal real roots | Parabola touches the x-axis once |
| D < 0 | No real roots | Parabola does not meet the x-axis |
These cases form part of the current CBSE Class 10 chapter requirements.
Consider:
x² − 5x + 6 = 0
Here:
a = 1
b = −5
c = 6
D = (−5)² − 4(1)(6)
D = 25 − 24
D = 1
Since D > 0, the equation has two distinct real roots.
Consider:
x² − 6x + 9 = 0
D = (−6)² − 4(1)(9)
D = 36 − 36
D = 0
Therefore, the roots are equal.
The repeated root is:
x = −b/(2a)
x = −(−6)/(2 × 1)
x = 6/2
x = 3
Consider:
2x² − 4x + 5 = 0
D = (−4)² − 4(2)(5)
D = 16 − 40
D = −24
Since D < 0, the equation has no real roots.
At Class 10 level, the answer should stop at “no real roots.” Complex-number solutions are outside the stated syllabus scope.
Value-of-k questions are solved by applying the discriminant condition stated in the question.
| Wording | Condition |
| Two equal roots | D = 0 |
| Two distinct real roots | D > 0 |
| Real roots | D ≥ 0 |
| No real roots | D < 0 |
Find k if:
2x² + kx + 8 = 0
has equal roots.
For equal roots:
D = 0
Here:
a = 2
b = k
c = 8
Therefore:
k² − 4(2)(8) = 0
k² − 64 = 0
(k − 8)(k + 8) = 0
Therefore:
k = 8 or k = −8
Find the values of k for which:
x² − 4x + k = 0
has real roots.
For real roots:
D ≥ 0
Here:
a = 1
b = −4
c = k
D = (−4)² − 4(1)(k)
D = 16 − 4k
Therefore:
16 − 4k ≥ 0
−4k ≥ −16
Dividing by −4 reverses the inequality:
k ≤ 4
Therefore, the equation has real roots when:
k ≤ 4
Find the values of k for which:
x² + 2x + k = 0
has no real roots.
For no real roots:
D < 0
D = 2² − 4(1)(k)
4 − 4k < 0
−4k < −4
Divide by −4 and reverse the inequality:
k > 1
Therefore, the equation has no real roots when:
k > 1
The graph of y = ax² + bx + c is a parabola, and its x-intercepts represent the real roots of ax² + bx + c = 0.
| Coefficient | Main effect |
| a | Controls opening direction and width |
| b | Influences horizontal position and axis of symmetry |
| c | Gives the y-intercept |
At x = 0:
y = a(0)² + b(0) + c
y = c
Therefore, the graph meets the y-axis at:
(0, c)
| Root condition | Graph |
| Two distinct real roots | Two x-intercepts |
| Equal roots | One touching point on the x-axis |
| No real roots | No x-intercept |
The axis of symmetry is:
x = −b/(2a)
This formula is useful for interpreting the graph, although it is enrichment rather than a separately prescribed solving method in the chapter.
A quadratic word problem is solved by defining a variable, forming an equation, solving it, and rejecting any root that is impossible in the context.
The current syllabus requires students to formulate and solve quadratic equations in real-life situations.
The length of a rectangle is 3 cm more than its breadth. Its area is 40 cm². Find its dimensions.
Let the breadth be x cm.
Then the length is:
x + 3 cm
Area of rectangle:
Length × Breadth = 40
x(x + 3) = 40
Expand:
x² + 3x = 40
Write in standard form:
x² + 3x − 40 = 0
Factorise:
(x + 8)(x − 5) = 0
Therefore:
x = −8 or x = 5
A breadth cannot be negative, so reject x = −8.
Therefore:
Breadth = 5 cm
Length = 5 + 3 = 8 cm
The product of two consecutive positive integers is 156. Find the integers.
Let the smaller integer be x.
The next integer is:
x + 1
According to the question:
x(x + 1) = 156
x² + x − 156 = 0
Factorise:
(x + 13)(x − 12) = 0
Therefore:
x = −13 or x = 12
The integers are positive, so reject −13.
Therefore, the integers are:
12 and 13
A vehicle travels 240 km. If its speed were 4 km/h faster, the journey would take one hour less. Find its original speed.
Let the original speed be x km/h.
Original time:
240/x hours
New speed:
x + 4 km/h
New time:
240/(x + 4) hours
The new journey takes one hour less:
240/x − 240/(x + 4) = 1
Take the LCM:
[240(x + 4) − 240x]/[x(x + 4)] = 1
960/[x(x + 4)] = 1
x(x + 4) = 960
x² + 4x − 960 = 0
Use the quadratic formula:
x = [−4 ± √(4² − 4(1)(−960))]/2
x = [−4 ± √3856]/2
Since:
√3856 = 4√241
x = [−4 ± 4√241]/2
x = −2 ± 2√241
The positive value is:
x = −2 + 2√241
x ≈ 29.05
Therefore, the original speed was approximately:
29.05 km/h
The negative root is rejected because speed cannot be negative.
A root may solve the algebraic equation but fail the real-life condition.
Negative roots are commonly rejected when the variable represents:
Factorisation is usually faster for equations with simple factors, while the quadratic formula works reliably even when the factors are not obvious.
| Feature | Factorisation | Quadratic formula |
| Best for | Simple integer or rational factors | Any quadratic equation within the real-root scope |
| Main skill | Finding factor pairs | Accurate substitution |
| Speed | Usually faster | Usually longer |
| Common error | Incorrect middle-term split | Sign or denominator error |
| Irrational roots | Usually not convenient | Suitable |
| Gives discriminant | No | Yes |
| Verification | Expand factors | Substitute roots |
Most errors result from incorrect standard form, lost signs, wrong factor pairs, or misuse of the quadratic formula.
| Mistake | Incorrect approach | Correct approach |
| Reading coefficients too early | Identifying a, b, and c before rearranging | Write ax² + bx + c = 0 first |
| Losing the sign of b | Taking b = 5 in x² − 5x + 6 = 0 | Use b = −5 |
| Using product c | Finding factors of 3 in 2x² − 7x + 3 | Find factors of ac = 6 |
| Squaring incorrectly | Writing −4² = 16 without brackets | Write (−4)² = 16 |
| Dividing incorrectly | Dividing only √D by 2a | Divide the complete numerator |
| Ignoring one formula result | Using only the plus sign | Calculate both ± cases |
| Using D = 0 for all real roots | Excluding distinct roots | Use D ≥ 0 for real roots |
| Accepting impossible values | Reporting a negative length | Check the context |
| Forgetting verification | Ending after calculation | Substitute roots into the equation |
| Ignoring cancellation | Classifying before simplifying | Simplify completely first |
A complete revision set should include identification, factorisation, formula, discriminant, parameter, and application questions.
Check whether:
(x + 1)² = 3x + 7
is a quadratic equation.
Solution:
Expand:
x² + 2x + 1 = 3x + 7
Rearrange:
x² − x − 6 = 0
The highest power is 2.
Therefore, it is a quadratic equation.
Solve:
x² − 9x + 20 = 0
Answer:
(x − 4)(x − 5) = 0
x = 4 or x = 5
Solve:
3x² − 8x + 4 = 0
Answer:
3x² − 6x − 2x + 4 = 0
3x(x − 2) − 2(x − 2) = 0
(3x − 2)(x − 2) = 0
x = 2/3 or x = 2
Find the roots of:
2x² + x − 7 = 0
Answer:
x = (−1 ± √57)/4
Find the nature of the roots of:
4x² − 4x + 1 = 0
Answer:
D = (−4)² − 4(4)(1)
D = 16 − 16
D = 0
The equation has two equal real roots.
Find k if:
x² + kx + 16 = 0
has equal roots.
Answer:
k = 8 or k = −8
Find the values of k for which:
x² − 4x + k = 0
has real roots.
Answer:
k ≤ 4
The product of two consecutive positive integers is 156. Find the integers.
Answer:
12 and 13
A rectangle has an area of 96 m², and its length is 4 m more than its breadth. Find its dimensions.
Answer:
Breadth = 8 m
Length = 12 m
A vehicle travels 240 km. If its speed were 4 km/h faster, it would take one hour less. Find its original speed.
Answer:
Original speed = −2 + 2√241 km/h
Approximately:
29.05 km/h
Also Check:
Quadratic-equation MCQs test standard form, roots, discriminant, and method selection.
Q1. Which is a quadratic equation?
A. 3x + 5 = 0
B. x² − 4x + 3 = 0
C. x³ − 2 = 0
D. 1/x + 2 = 0
Answer: B
Q2. The discriminant of ax² + bx + c = 0 is:
A. b² + 4ac
B. b² − 4ac
C. 4ac − b²
D. b − 4ac
Answer: B
Q3. If D = 0, the equation has:
A. Two distinct real roots
B. No real roots
C. Two equal real roots
D. Three roots
Answer: C
Q4. The roots of x² − 5x + 6 = 0 are:
A. 1 and 6
B. 2 and 3
C. −2 and −3
D. 5 and 6
Answer: B
Q5. For 2x² − 7x + 3 = 0, the value of ac is:
A. 3
B. 5
C. 6
D. −6
Answer: C
Q6. If D < 0, the equation has:
A. Two equal real roots
B. Two distinct real roots
C. No real roots
D. One positive root
Answer: C
Assertion: A quadratic equation with equal roots has D = 0.
Reason: Its parabola touches the x-axis at exactly one point.
Answer: Both statements are true, and the reason correctly explains the assertion.
The essential formulas are the standard form, quadratic formula, discriminant conditions, and repeated-root formula.
| Concept | Formula or rule |
| Standard form | ax² + bx + c = 0, a ≠ 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 |
| Real roots | D ≥ 0 |
| Repeated root | x = −b/(2a) |
| y-intercept | (0, c) |
| Axis of symmetry | x = −b/(2a) |
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A quadratic equation can be written as ax² + bx + c = 0, where a ≠ 0. Its highest power after simplification is 2.
Find two numbers whose product is ac and whose sum is b. Split the middle term, factorise by grouping, and set each linear factor equal to zero.
The formula is:
x = (−b ± √(b² − 4ac))/(2a)
It should be used only after writing the equation in standard form.
Calculate D = b² − 4ac. The roots are distinct and real when D > 0, equal and real when D = 0, and not real when D < 0.
A quadratic equation has two equal roots when:
D = b² − 4ac = 0
The repeated root is x = −b/(2a).
Find two numbers whose product is ac and whose sum is b. Replace bx with the two corresponding terms and factorise by grouping.
The CBSE 2026–27 syllabus specifies factorisation and the quadratic formula for equations with real roots.
Completing the square is not listed as a separate required solving method in the 2026–27 CBSE syllabus. It may be studied as enrichment.
No. The syllabus specifies solutions with real roots, so an equation with D < 0 should be reported as having no real roots.
Download the Quadratic Equations Class 10 notes PDF from Infinity Learn website's this page to access printable notes containing formulas, examples, common mistakes, and revision questions.
The main formulas are ax² + bx + c = 0, x = (−b ± √(b² − 4ac))/(2a), and D = b² − 4ac.
Define a variable, translate the situation into an equation, solve it, and test both roots against the original context. Reject any value that gives an impossible age, length, speed, time, or quantity.