Quadratic Equations is one of the most important chapters in Class 10 Maths because it connects algebra, factorisation, formulas, graphs, and word problems in one scoring topic. This chapter is very important not only for board exams, but also for building a strong base for higher classes. It also supports the chapter on polynomials because many concepts like roots and factor-based reasoning are closely connected.
Many students feel that Quadratic Equations is difficult because it has formulas, sign changes, and different question patterns. In reality, the chapter becomes much easier once students understand the standard form of a quadratic equation, learn the main solving methods, and clearly remember the role of the discriminant.
At Deeksha Vedantu, we always encourage students to learn this chapter in a sequence: first understand the form, then solve equations, then study the nature of roots, and finally practise application-based questions. This approach makes the chapter easier to retain and more useful in exams.
Why Quadratic Equations Is Important in Class 10
Quadratic Equations is one of the most useful and repeated chapters in board exams.
Why Students Should Prepare This Chapter Well
- it is a regular board-exam chapter
- it is used in direct, application-based, and competency-based questions
- it includes MCQs, short answers, long answers, and case-based formats
- it improves algebraic accuracy and formula usage
- it builds a foundation for higher Maths topics
Chapter Overview at a Glance
This quick table helps students revise the full chapter faster.
Quick Concept Table
| Topic | Key idea |
| Quadratic equation | Equation of degree 2 |
| Standard form | ax² + bx + c = 0, where a ≠ 0 |
| Roots | Values of x that satisfy the equation |
| Discriminant | D = b² – 4ac |
| Equal roots condition | D = 0 |
| Sum of roots | α + β = -b/a |
| Product of roots | αβ = c/a |
| Main solving methods | Factorisation and quadratic formula |
What Is a Quadratic Equation
A quadratic equation is an equation of degree 2.
Standard Form of a Quadratic Equation
ax² + bx + c = 0
Where:
- a, b, and c are real numbers
- a is not equal to 0
Why a Cannot Be 0
If a = 0, then the x² term disappears and the equation becomes linear, not quadratic.
So for an equation to be quadratic, the highest power of the variable must be 2.
How to Identify Whether an Equation Is Quadratic or Not
Many students make mistakes in this type of question because they only look at the equation once without simplifying it.
Important Rule
Before deciding whether an equation is quadratic or not, simplify it completely.
Solved Example 1: Check Whether an Equation Is Quadratic
Check whether the following equation is quadratic or not:
x² + 3x + 1 = (x – 2)²
Given
Equation:
x² + 3x + 1 = (x – 2)²
Step 1: Expand the Right Side
(x – 2)² = x² – 4x + 4
Step 2: Compare Both Sides
x² + 3x + 1 = x² – 4x + 4
Step 3: Simplify
The x² terms cancel:
7x – 3 = 0
Answer
This is a linear equation, not a quadratic equation.
Solved Example 2: Simplify Before Identifying
Check whether the following equation is quadratic or not:
x³ – 4x² – x + 1 = (x – 2)³
Given
Equation:
x³ – 4x² – x + 1 = (x – 2)³
Step 1: Expand the Right Side
(x – 2)³ = x³ – 6x² + 12x – 8
Step 2: Compare Both Sides
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
Step 3: Simplify
The x³ terms cancel:
2x² – 13x + 9 = 0
Answer
This is a quadratic equation because the highest remaining power is 2.
Terms Related to a Quadratic Equation
Students should know the meaning of these terms clearly.
Terms Summary Table
| Term | Meaning |
| Coefficient of x² | a |
| Coefficient of x | b |
| Constant term | c |
| Roots | Values of x that satisfy the quadratic equation |
| Standard form | ax² + bx + c = 0 |
Coefficients
In ax² + bx + c = 0:
- a is the coefficient of x²
- b is the coefficient of x
- c is the constant term
Roots of a Quadratic Equation
The values of x that satisfy the quadratic equation are called its roots.
If α and β are the roots of a quadratic equation, then they make the equation true.
Methods to Solve a Quadratic Equation
There are two main methods students use in Class 10.
Method Summary Table
| Method | Best use |
| Factorisation method | Useful when the equation splits into simple factors |
| Quadratic formula method | Useful when factorisation is not easy |
Method 1: Factorisation Method
This method works well when the equation can be split into simple factors.
Method 2: Quadratic Formula Method
This method is useful when factorisation is not easy or when the equation is more complex.
Solving by Factorisation Method
This is one of the most common and important methods.
General Idea
First write the equation in the form:
ax² + bx + c = 0
Then split the middle term and factorise the expression.
Step-by-Step Rule for Factorisation
| Step | What to do |
| Step 1 | Multiply the coefficient of x² and the constant term |
| Step 2 | Find two numbers whose product is that value and whose sum is the coefficient of x |
| Step 3 | Split the middle term using those two numbers |
| Step 4 | Factorise by grouping |
| Step 5 | Set each factor equal to 0 and find the roots |
Solved Example 3: Factorisation Method
Solve:
x² – 9x + 18 = 0
Given
Equation:
x² – 9x + 18 = 0
Step 1: Product of First and Last Coefficients
1 × 18 = 18
Step 2: Find Two Numbers
We need two numbers whose product is 18 and sum is -9.
Those numbers are -3 and -6.
Step 3: Split the Middle Term
x² – 3x – 6x + 18 = 0
Step 4: Factorise by Grouping
x(x – 3) – 6(x – 3) = 0
(x – 3)(x – 6) = 0
Step 5: Find the Roots
x – 3 = 0 gives x = 3
x – 6 = 0 gives x = 6
Answer
The roots are 3 and 6.
Solving by Quadratic Formula Method
This is the second major solving method.
Quadratic Formula
x = [-b ± √(b² – 4ac)] / 2a
Students should memorise this formula very carefully.
What Is the Discriminant
The expression inside the square root is called the discriminant.
Formula for Discriminant
D = b² – 4ac
This value helps us understand the nature of roots and is also used directly in many questions.
Solved Example 4: Formula Method
Solve:
2x² – 5x + 3 = 0
Given
Equation:
2x² – 5x + 3 = 0
Step 1: Identify a, b, and c
- a = 2
- b = -5
- c = 3
Step 2: Find the Discriminant
D = b² – 4ac
D = (-5)² – 4 × 2 × 3
D = 25 – 24
D = 1
Step 3: Use the Formula
x = [-(-5) ± √1] / (2 × 2)
x = [5 ± 1] / 4
Step 4: Find the Two Roots
x = (5 + 1)/4 = 6/4 = 3/2
x = (5 – 1)/4 = 4/4 = 1
Answer
The roots are 3/2 and 1.
Which Method Should Students Use
Students often ask whether they should use factorisation or the quadratic formula.
Method Choice Table
| Method | Use it when |
| Factorisation | The numbers are simple and the equation splits quickly |
| Quadratic formula | The equation looks complex or does not factorise easily |
Relation Between Roots and Coefficients
This is a very important concept for board exams.
If α and β are the roots of the quadratic equation ax² + bx + c = 0, then:
Root-Coefficient Relation Table
| Concept | Formula |
| Sum of roots | α + β = -b/a |
| Product of roots | αβ = c/a |
These formulas are used in direct questions and in application-based questions.
Solved Example 5: Use Root-Coefficient Relation
If α and β are the roots of 5x² – 6x + 1 = 0, find:
α + β + αβ
Given
Polynomial:
5x² – 6x + 1 = 0
Step 1: Find α + β
α + β = -b/a = -(-6)/5 = 6/5
Step 2: Find αβ
αβ = c/a = 1/5
Step 3: Add
α + β + αβ = 6/5 + 1/5 = 7/5
Answer
The required value is 7/5.
Reciprocal Root Questions
These are frequently asked and very important.
If one root is the reciprocal of the other, then the roots can be written as:
α and 1/α
Important Result
Their product is:
α × 1/α = 1
This idea helps students solve such questions quickly.
Solved Example 6: Reciprocal Roots
If one zero of the polynomial 6x² + 37x – (k – 2) is the reciprocal of the other, find the value of k.
Given
Polynomial:
6x² + 37x – (k – 2)
Step 1: Use the Product of Roots
If roots are reciprocal, then their product is 1.
For a quadratic equation:
Product of roots = c/a
Here:
c = -(k – 2) = -k + 2
a = 6
So:
(-k + 2)/6 = 1
Step 2: Solve
-k + 2 = 6
-k = 4
k = -4
Answer
The value of k is -4.
Nature of Roots
This is one of the most important parts of the chapter.
The discriminant tells us the nature of roots without solving the full equation.
Case 1: D > 0
If D > 0, the equation has two distinct real roots.
Meaning
The two roots are real and different from each other.
Case 2: D = 0
If D = 0, the equation has two equal real roots.
Meaning
The two roots are real and equal.
Case 3: D < 0
If D < 0, the equation has no real roots.
Meaning
The roots are not real because the square root of a negative number does not give a real value in Class 10 Maths.
Nature of Roots Summary Table
| Condition on D | Nature of roots |
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots |
| D < 0 | No real roots |
Graphical Meaning of Nature of Roots
This is important for competency-based questions.
Case 1: When There Are Two Distinct Real Roots
The graph of the quadratic equation cuts the x-axis at two different points.
Case 2: When There Are Two Equal Real Roots
The graph touches the x-axis at one point.
Case 3: When There Are No Real Roots
The graph does not touch the x-axis at all.
This graphical understanding helps students answer visual and assertion-reason questions.
Graphical Meaning Table
| Nature of roots | Graph behaviour |
| Two distinct real roots | Cuts the x-axis at two points |
| Two equal real roots | Touches the x-axis at one point |
| No real roots | Does not touch the x-axis |
Solved Example 7: Nature of Roots from Discriminant
Find the nature of roots of:
4x² – 5 = 0
Given
Equation:
4x² – 5 = 0
Step 1: Write in Standard Form
4x² + 0x – 5 = 0
So:
- a = 4
- b = 0
- c = -5
Step 2: Find the Discriminant
D = b² – 4ac
D = 0² – 4 × 4 × (-5)
D = 80
Step 3: Interpret the Result
Since D > 0, the equation has two distinct real roots.
Answer
The equation has two distinct real roots.
Equal Root Condition Questions
In these questions, students are asked to find a value so that the roots become equal.
Important Rule
If the roots are equal, then:
D = 0
This is the main condition used.
Solved Example 8: Equal Roots Condition
Find the value of k for which the quadratic equation:
(k + 1)x² – 6(k + 1)x + 3(k + 9) = 0
has equal roots.
Given
Equation:
(k + 1)x² – 6(k + 1)x + 3(k + 9) = 0
Step 1: Use D = 0
Here:
- a = k + 1
- b = -6(k + 1)
- c = 3(k + 9)
So:
b² – 4ac = 0
Step 2: Simplify
[-6(k + 1)]² – 4(k + 1) × 3(k + 9) = 0
36(k + 1)² – 12(k + 1)(k + 9) = 0
Take 12(k + 1) common:
12(k + 1)[3(k + 1) – (k + 9)] = 0
12(k + 1)(2k – 6) = 0
(k + 1)(k – 3) = 0
So:
k = -1 or k = 3
Step 3: Check Validity
If k = -1, then the coefficient of x² becomes 0, so the equation is no longer quadratic.
So the valid value is:
k = 3
Answer
The valid value of k is 3.
Word Problems Based on Quadratic Equations
This is a highly important exam area.
Students often find word problems difficult because they rush the language and do not define the variable carefully.
Simple Strategy for Word Problems
| Step | What to do |
| Step 1 | Identify what the question is asking |
| Step 2 | Assume that unknown quantity as x |
| Step 3 | Use the given condition to form an equation |
| Step 4 | Convert it into a quadratic equation |
| Step 5 | Solve and reject invalid values if needed |
Solved Example 9: Natural Number Problem
A natural number is such that when increased by 12, it becomes 160 times its reciprocal. Find the number.
Given
Condition:
x + 12 = 160/x
Step 1: Let the Number Be x
Take the required natural number as x.
Step 2: Cross Multiply
x(x + 12) = 160
x² + 12x = 160
x² + 12x – 160 = 0
Step 3: Factorise
x² + 20x – 8x – 160 = 0
x(x + 20) – 8(x + 20) = 0
(x + 20)(x – 8) = 0
Step 4: Find the Values
x = -20 or x = 8
Step 5: Accept the Valid Value
Since the question asks for a natural number, x = 8 is the valid answer.
Answer
The required natural number is 8.
Solved Example 10: Time and Work Problem
A and B working together can complete a piece of work in 6 days. A takes 5 days less than B to complete the work alone. Find the time taken by B alone.
Given
- Together, A and B finish the work in 6 days
- A takes 5 days less than B alone
Step 1: Let B Take x Days
Then A takes:
x – 5 days
Step 2: Write One-Day Work
B’s one-day work = 1/x
A’s one-day work = 1/(x – 5)
Together they finish the work in 6 days, so one-day work together is:
1/6
Step 3: Form the Equation
1/x + 1/(x – 5) = 1/6
Step 4: Simplify
[(x – 5) + x] / [x(x – 5)] = 1/6
(2x – 5) / [x(x – 5)] = 1/6
Cross multiply:
6(2x – 5) = x(x – 5)
12x – 30 = x² – 5x
x² – 17x + 30 = 0
Step 5: Factorise
x² – 15x – 2x + 30 = 0
x(x – 15) – 2(x – 15) = 0
(x – 15)(x – 2) = 0
So:
x = 15 or x = 2
Step 6: Accept the Valid Value
Since B alone cannot take 2 days when A and B together take 6 days, the valid value is:
x = 15
Answer
B alone takes 15 days.
Important Question Types from Quadratic Equations
Board exams often repeat these question patterns.
Question Type Summary Table
| Type | Focus area |
| Type 1 | Simplify and identify degree correctly |
| Type 2 | Split middle term and factorise |
| Type 3 | Use quadratic formula carefully |
| Type 4 | Use discriminant for nature of roots |
| Type 5 | Apply D = 0 |
| Type 6 | Use product of roots = 1 for reciprocal roots |
| Type 7 | Form equation carefully from language |
| Type 8 | Connect graph behaviour with nature of roots |
Common Mistakes Students Make in Quadratic Equations
Common Mistakes Table
| Mistake | Correct idea |
| Forgetting the standard form | Always bring the equation to ax² + bx + c = 0 |
| Sign errors in factorisation | Split and group carefully |
| Wrong use of quadratic formula | Keep the full expression under the square root |
| Wrong discriminant calculation | Calculate b² – 4ac carefully |
| Accepting invalid values in word problems | Reject impossible values like negative time or invalid natural numbers |
| Ignoring that a ≠ 0 | If the coefficient of x² becomes 0, the equation is not quadratic |
Quick Formula Sheet for Quadratic Equations
This section is useful for final board revision.
Quick Formula Table
| Formula type | Formula |
| Standard form | ax² + bx + c = 0 |
| Quadratic formula | x = [-b ± √(b² – 4ac)] / 2a |
| Discriminant | D = b² – 4ac |
| Sum of roots | α + β = -b/a |
| Product of roots | αβ = c/a |
Nature of Roots Quick Table
| Condition | Result |
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots |
| D < 0 | No real roots |
Practice Questions
This section helps students revise the chapter through important question types.
Basic Practice Questions
- Check whether the following equation is quadratic or not:
x² + 3x + 1 = (x – 2)²
- Solve:
x² – 9x + 18 = 0
- Solve using the quadratic formula:
2x² – 5x + 3 = 0
- Find the nature of roots of:
3x² + 5x + 2 = 0
- Find the nature of roots of:
x² + 4x + 8 = 0
Root-Based Practice Questions
- If α and β are the roots of 5x² – 6x + 1 = 0, find α + β and αβ.
- If one root is the reciprocal of the other in 6x² + 37x – (k – 2) = 0, find k.
- Find the value of k for which x² + kx + 9 = 0 has equal roots.
Word Problem Practice Questions
- A natural number is such that when increased by 10, it becomes 39 times its reciprocal. Find the number.
- A and B working together can complete a work in 8 days. A takes 6 days less than B to complete it alone. Find the time taken by B.
Best Study Strategy for Quadratic Equations
This chapter becomes much easier when students revise it in a fixed order.
Step-by-Step Revision Table
| Step | What to do |
| Step 1 | Revise the standard form and core formulas |
| Step 2 | Practise factorisation questions |
| Step 3 | Practise quadratic formula questions |
| Step 4 | Revise discriminant and nature of roots |
| Step 5 | Solve root-based questions with α and β |
| Step 6 | Practise word problems carefully |
FAQs
Q1. What is a quadratic equation in Class 10 Maths?
A quadratic equation is an equation of degree 2 written in the form ax² + bx + c = 0, where a is not equal to 0.
Q2. What is the standard form of a quadratic equation?
The standard form is ax² + bx + c = 0.
Q3. What are the methods to solve a quadratic equation?
The main methods are factorisation and the quadratic formula method.
Q4. What is the discriminant?
The discriminant is D = b² – 4ac. It helps determine the nature of roots.
Q5. When does a quadratic equation have equal roots?
A quadratic equation has equal roots when D = 0.
Q6. When does a quadratic equation have no real roots?
It has no real roots when D < 0.
Q7. What is the formula for the sum and product of roots?
If α and β are the roots of ax² + bx + c = 0, then α + β = -b/a and αβ = c/a.
Q8. How can I score well in Quadratic Equations?
You can score well by learning the standard form, practising both solving methods, revising the discriminant cases, and solving word problems carefully.
Conclusion
Quadratic Equations is one of the most important and scoring chapters in Class 10 Maths because it brings together concepts, formulas, logic, and application-based solving in one place. Once students understand the standard form, factorisation, quadratic formula, and discriminant properly, the chapter becomes much more manageable.
The best way to prepare this chapter is to practise every question type step by step and not rely only on memorisation. At Deeksha Vedantu, we always remind students that strong conceptual understanding makes formulas easier to use and questions much faster to solve.
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