This comprehensive guide to CBSE Class 10 Maths Chapter 4, Quadratic Equations, has everything you need to master the topic. Learn the standard form, all three methods for solving equations (factorization, completing the square, and the quadratic formula), and how to find the nature of roots using the discriminant. We’ll also cover word problems and let you see equations in action with an interactive parabola visualizer.

Chapter 4

Quadratic Equations

Your complete guide to understanding, solving, and visualizing quadratic equations. Master the concepts for your CBSE Class 10 exams.

1. What is a Quadratic Equation?

The Standard Form

Any equation that can be written in the form:

ax² + bx + c = 0

…is a quadratic equation. Here’s what the letters mean:

a
The Quadratic Coefficient. It’s the number next to x².
Important: a cannot be zero. If a=0, the x² term disappears, and it becomes a linear equation (bx + c = 0).
b
The Linear Coefficient. It’s the number next to x.
c
The Constant Term. It’s the number with no x attached.

The values a, b, and c are all real numbers.

Identifying Quadratic Equations

Sometimes, an equation doesn’t look quadratic at first. You must simplify it to its standard form.

Example: YES

x(x + 1) + 8 = (x + 2)(x – 2)

Simplifies to:
x² + x + 8 = x² – 4
x + 12 = 0
No. The x² terms cancel out. This is a linear equation.

Correct Example: YES

(x + 2)³ = x³ – 4

Simplifies to:
x³ + 6x² + 12x + 8 = x³ – 4
6x² + 12x + 12 = 0
Yes. The highest power of x is 2. (Here, a=6, b=12, c=12).

Example: NO

x²(x + 1) = 5

Simplifies to:
x³ + x² – 5 = 0
No. The highest power of x is 3. This is a cubic equation.

2. How to Solve Quadratic Equations

“Solving” an equation means finding the values of x that make the equation true. These values are called the roots of the equation. A quadratic equation can have two roots, one root, or no real roots.

Method 1

Solution by Factorization

This method involves splitting the middle term (bx) into two terms.

Example: Solve `2x² - 5x + 3 = 0`

Find a × c: 2 × 3 = 6
Find two numbers that multiply to 6 and add up to ‘b’ (which is -5).
These numbers are -2 and -3. (-2 × -3 = 6 and -2 + -3 = -5)
Split the middle term:
2x² - 2x - 3x + 3 = 0
Factor by grouping:
2x(x - 1) - 3(x - 1) = 0
Extract the common factor:
(2x - 3)(x - 1) = 0
Solve for x:
Either 2x - 3 = 0 ➔ x = 3/2
Or x - 1 = 0 ➔ x = 1

Roots are: 1 and 3/2

Method 2

Solution by Quadratic Formula

This formula gives the roots of ax² + bx + c = 0 directly. It always works.

x = [ -b ± √(b² - 4ac) ] / 2a

Example: Solve `2x² - 5x + 3 = 0`

Identify a, b, c:
a = 2, b = -5, c = 3
Plug into the formula:
x = [ -(-5) ± √((-5)² - 4*2*3) ] / (2*2)
Simplify:
x = [ 5 ± √(25 - 24) ] / 4
x = [ 5 ± √1 ] / 4
x = ( 5 ± 1 ) / 4
Find the two roots:
Root 1: (5 + 1) / 4 = 6 / 4 = 3/2
Root 2: (5 - 1) / 4 = 4 / 4 = 1

Roots are: 1 and 3/2 (Same as before!)

Method 3

Solution by Completing the Square

This method converts the equation into the form (x + k)² = m. It’s also the method used to *prove* the quadratic formula.

Example: Solve `2x² - 5x + 3 = 0`

Make a = 1 (Divide by 2):
x² - (5/2)x + 3/2 = 0
Move constant c to the right:
x² - (5/2)x = -3/2
Add (b/2)² to both sides.
Here, b = -5/2, so (b/2) = -5/4, and (b/2)² = 25/16.
x² - (5/2)x + 25/16 = -3/2 + 25/16
Factor the left side (it’s a perfect square):
(x - 5/4)² = -24/16 + 25/16
(x - 5/4)² = 1/16
Take the square root of both sides:
x - 5/4 = ±√(1/16)
x - 5/4 = ± 1/4
Solve for x:
Root 1: x = 5/4 + 1/4 = 6/4 = 3/2
Root 2: x = 5/4 - 1/4 = 4/4 = 1

3. Nature of Roots (The Discriminant)

We can know the *type* of roots an equation has without actually solving it, using a value called the Discriminant.

The discriminant is the part inside the square root from the quadratic formula.

D = b² - 4ac

What ‘D’ tells us:

D > 0 Positive

The equation has TWO distinct, real roots.
(The parabola crosses the x-axis twice).

Example: x² – 5x + 6 = 0
D = (-5)² – 4(1)(6) = 25 – 24 = 1

D = 0 Zero

The equation has ONE real root (or two equal, real roots).
(The parabola just touches the x-axis at one point).

Example: x² – 4x + 4 = 0
D = (-4)² – 4(1)(4) = 16 – 16 = 0

D < 0 Negative

The equation has NO real roots.
(The parabola never crosses the x-axis).

Example: x² + 2x + 3 = 0
D = (2)² – 4(1)(3) = 4 – 12 = -8

4. Relationship Between Roots & Coefficients

If the roots (solutions) of the equation ax² + bx + c = 0 are α (alpha) and β (beta), there is a direct relationship between these roots and the coefficients a, b, and c.

Sum of Roots (α + β)

The sum of the two roots is always equal to -b/a.

α + β = -b / a

Product of Roots (α × β)

The product of the two roots is always equal to c/a.

α × β = c / a

Forming a Quadratic Equation from its Roots

If you are given the roots α and β, you can construct the equation directly.

x² - (α + β)x + (α × β) = 0

or simply: x² – (Sum of roots)x + (Product of roots) = 0

Example: Form an equation whose roots are `4` and `-2`.

Find the Sum (α + β): 4 + (-2) = 2
Find the Product (α × β): 4 × (-2) = -8
Use the formula x² - (Sum)x + (Product) = 0:
x² - (2)x + (-8) = 0
Final Equation: x² - 2x - 8 = 0

5. Interactive Parabola Visualizer

A quadratic equation y = ax² + bx + c forms a U-shaped curve called a parabola. The “roots” are simply the points where this curve crosses the x-axis (where y=0).

Use the sliders to change a, b, and c and see how the graph and roots change in real-time.

Coefficient a 1.0

If a > 0, parabola opens up (Smiley face 😊). If a < 0, it opens down (Frowny face 😟).

Coefficient b -2.0

b and a together shift the vertex (the lowest/highest point) left or right.

Constant c -3.0

c is the y-intercept. It's where the curve crosses the vertical y-axis.

6. Applications & Word Problems

Quadratic equations appear in many real-world scenarios. The key is to translate the words into a mathematical equation in the standard form ax² + bx + c = 0.

Strategy for Solving Word Problems

Represent: Read the problem. Assign a variable x to the primary unknown quantity.
Relate: Express all other unknown quantities in terms of x.
Formulate: Use the conditions given in the problem to write an equation.
Standardize: Simplify the equation into the standard form ax² + bx + c = 0.
Solve: Find the value(s) of x using factorization or the quadratic formula.
Verify: Check if the solutions make sense. For example, age, speed, or length cannot be negative. Discard any invalid solutions.

Common Problem Types

Type 1: Geometry & Area

Involves area of shapes, often rectangles (Area = l × w) or triangles (Area = ½ × b × h).

Problem:

"The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides."

Setup:

Let base = x. Then altitude = x - 7.
Using Pythagoras' theorem (a² + b² = c²):
(x)² + (x - 7)² = 13²
x² + x² - 14x + 49 = 169
2x² - 14x - 120 = 0
x² - 7x - 60 = 0 (Solve this)

Type 2: Speed, Distance, Time

Uses the formula Time = Distance / Speed. Often compares two scenarios.

Problem:

"A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less."

Setup:

Let speed = x.
Original Time (T1) = 360/x
New Speed = x + 5
New Time (T2) = 360/(x+5)
Given: T1 - T2 = 1
(360/x) - (360/(x+5)) = 1
This simplifies to x² + 5x - 1800 = 0.

Type 3: Age Problems

Involves relationships between ages in the past, present, or future.

Problem:

"The sum of the reciprocals of Rehman's age 3 years ago and 5 years from now is 1/3. Find his present age."

Setup:

Let present age = x.
Age 3 years ago = x - 3
Age 5 years from now = x + 5
Given: 1/(x-3) + 1/(x+5) = 1/3
This simplifies to x² - 4x - 21 = 0.

7. Practice Q&A (Check Your Knowledge)

Q1: Is (x - 2)² + 1 = 2x - 3 a quadratic equation? Click to reveal

Yes, it is.

Step 1: Expand (x - 2)² to x² - 4x + 4.
Step 2: The equation becomes x² - 4x + 4 + 1 = 2x - 3.
Step 3: Simplify: x² - 4x + 5 = 2x - 3.
Step 4: Move all terms to one side: x² - 4x - 2x + 5 + 3 = 0.
Final Form: x² - 6x + 8 = 0.

Since the highest power of x is 2 (and a=1, b=-6, c=8), it is a quadratic equation.

Q2: Find the discriminant (D) for the equation 3x² - 2x + 1/3 = 0. Click to reveal

The discriminant (D) is 0.

Step 1: Identify a, b, and c.
a = 3, b = -2, c = 1/3
Step 2: Use the formula D = b² - 4ac.
Step 3: Substitute the values: D = (-2)² - 4(3)(1/3)
Step 4: Calculate: D = 4 - 4
Final Answer: D = 0.

Bonus: Since D = 0, this equation has exactly one real root (or two equal real roots).

Q3: Find the roots of x² - 7x + 10 = 0 using factorization. Click to reveal

The roots are x = 2 and x = 5.

Step 1: We need two numbers that multiply to a*c = 10 and add up to b = -7.
Step 2: The two numbers are -2 and -5.
Step 3: Split the middle term: x² - 2x - 5x + 10 = 0
Step 4: Factor by grouping: x(x - 2) - 5(x - 2) = 0
Step 5: Extract common factor: (x - 5)(x - 2) = 0
Final Answer: Either x - 5 = 0 (so x = 5) or x - 2 = 0 (so x = 2).

Q4: If the roots of 2x² - 8x + k = 0 are equal, what is the value of k? Click to reveal

The value of k is 8.

Step 1: "Equal roots" means the discriminant (D) must be 0.
Step 2: The formula is D = b² - 4ac = 0.
Step 3: Identify a, b, and c from 2x² - 8x + k = 0.
a = 2, b = -8, c = k
Step 4: Substitute into the formula: (-8)² - 4(2)(k) = 0
Step 5: Calculate: 64 - 8k = 0
Step 6: Solve for k: 64 = 8k ➔ k = 8.

Q5: Find the sum and product of the roots for 3x² + 7x - 5 = 0 without solving. Click to reveal

Sum = -7/3 and Product = -5/3.

Step 1: Identify a, b, and c.
a = 3, b = 7, c = -5
Step 2: Use the Sum of Roots formula: α + β = -b / a
Sum: - (7) / 3 = -7/3
Step 3: Use the Product of Roots formula: α × β = c / a
Product: (-5) / 3 = -5/3

Quadratic Equations Class 10 Maths: Chapter 4 Guide (CBSE) | Solve, Visualize & Master Roots