Quadratic-2

Quadratic Equations

Quadratic Equations - Interactive Presentation

Introduction to Quadratic Equations

What is a Quadratic Equation?

A quadratic equation in variable x is an equation of the form:

ax² + bx + c = 0

where a, b, c are real numbers and a ≠ 0

Historical Background

Babylonians were the first to solve quadratic equations. Indian mathematicians like Brahmagupta (598-665 CE) and Sridharacharya (1025 CE) developed the quadratic formula. The method of completing the square was known to ancient civilizations.

Real-life Example

A charity trust wants to build a prayer hall with carpet area 300 m². If length is one more than twice the breadth, find the dimensions.

Let breadth = x metres

Then length = (2x + 1) metres

Area equation: x(2x + 1) = 300

This gives us: 2x² + x - 300 = 0

Standard Form and Identification

Standard Form

ax² + bx + c = 0 (a ≠ 0)

Terms are written in descending order of their degrees.

Examples of Quadratic Equations

2x² + x - 300 = 0
2x² - 3x + 1 = 0
4x - 3x² + 2 = 0 → -3x² + 4x + 2 = 0
x² + 3x + 1 = 0

Quick Practice

Question 1: Which of the following is NOT a quadratic equation?

a) x² - 4x + 4 = 0

b) 2x - 5 = 0

c) x² + 1 = 0

d) 3x² = 2x - 1

Answer: b) 2x - 5 = 0 (This is a linear equation, degree 1)

Identifying Quadratic Equations

Important Rules

Simplify the equation first
Check if the highest degree is 2
Coefficient of x² must not be zero
The equation might appear cubic but simplify to quadratic

Example: Check if (x + 2)³ = x³ - 4 is quadratic

Expand LHS: (x + 2)³ = x³ + 6x² + 12x + 8

Set up equation: x³ + 6x² + 12x + 8 = x³ - 4

Simplify: 6x² + 12x + 12 = 0

Divide by 6: x² + 2x + 2 = 0

Result: This IS a quadratic equation!

Practice Problems

Question: Is x(x + 1) + 8 = (x + 2)(x - 2) a quadratic equation?

Solution:
LHS: x² + x + 8
RHS: x² - 4
Equation becomes: x² + x + 8 = x² - 4
Simplifying: x + 12 = 0
Answer: No, it's a linear equation.

Roots and Solutions

Definition of Root

A real number α is called a root of the quadratic equation ax² + bx + c = 0 if:

aα² + bα + c = 0

We also say that x = α is a solution of the quadratic equation.

Important Facts

Zeroes of polynomial ax² + bx + c = roots of equation ax² + bx + c = 0
A quadratic equation can have at most 2 roots
Roots can be found by factorization, completing square, or quadratic formula

Example: Verify that x = 1 is a root of 2x² - 3x + 1 = 0

Substitute x = 1:

2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0 ✓

Therefore, x = 1 is indeed a root.

Solution by Factorization

Method

Express the quadratic as a product of two linear factors
Set each factor equal to zero
Solve the linear equations

Example: Solve 2x² - 5x + 3 = 0

Split middle term: -5x = -2x - 3x

Rewrite: 2x² - 2x - 3x + 3 = 0

Group: 2x(x - 1) - 3(x - 1) = 0

Factor: (2x - 3)(x - 1) = 0

Solve: 2x - 3 = 0 or x - 1 = 0

Roots: x = 3/2 or x = 1

Practice Problem

Question: Solve 6x² - x - 2 = 0 by factorization

Solution:
6x² - x - 2 = 6x² + 3x - 4x - 2
= 3x(2x + 1) - 2(2x + 1)
= (3x - 2)(2x + 1) = 0
Roots: x = 2/3 or x = -1/2

The Quadratic Formula

Universal Solution

For any quadratic equation ax² + bx + c = 0, the roots are given by:

x = (-b ± √(b² - 4ac)) / (2a)

This formula works for all quadratic equations where b² - 4ac ≥ 0

This formula was developed by Sridharacharya (1025 CE) using the method of completing the square and is sometimes called "Shreedhara's formula".

Example: Solve x² + 7x - 60 = 0 using quadratic formula

Identify: a = 1, b = 7, c = -60

Calculate discriminant: b² - 4ac = 49 - 4(1)(-60) = 289

Apply formula: x = (-7 ± √289) / 2

Simplify: x = (-7 ± 17) / 2

Roots: x = 5 or x = -12

Nature of Roots - The Discriminant

The Discriminant

The discriminant Δ = b² - 4ac determines the nature of roots:

Discriminant (Δ)	Nature of Roots	Example
Δ > 0	Two distinct real roots	x² - 5x + 6 = 0 Δ = 25 - 24 = 1 > 0
Δ = 0	Two equal real roots	x² - 6x + 9 = 0 Δ = 36 - 36 = 0
Δ < 0	No real roots	x² + x + 1 = 0 Δ = 1 - 4 = -3 < 0

Quick Check

Question: Find the nature of roots of 2x² - 4x + 3 = 0

Solution:
a = 2, b = -4, c = 3
Δ = (-4)² - 4(2)(3) = 16 - 24 = -8 < 0
Answer: No real roots

Real-Life Applications

Common Applications

Area and perimeter problems
Age-related problems
Motion and distance problems
Business and profit problems
Number problems

Example: Prayer Hall Problem (from Introduction)

Problem: Prayer hall with area 300 m², length = 2 × breadth + 1

Let breadth = x metres

Length = (2x + 1) metres

Area equation: x(2x + 1) = 300

Expand: 2x² + x = 300

Standard form: 2x² + x - 300 = 0

Factor: (x - 12)(2x + 25) = 0

Solutions: x = 12 or x = -12.5

Since x > 0, breadth = 12m, length = 25m

More Application Examples

Example 1: Consecutive Integers

Problem: Product of two consecutive positive integers is 306. Find them.

Let the integers be x and (x + 1)

Product equation: x(x + 1) = 306

Expand: x² + x = 306

Standard form: x² + x - 306 = 0

Using quadratic formula or factoring: x = 17 or x = -18

Since positive integers: x = 17

Answer: The integers are 17 and 18

Example 2: Age Problem

Problem: Rohan's mother is 26 years older. Product of their ages 3 years from now = 360.

Let Rohan's present age = x years

Mother's present age = (x + 26) years

After 3 years: Rohan = (x + 3), Mother = (x + 29)

Product equation: (x + 3)(x + 29) = 360

Expand: x² + 32x + 87 = 360

Standard form: x² + 32x - 273 = 0

Practice Problems - Set 1

1. Check whether (x - 2)(x + 1) = (x - 1)(x + 3) is a quadratic equation.

Solution:
LHS: x² - x - 2
RHS: x² + 2x - 3
Equation: x² - x - 2 = x² + 2x - 3
Simplifying: -3x + 1 = 0
Answer: No, it's linear.

2. Solve by factorization: x² - 3x - 10 = 0

Solution:
x² - 3x - 10 = (x - 5)(x + 2) = 0
Roots: x = 5 or x = -2

3. Find the discriminant of 3x² - 4√3x + 4 = 0 and nature of roots.

Solution:
a = 3, b = -4√3, c = 4
Δ = (-4√3)² - 4(3)(4) = 48 - 48 = 0
Answer: Two equal real roots

Practice Problems - Set 2

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

Solution:
Let base = x cm, altitude = (x - 7) cm
By Pythagoras: x² + (x - 7)² = 13²
x² + x² - 14x + 49 = 169
2x² - 14x - 120 = 0
x² - 7x - 60 = 0
(x - 12)(x + 5) = 0
x = 12 (taking positive value)
Answer: Base = 12 cm, Altitude = 5 cm

2. Find two numbers whose sum is 27 and product is 182.

Solution:
Let numbers be x and (27 - x)
Product: x(27 - x) = 182
27x - x² = 182
x² - 27x + 182 = 0
(x - 13)(x - 14) = 0
Answer: The numbers are 13 and 14

Summary and Key Points

Key Concepts

Quadratic Equation: ax² + bx + c = 0 (a ≠ 0)
Roots: Values of x that satisfy the equation
Maximum roots: 2 for any quadratic equation

Solution Methods

Factorization: Express as product of linear factors
Quadratic Formula: x = (-b ± √(b² - 4ac)) / (2a)
Completing the Square: Alternative algebraic method

Quadratic Formula: x = (-b ± √(b² - 4ac)) / (2a)

Discriminant (Δ = b² - 4ac)

Δ > 0: Two distinct real roots
Δ = 0: Two equal real roots
Δ < 0: No real roots

Applications

Area and geometric problems
Age-related word problems
Motion and time problems
Business and optimization problems
Number theory problems

Final Challenge Problems

Challenge 1: Is it possible to design a rectangular park with perimeter 80m and area 400m²? If so, find dimensions.

Solution:
Let length = l, breadth = b
Perimeter: 2(l + b) = 80 → l + b = 40 → l = 40 - b
Area: lb = 400 → b(40 - b) = 400
40b - b² = 400
b² - 40b + 400 = 0
Δ = 1600 - 1600 = 0
b = 20, l = 20
Answer: Yes, it's a square with sides 20m each.

Challenge 2: For what values of k does 2x² + kx + 3 = 0 have two equal roots?

Solution:
For equal roots, Δ = 0
b² - 4ac = 0
k² - 4(2)(3) = 0
k² - 24 = 0
k² = 24
Answer: k = ±2√6

Remember

Quadratic equations appear everywhere in mathematics and real life. Master the three solution methods and always check your answers by substitution!