Infographics: Quadratic Equations

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Quadratic Equations: A Complete Guide
A quadratic equation is a second-degree polynomial equation involving 
the square of a variable. Understanding quadratic equations is 
fundamental to algebra, as they appear in numerous real-world 
applications from physics to finance. These equations can have up to two 
solutions, which may be real or complex numbers.
Standard Form
ax² + bx + c = 0
where a, b, c are 
real numbers and a 
b 0
Roots
Values of x that 
satisfy the equation
Can have two, one, 
or no real roots
Discriminant
D = b² 2 4ac
Determines nature 
of roots
What is a Root?
Let x = ³ where ³ is a real number. If ³ satisfies the quadratic equation 
ax² + bx + c = 0 such that a³² + b³ + c = 0, then ³ is the root of the 
equation.
Since quadratic polynomials have degree 2, quadratic equations can have 
two roots. The zeros of quadratic polynomial p(x) = ax² + bx + c are the 
same as the roots of the equation ax² + bx + c = 0.
Methods to Solve Quadratic Equations
1
Factorisation Method
Step 1: Write equation as ax² + bx + c = 0
Step 2: Split middle term bx as mx + nx where m + n = b and 
m × n = ac
Step 3: Factorise to get (x + p)(x + q) = 0
Step 4: Equate each factor to zero to find x
2
Quadratic Formula Method
x = 
2 a
2 b ± b 2 4 a c
2
Where D = b² 2 4ac is the discriminant
Simply substitute values of a, b, and c to find roots directly
Nature of Roots Based on Discriminant
D > 0
Two distinct real roots
Example: x² 2 7x + 
10 = 0
D = 49 2 40 = 9 > 0
Roots: x = 2, x = 5
D = 0
One repeated real root
Example: x² 2 6x + 9 
= 0
D = 36 2 36 = 0
Root: x = 3 (twice)
D < 0
No real roots
Example: x² + 2x + 5 
= 0
D = 4 2 20 = 216 < 0
Complex roots only
Solved Example: Factorisation Method
Problem
Solve: 4:3x² + 5x 2 2:3 = 0
01
Identify coefficients: a 
= 4:3, b = 5, c = 22:3
Product ac = 4:3 × 
(22:3) = 224
02
Find factors of 224 
where sum equals 5
Factors: 8 and 23 
(since 8 2 3 = 5)
03
Split middle term: 
4:3x² + 8x 2 3x 2 2:3 
= 0
04
Factorise: 4x(:3x + 2) 2 :3(:3x + 
2) = 0
Therefore: (4x 2 :3)(:3x + 2) = 0
05
Solve: 4x 2 :3 = 0 or :3x + 2 = 0
Roots: x = :3/4 and x = 22/:3
Solved Example: Quadratic Formula
P r obl em
Find roots of x² 2 7x + 10 = 0 using the quadratic fo rmula
Given Information
a = 1
b = 27
c = 10
Apply Formula
x = 
2 a
2 b ± b 2 4 a c
2
Step-by-Step Solution
Substitute values:
x = 
2(1)
2(27) ± (27) 2 4(1)(10)
2
x = 
2
7 ± 49 2 40
x = =
2
7 ± 9
 
2
7 ± 3
Two solutions:
x = (7 + 3)/2 = 10/2 = 5
x = (7 2 3)/2 = 4/2 = 2
Real-World Application
Geometry Problem
The altitude of a right-
angled triangle is 7 cm 
less than its base. If 
the hypotenuse is 13 
cm, find the other two 
sides.
Solution Approach
Let base = x cm, then 
altitude = (x 2 7) cm
Using Pythagoras 
theorem: 13² = x² + (x 
2 7)²
This gives: x² 2 7x 2 
60 = 0
Final Answer
Factorising: (x + 5)(x 2 
12) = 0
x = 12 cm (base cannot 
be negative)
Base = 12 cm, Altitude 
= 5 cm
Key Examples to Remember
Checking Quadratic 
Form
(x + 1)² = 2(x 2 3)
Expanding: x² + 2x 
+ 1 = 2x 2 6
Simplifying: x² + 7 
= 0
7 This is quadratic 
(b = 0)
Not Quadratic
x(x + 1)(x + 8) = (x 
+ 2)(x 2 2)
Expanding: x³ + 
9x² + 8x = x² 2 4
Simplifying: x³ + 
8x² + 8x + 4 = 0
7 This is not 
quadratic (degree 
3)
Finding Equal Roots
For px(x 2 2) + 9 = 
0 to have equal 
roots
Discriminant D = 0
4p² 2 36p = 0
p = 9 (p b 0)
Quick Reference: Discriminant Values
36
Discriminant for x² 2 
7x + 10 = 0
Leads to two distinct 
real roots
25
Lucky Number 
Example
Discriminant of x² + 9x 
+ 14 = 0
0
Equal Roots Condition
When b² = 4ac exactly
Practice Problems Summary
Identify
Determine if given equations are quadratic
Solve
Use factorisation or formula method
Analyse
Find discriminant and nature of roots
Apply
Solve real-world problems
Remember: 
The discriminant D = b² 2 4ac is the key to understanding quadratic equations. It tells you 
everything about the nature of roots before you even solve the equation!