The document presents an exploration of quadratic equations in one variable, detailing their definition, types, forms, and solutions, including methods such as factoring and completing the square. It highlights the differences between quadratic and linear equations and provides practical scenarios involving quadratic equations. Additionally, examples demonstrate solving quadratic equations using various methods and the derivation of the quadratic formula.

1. PRESENTED TO: SIR ALI RAZA
SUBJECT:EXPLORING QUANTITATIVE SKILLS(MATHS1126)
PRESENTED BY:GROUP 2
QUADRATIC EQUATION IN ONE VARIABLE
&
PRACTICAL SNERIO INVOLVING EXPRESSIONS
GROUP PARTICIPANTS:
RABIA SANI, MEHAK FATIMA
KISHAAF MAQBOOL, AREESHA RAZA

2. MAIN POINTS TO DISCUSS
What is quadratic equation in one variable?​
How it differs from linear equation?​
Forms of quadratic equation ( standard & pure).​
Solution of quadratic equation​
Types of solution​
Practical snerio involving expressions

3. QUADRATIC EQUATION
DEFINITION: An equation which contains the square of variable quantity, but no
high power , is called a quadratic equation OR an equation of the second degree.
DIFFERENCE BETWEEN LINEAR AND QUADRATIC EQUATION:
It differes from linear in such a way that :
 The linear euation is with highest power 1 of its variable wherease quadratic
equation is with variable of second degree.
 There is single answer( 1 root) in solution set of linear equation ,however, the
solution set of quadratic equation contains two possible values(2 roots).
FORMS OF QUADRATIC EQUATION:
I. Standard form: i.e. ax2 + bx + c = 0, (where a, b, and c are constants)(a is not 0)
II. Pure form: i.e. ax2+c=0, (where b is 0)

4. SOLUTION OF QUADRATIC EQUATION
 Rewrite in standard
form i.e.ax2 + bx + c =
0
(splitting)
 Find factors of (c)(a)
that add to “b”
(expanding)
 Grouping & solution
(grouping)
FACTORIZATION
 Rewriting in formate
i.e.ax2 + bx= -c
 Complete the square:
square=(b/2)2
add on bothsides
 Write in whole square
and solve by taking sq
roots
COMPLETING
SQUARE
METHOD
 Rewrite in standard form
 Put values in quadratic
formula:
x=(-b±√(b²-4ac))/(2a)
 Solve
USING
QUADRATIC
FORMULA
01 02 03

5. FACTORIZATION METHOD
RULES TO SOLVE IT:
 Rewrite in standard form i.e.ax2 + bx + c = 0(splitting)
 Find factors of (c)(a) that add to “b” (expanding)
 Grouping & solution(grouping)
EXAMPLE NO. 1
Solve the equation, x2−3x−10=0
Factor the left side: (x−5)(x+2)=0
Set each factor to zero: x−5=0 or x+2=0
Solve each equation: x=5 or x=−2
The solution set is ss= {5,−2}
.

6. COMPLETING SQUARE METHOD
○ RULES TO SOLVE:
○ Rewriting in formate i.e.ax2 + bx= -c
○ Complete the square:
○ square=(b/2)2
○ add on bothsides
○ Write in whole squareand solve by taking sq roots
○ EXAMPLE no.2 x2 + 4x – 5 = 0
b = 4, c = -5
(x + b/2)2 = -(c – b2/4)
So, [x + (4/2)]2 = -[-5 – (42/4)]
(x + 2)2 = 5 + 4
⇒ (x + 2)2 = 9
⇒ (x + 2) = ±√9
⇒ (x + 2) = ± 3
⇒ x + 2 = 3, x + 2 = -3
⇒ x = 1 , -5
ss={1, -5}

7. DERIVATION OF QUADRATIC FORMULA
1. The quadratic formula, which is used to find the solutions of a quadratic equation in the form
ax^2 + bx + c = 0, can be derived by completing the square. Here's a step-by-step derivation:
2. Starting with the equation: ax^2 + bx + c = 0
3. Divide both sides of the equation by 'a' x^2 + (b/a)x + c/a = 0
4. Move the constant term 'c/a' to the right side x^2 + (b/a)x = -c/a
5. To complete the square on the left side of the
equation, you need to add and subtract (b/2a)^2.
This is because (b/2a)^2 is the term that, will make
the left side a perfect square trinomial. x^2 + (b/a)x + (b/2a)^2 = -c/a
+(b/2a)^2
1. left side can be factored as a perfect square: (x + b/2a)^2 = -c/a + (b/2a)^2
2. Take the square root of both sides: x + b/2a = ±√(-c/a + (b/2a)^2)
3. Solve for 'x' by isolating it on the left side: x = -b/2a ± √(-c/a + (b/2a)^2)
4. Simplify the right side:
5. x = (-b ± √(b^2 - 4ac)) / (2a)

8. USING QUADRATIC FORMULA
○ RULES TO SOLVE IT:
○ Rewrite in standard form
○ Put values in quadratic formula:
○ x=(-b±√(b²-4ac))/(2a)
○ Solve
○ Example no 3; 2x2 - 5x + 3 = 0 where a=2,b=-5,c=3
putting values in formula, x=(-b±√(b²-4ac))/(2a)
x = (-(-5) ± √((-5)^2 - 4(2)(3))) / (2(2))
x = (5 ± √(25 - 24)) / 4
x = (5 ± √1) / 4
Calculate the two possible solutions:
x₁ = (5 + 1) / 4 = 6 / 4 = 3/2
x₂ = (5 - 1) / 4 = 4 / 4 = 1
ss={1, 3/2}

9. PRACTICAL SNERIO
INVOLVING
EXPRESSIONS

10. EXAMPLE NO. 1
You are asked to built a rectangular field with area=7000m2 .The length of the fieldshould be 30
mmore than its width.How will you build the field?
Solution:
Let suppose that,width= x
then according to the given condition; length =x+30
for area of a rectangle,
A=(l)(b)
7000=(x+30)(x)
7000=x2+30x l=x+30m
x2 +30x-7000=0
A=7000m2

11. Three less than a certain number multiplied by 9 less than
twice the number is 104. Find the number.
SOLUTION
Let the required number= x. Then 3 less then that =x-3
and nine less than twice the number=2x-9
According to the given condition:
(x-3)(2x-9)=104
2x2 -15x+27=104
2x2-15x-77=0
EXAMPLE NO.2

12. The length of a base and the corresponding height of a triangle are (x+3) and (2x-5)
respectively.given the area of the triangle is 20cm2 .Find the value of x.
We are given that; Area=20cm2 Base=x+3 Height=2x-5
as Area of a triangle =1/2(b)(h)
putting values we get:
20=1/2(x+3)(2x-5)=(2x2-5x+6x-15)/2
20=2x2+x-15/2
20 X 2=2x2+x-15
0=2x2+x-40-15
2x2+x-25=0
b=x+3
EXAMPLE NO.3
h=(2x-5)