An expression in the form of ax² + bx + c, where a,b,c are real number and a ≠ 0, is called a quadratic polynomial.
For example- 3x² + 5x + 7, √2x² + 5x + 2 etc.

When a quadratic expression equated to zero is called quadratic equations. Hence an equation in the form of ax² + bx + c = 0,
where a and b are coefficients of x² and x, respectively and c is a constant.
For example- 3x2 + 5x + 7 = 0, √2x2 + 5x + 2 = 0 etc.

Example: Which of the following is not a quadratic equation?
(a) x² + 2x + 2 ( 3 - x ) = 0
(b) x ( x + 1 ) + 1 = ( x - 2 ) ( x - 5 )
(c) ( 2x - 1 ) ( x - 3 ) = ( x + 5 ) ( x - 1 )
(d) x³ + 4x² - x + 1 = ( x - 2 )³
Solution:- (b)
x ( x + 1 ) + 1 = ( x - 2 ) ( x - 5 )
or, x² + x + 1 = x ( x - 5 ) - 2 ( x - 5 )
or, x² + x + 1 = x² - 5x - 2x + 10
or, x² + x + 1 = x² - 7x + 10
or, x + 1 = -7x + 10
or, 8x - 9 = 0, which is not a quadratic equation.

Derivation of Quadratic Formula

we know that, ax² + bx + c = 0 is the standard form of quadratic equation
Dividing on both sides by 'a'

⇒ x2 +			b		x +		c		= 0
			a				a

⇒ x2 +			b		x = -		c
			a				a

Adding on both sides		b		2
		2a

⇒ x2 +

b

x +

b

2

= -

c

+

b

2

a

2a

a

2a

⇒ x2 + 2 × x ×

b

+

b

2

= -

c

+

b

2

2a

2a

a

2a

⇒

x +

b

2

=

b

2

-

c

2a

2a

a

⇒

x +

b

2

=

b2

-

c

2a

4a2

a

⇒		x +		b		2	=		b2 - 4ac
				2a					4a2

⇒		x +		b		= ±	√( b² - 4ac )
				2a			2a

Discriminant (D)

For the quadratic equation ax² + bx + c = 0,
D = b² - 4ac
Where, D is the symbol of discriminant and a and b is the coefficients of x² and x and c is a constant.
Example: If x² - 3x + 1 = 0, find the value of D.
Solution:- a = 1 , b = -3 and c = 1
D = b² - 4ac
= (-3)² - 4 × 1 × 1
= 9 - 4
= 5

Roots of a quadratic Equation :-

1.If D > 0, then quadratic equation has two distinct real roots given by

∴ α =		-b + √D
		2a

and β =		-b - √D
		2a

Example: Find the roots of the given equation, x² + 8x + 4 = 0
Solution: given equation,
x² + 8x + 4 = 0
Comparing with ax² + bx + c = 0 , we get a = 1 , b = 8 and c = 4
then, D = b² - 4ac
= 8² - 4 × 1 × 4
= 64 - 1
∴ D = 63

∴ α =		-b + √D		=		-8 + √63
		2a				2 × 1

∴ β =		-b - √D		=		-8 - √63
		2a				2 × 1

2. If D = 0, then quadratic equation has two equal roots given by

Example: If ax² + bx + c = 0 has equal roots, then find the value of c.
Solution: ax² + bx + c = 0
D = 0
b² - 4ac = 0
b² = 4ac

3. If D < 0, then quadratic equation has no real roots.
Example: 3x² + 7x + 5 = 0
Solution: 3x² + 7x + 5 = 0
Comparing with ax² + bx + c = 0, we get a = 3, b = 7 and c = 5
D = b² - 4ac
= 7² - ( 4 × 3 × 5 )
= 49 - 60
= -11
∴ No real roots.

Sum and Product of Roots :-

Sum of roots :-

Example: Find the sum and product of the roots in the equation x² + 6x + 9 = 0
Solution: given equation x² + 6x + 9 = 0
Here, a = 1, b = 6 and c = 9

Important points

1.A quadratic equation has two and only two roots.

2. A quadratic equation can not have more than two different roots.

3. If D is a perfect square then roots are rational otherwise irrational.

4. If a + √b is one root of a quadratic equation, then its conjugate a -√b must be the other root.

Formation of an Equation with given roots

If α and β are the roots of a quadratic equation ax² + bx + c = 0, then the quadratic equation will be

⇒ x2 +			b		x +		c		= 0
			a				a

⇒ x2 -			-b		x +		c		= 0
			a				a

Example: If α and β are the roots of the equation ax² + bx + c = 0, then find the quadratic equation whose root are ( 2 + √3 ) and ( 2 - √3 ).
Solution: given roots are 2 + √3 and 2 - √3
Sum of roots = 2 + √3 + 2 - √3 = 4
Product of roots = ( 2 + √3 ) × ( 2 - √3 ) = (2)² - ( √3 )²
= 4 - 3 = 1
Required equation,
x² - ( α + β ) x + α . β = 0
x² - 4x + 1 = 0

Example: If α and β are the roots of the equation 5x² - 2x + 8 = 0, then find the quadratic equation whose roots are 1/α and 1/β.
Solution: given equation,
5x² - 2x + 8 = 0
a = 5 , b = -2 and c = 8

⇒ x2 -			2		x +		8		= 0
			5				5

Reference