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
Taking square root on both sides ,
 ⇒ x + b = ± √( b² - 4ac ) 2a 2a
 ∴ x = - b ± √( b² - 4ac ) 2a
Where, a = coefficient of x2
b = coefficient of x
c = constant term

### 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
Where, α and β are symbol of the roots of quadratic equation.

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
 ⇒ α = ( -8 + √63 ) 2
 ∴ β = -b - √D = -8 - √63 2a 2 × 1
 ⇒ β = ( -8 - √63 ) 2

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

 ∴ α = β = -b 2a

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

 ∴ c = b2 4a

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 :-

 α = - b + √( b² - 4ac ) 2a
 β = - b - √( b² - 4ac ) 2a

Sum of roots :-

 α + β = - b + √( b² - 4ac ) + - b - √( b² - 4ac ) 2a 2a
 α + β = -2b 2a
 ∴ α + β = -b = -coefficient of x a coefficient of x2
Product of roots :-

 α × β = - b + √( b² - 4ac ) × - b - √( b² - 4ac ) 2a 2a
 α × β = (-b)2 - { √( b² - 4ac ) }2 4a2
 αβ = b2 - b² + 4ac 4a2
 ∴ α.β = c = constamnt term a coefficient of x2

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

 ∴ Sum of roots = -b = -6 = -6 a 1
 ∴ Product of roots = c = 9 = 9 a 1

### 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

 ∴ Sum of roots = α + β = -b ........( 1 ) a
 ∴ Product of roots = α.β = c ........( 2 ) a
Given equation,
ax2 + bx + c = 0
Dividing on both sides by a, we get
 ⇒ x2 + b x + c = 0 a a
 ⇒ x2 - -b x + c = 0 a a
From eq. (1) and (2), we get
x2 - ( sum of the roots ) x + Product of roots = 0
x2 - ( α + β ) x + α . β = 0

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

 ∴ α + β = -b = - ( -2 ) = 2 a 5 5
 ∴ α.β = c = 8 a 5
 Now, 1 + 1 = α + β = 2 / 5 α β α.β 8 / 5
 1 + 1 = 2 = 1 α β 8 4
 1 × 1 = 1 = 1 = 5 α β α.β 8 / 5 8
Required equation, x2 - ( α + β ) x + α . β = 0
 ⇒ x2 - 2 x + 8 = 0 5 5