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5: ROOTS OF A QUADRATIC EQUATION The general form of a quadratic equation We have grown accustomed to recognising a quadratic equation in the form 𝑎𝑥# + 𝑏𝑥 + 𝑐 = 0. In this section, we will be introduced to a new format for such a quadratic equation. This format would express the quadratic in the form of its roots. It is a convenient form to know and it allows us the flexibility to switch from this form to the standard form. Roots of a quadratic equation (∝ 𝒂𝒏𝒅𝜷) A quadratic equation in x is of the general form

, where a, b and c are constants. If we divide each term by a, then the quadratic equation can be expressed in an equivalent form with the coefficient of x2 is equal to one as shown below.

Now consider ∝ and 𝛽 as the roots of the quadratic. We can now rewrite the quadratic in the form:

.

By expanding we get, . (2)

Equation (2) is an equivalent form of equation (1). In fact, any quadratic equation, in x, can always be expressed in the form of its roots. We can replace ( ) by the ‘sum of the roots’ and

by the ‘product of the roots’, to obtain the following form for a quadratic equation.

Sum and product of the roots of a quadratic equation Equations (1) and (2) above are two equivalent forms of a quadratic equation. Equating both forms we get:

When we equate coefficients, the following is obtained:

and .

We can now make a general statement about the roots of a quadratic. For the quadratic equation

,

the sum of the roots and

the product of the roots .

Example 1 If and are the roots of the quadratic

equation , determine (i) the sum of the roots and (ii) the product of the roots.

Solution In the quadratic equation a =1, b = −3 and c = 2.

(i) The sum of the roots,

(ii) The product of the roots,

.

Example 2 The quadratic equation has roots and . a) Obtain the equation whose roots are and

.

b) Obtain the equation whose roots are and .

Solution If the equation has roots and , then 𝑎 = 1, 𝑏 = −4and𝑐 = 3. Hence, and To obtain an equation whose roots are 𝛼 + 1 and

2 0ax bx c+ + =

2

2

0

0 (1)

ax bx cb cx xa a

+ + =

+ + =

( )( ) 0x xa b- - =

( )2 0x xa b ab- + + =

a b+ab

( )2 sum of roots product of roots 0x x- + =

( )2 2b cx x x xa a

a b ab+ + = - + +

ba

a b+ = -ca

ab =

2 0ax bx c+ + =

ba

= -

ca

=

a b2 3 2 0x x- + =

2 3 2 0x x- + =

ba

a b+ = -

( )33

1a b

- -+ = =

ca

ab =

2 21

ab = =

2 4 3 0x x- + = ab

1a +1b +

2a 2b

2 4 3 0x x- + = a b

( ) 4a b+ = 3ab =

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𝛽 + 1, we can substitute these roots in the following equation:

This is the required equation. Part b) To obtain an equation whose roots are 𝛼# and 𝛽#, we substitute these roots in:

[Recall: ]

This is the required equation. Example 3 Given that has real roots which differ by 4, determine

i. the value of each root ii. the value of k.

Solution If we let α be the smaller real root, then the other will be (α + 4). Then the sum of the roots is : α + (α + 4) = 2α +4 The product of the roots is α(α+4). From the given equation , The sum of the roots is: -(k-5) The product of the roots is: -k Equating coefficients, we have: Sum of roots 2α +4 = -(k-5) 2α +4 = - k +5 k = 1- 2α (1)

Product of roots α(α+4) = -k 𝑘 = −𝛼(𝛼 + 4) (2)

Equating equations (1) and (2) to eliminate k, we have:

−∝#− 4 ∝= 1 − 2 ∝ ∝#+ 2 ∝ +1 = 0 (∝ +1)(∝ +1) = 0

∝= -1 The value of k: k = 1- 2α = 1-2(-1)= 3

Roots are -1 and -1 + 4 The roots are -1 and 3. Alternative Method If we let α be the smaller real root, then the other will be (α + 4). Hence the quadratic equation may be expressed as

Equating coefficient of x, we obtain

Equating constant terms, we obtain

Roots are -1 and -1 + 4 The roots are -1 and 3.

( )2 sum of roots product of roots 0x x- + =

( ) ( )

2

2

2

2

[( 1) ( 1)] [( 1) ( 1)] 0[( ) 2)] [ ( ) 1] 0

4 2 3 4 1 0

6 8 0

x xx x

x x

x x

a b a b

a b ab a b

- + + + + + + =

- + + + + + + =

- + + + + =

- + =

( )2 sum of roots product of roots 0x x- + =

( ) ( )( ) ( )

2 2 2 2 2

2 2 2 2 2

0

0

x x

x x

a b a b

a b a b

- + + ´ =

- + + =

( )2 2 2 2a b a b ab+ = + +

( )( ) ( )

( ) ( ){ } ( )

2 22

2 22

2

2 0

4 2 3 3 0

10 9 0

x x

x x

x x

a b ab ab- + - + =

- - + =

- + =

( )2 5 0x k x k+ - - =

( )2 5 0x k x k+ - - =

\

( ) ( )( )( )( )

2 2

4 0

4 0

4 4 0

x a x

x a x

x x ax x

a

a

a a a

- - + =

- - - =

- - + - + =

( ) ( )2 22 4 4 0x xa a a+ - - + + =

2 4 5ka- - = -

2 4 51 212

kk

k

aa

a

- - = -= -

-\ =

( )

2

2

2

2

2

2

4

1 142 2

1 2 2 2 04

1 2 8 8 4 06 9 0

3 03

k

k k k

k k k k

k k k kk k

kk

a a+ = -

- -æ ö æ ö\ + = -ç ÷ ç ÷è ø è ø

- ++ - + =

- + + - + =

- + =

- =

=1 3When, 3, 1

2k a -= = = -

\

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