QUADRATIC EQUATIONS

CONTENT

Construction of Quadratic Equations from Sum and Product of Roots.
Word Problem Leading to Quadratic Equations.

CONSTRUCTION OF QUADRATIC EQUATIONS FROM SUM AND PRODUCT OF ROOTS

We can find the sum and product of the roots directly from the coefficient in the equation. It is usual to call the roots of the equation α and β If the equation

ax² +bx + C = 0 ……………. I

has the roots α and β then it is equivalent to the equation

(x – α )( x – β ) = 0

x² – βx – βx + αβ = 0 ………… 2

Divide equation (i)by the coefficient of x²

ax²+ bx + C = 0 ………… 3

aaa

Comparing equations (2) and (3)

x² + b x + C = 0

x² – ( α +β)x + αβ = 0

then

α+ β= -b

and αβ = C

For any quadratic equation, ax² +bx + C = 0 with roots α and β

α + β = –b

αβ = C

Examples

If the roots of 3x² – 4x – 1 = 0 are αand β, find α + β and αβ
if α and βare the roots of the equation

3x² – 4x – 1 = 0 , find the value of

(a) α + β

β α

(b) α – β

Solutions

Since α + β = -b

Comparing the given equation 3x² – 4x – 1= 0 with the general form

ax² + bx + C = 0

a = 3, b = -4, C = 1.

Then

α + β = –b = -(-4)

a 3

= + 4 = +1 ¹/₃

αβ =C = –1 = -1

a 3 3

2.aα + β = α² +β²

β α αβ

= (α + β )² – 2αβ

αβ

Here, comparing the given equation, with the general equation,

a = 3, b = -4, C = – 1

from the solution of example 1 (since the given equation are the same ),

α + β = –b = – (-4) = +4

3 3

αβ = C = – 1

a 3

then

α + β = ( α+ β )² – 2 αβ

β α αβ

= (4/3 )^.2 – 2 ( – 1/3 )

= 16 ± 2

9 3

– 1

= 16 + 6 ÷ -1/3

22 x -3

9 1

= -22

or α + β = – 22 = – 7 1/3

β α 3

b) Since

(α-β) ² =α² + β^{– 2} α β

but

α² + β² = ( α + β)² -2 α β

:.(α- β)² = ( α+ β )² – 2αβ -2αβ

(α – β)² = (α + β )² – 4α β

:.( α – β) = √(α + β )² – 4αβ

( α – β) =√ (4/3 )² – 4 ( – ¹/3 )

= √ 16/9 +⁴/3

= √16 + 12

= √28 = √28

9 3

:. α – β = √28

Evaluation

If α and β are the roots of the equation

2x² – 11x + 5 = 0, find the value of

α – β
1 + 1

α + 1 β+ 1

α²+ β²

WORD PROBLEM LEADING TO QUADRATIC EQUATIONS

Examples

Find two numbers whose difference is 5 and whose product is 266.

Solution

Let the smaller number be x.

Then the smaller number be x+5.

Their product is x(x+5) .

Hence,

x(x+5) = 266

x²+5x- 266 = 0

(x-14)(x+19)=0

x=14 or x= -19

The other number is 14+5 or -19+5 i.e 19 or -14

:. The two numbers are 14 and 19 or -14 and -14.

Tina is 3 times older than her daughter. In four years time, the product of their ages will be 1536. How old are they now?

Solution

Let the daughter’s age be x.

Tina’s age = 3x

In four years’ time,

Daughter’s age = (x+4)years

Tina’s age = (3x+4)years

The product of their ages :

(x+4)(3x+4)= 1536

3x²+ 16x – 1520 = 0

(x-20)(3x+76) = =0

x=20 or x=-25.3

Since age cannot be negative, x=20years.

:. Daughter’s age = 20years.

Tina’s age = 20×3=60years.

Evaluation

Think of a number, square it, add 2 times the original number. The result is 80. Find the number.
The area of a square is 144cm² and one of its sides is (x+2)cm. Find x and then the side of the square.
Find two consecutive odd numbers whose product is 224.

GENERAL EVALUATION/REVISION QUESTIONS

The area of a rectangle is 60cm². The length is 11cm more than the width. Find the width.
A man is 37years old and his child is 8. How many years ago was the product of their ages 96?
If α and β are the roots of the equation 2x² – 9x+4=0, find
a) α + β (b) αβ (c) α – β (d) αβ/ α + β

WEEKEND ASSIGNMENT

If α and β are the roots of the equation 2x² + 9x+9=0:

Find the product of their roots. A. 4 B. 4.5 C. 5.5 B. -4.5
Find the sum of their roots. A. 4 B. 4.5 C. 5.5 B. -4.5
Find α²+β² A. 11.5 B. -11.25 C. 11.25 D. -11.5
Find αβ/ α + β A. 1 B.-1 C. 1.5 D. 4.5

THEORY

The base of a triangle is 3cm longer than its corresponding height. If the area is 44cm², find the length of its base.
Find the equation in the form ax²+bx+c=0 whose sum and products of roots are respectively:
a) 3,4 (b) -7/3 , 0 (c) 1.2,0.8