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Mathematics Notes

Form 2 Mathematics – QUADRATIC EQUATION

msomimaktaba, November 13, 2018August 17, 2024

QUADRATIC EQUATION


QUADRATIC EQUATION

Is any equation which can be written in the form of ax2 + bx + c=0 where a ≠ 0 and a, b and c are real numbers.

SOLVING QUADRATIC EQUATION

i) BY FACTORIZATION

Example 1

solve x2 + 3x – 10 = 0

Solution:

x2 + 3x – 10 = 0

(x2 – 2x) + 5 (x – 2) = 0

x (x – 2) + 5 (x – 2) = 0

(x + 5) (x – 2) = 0

Now x + 5 = 0 or x – 2 = 0

x = -5 or x = 2

x = -5 or 2

 

Example 2

Solve for x

i) 2x2 + 9x + 10 = 0

Solution:

Sum = 9

Product = 2 x 10 = 20

20 = 1 x 20

= 2 x 10

= 4 x 5

(2x2 +4x) + (5x + 10) = 0

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

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

Now,

2x + 5 = or x + 2 = 0

x= -2.5 or -2

 

ii) 2x2 – 12x = 0

Solution:

2x(x – 6) = 0

2x = 0 or x – 6 = 0

X = 0, or x = 6

X = 0 or 6

 

iii) x2 – 16 = 0

Solution:

x2 – 16 = 0

(x2) – (4)2 = 0

(x + 4) (x – 4) = 0

Now, x + 4 = 0 or x – 4 = 0

x = -4 or x = 4

 

EXERCISE

1. Solve for x from

X2 – 7x + 12=0

Solution:

x2 – 3x – 4x + 12 = 0

(x2 – 3x) – (4x – 12) = 0

x(x – 3) – 4(x – 3) = 0

(x – 4) (x – 3) = 0

Now, x – 4 = 0 or x – 3 = 0

x= 4 or x = 3

 

ii) 4x2 – 20x + 25 = 0

Solution:

4x2 – 10x – 10x – 25 = 0

(4x2 – 10x) – (10x – 25) = 0

2x(2x – 5) – 5(2x – 5) = 0

(2x – 5) (2x – 5) = 0

Now, 2x – 5 = 0 or 2x – 5 = 0

x =

iii) 4x2 – 1 = 0

Solution:

4x2 – 1 = 0

22x2 – 1 = 0

(2x)2 – (1)2 = 0

(2x + 1) (2x – 1) = 0

Now, 2x + 1 = 0, or 2x – 1 = 0

X = or x =

iv) (x – 1)2 – 81 = 0

Solution:

(x – 1)2 – 92 = 0

(x – 1 – 9)(x – 1 + 9) = 0

Now, x – 1 – 9 = 0, or x – 1 + 9

x – 10 = 0, x + 8 = 0

x = 10 or x = – 8

v) 2x2 = 10x

Solution:

2x2 – 10x = 0

2x(x – 5) = 0

2x = 0 or x – 5 = 0

x = 0, or x = 5

 

SOLVING BY COMPLETING THE SQUARE

Example 1

Solve i) 2x2 + 8x – 24 = 0

Solution:

x2 + 4x – 12 = 0

x2 + 4x = 12

x2 + 2x + 2x + 4 = 12 + 4

(x2 + 2x) + (2x + 4) = 16

x(x+ 2) + 2(x +2) = 16

(x +2) (x +2) = 16

(x +2)2 = 16

=

x + 2 = 4

X = 4 2

X = 2 or x =6

X = 2 or 6

 

ii) x2 + 5x – 14 = 0

solution:

x2 + 5x = 14

(x2 + ) + (+ ) = 14 +

x(x + ) +(x + ) =

(x + )(x + ) =

=

x + =

x= or x =

x = 2 or 7

iii) 3x2 – 7x– 6 = 0

Solution:

x2 – – 2 = 0

x2 – = 2

x2 – – + = 2 +

(x2 – ) –( – )=

x(x – ) – (x– )=

(x– )(x– )=

=

x – =

Now,

x – = , x – =

x = 3 or x =

iv) x2 – 5x + 2 = 0

x2 – 5x = -2

x2 – – + = -2 +

x(x – –) – (x – –) =

(x – )2 =

 

=

x – = ±

x = ±

 

x = or

GENERAL FORMULA

1. Solve ax2 + bx + c = 0

Solution:

x2 + + = 0

x2 + =

x2 + + + = +

(x2 + ) +( + ) =

x(x + ) + ( x + ) =

(x + )2 =

=

x + =

x =

Generally,

Example 1.

Solve for x by using generally formula

i) 6x2 + 11x + 3 = 0

Solution: a = 6, b = 11, c = 3

From the general equation,

and

and

 

ii) 5x2 – 6x – 1 = 0

Solution:

a= 5, b = -6, c =1

From the general equation,

and

and

 

 

iii) 0 = 400 + 20t – t2

solution:

t2 20t 400 = 0

a = 1, b = -20, c = -400

From the general equation

 

GRAPHICAL SOLUTION OF QUADRATIC EQUATION

– The general quadratic equation ax2 + bx + c =0 can be solved graphically

– First draw the graph by setting ax2 + bx + c = y and then

Drawing graphs

Example 1

Draw the graph of the following equation

i) y = x2 – 3

ii) y = 2 – x2

iii) y = x2 + x – 1

Solution:

i) y = x2 – 3

TABLE VALUE


x-3-2-10123
y61-2-3-216

 

ii) y = 2 – x2

 


 

 

iii) y = x2 + x – 1

Solution:

APPLICATION OF GRAPHS IN SOLVING QUADRATIC EQUATION

a) Solve graphically the equation x2 – x – 6 = 0

b) Use the graph in a to solve the equation

x2 – x – 2 = 0

Solution:
x2 – x – 6 = 0

Let y=x2 – x – 6…………………….(i) Then
y=0………………….(ii)


(b)From x2 – x – 6 = 0

Then
x2 – x – 2 = 0 can be written as
x2 – x –2-4 = 0-4
x2 – x – 6 = -4 But y=x2 – x – 6
∴y=-4

∴x=-1 or x=2

More examples

1. A man is 4 times as old as his son. In 4 years the product of their ages will be 520.

Find the sons present age

Now

(x + 4) (4x + 4) = 520

4x2 + 4x + 16x + 16 = 520

+ – =

x2 + 5x – 126 = 0

a=1, b = 5, c = -126

From the general equation,

and

and

 

x = 9 or -14.

The present age of the son is 9

2.Find the consecutive numbers such that the sum of their squares is equal to 145

Solution:

Let x be the first number and x + 1 be the second number

Sum of x2 + (x + 1)2 = 145 their squares

Now, x2 + (x + 1)2 = 145

x2 + x2 + 2x + 1 = 145

2 x2 + 2x – 144 = 0

Divide by 2 both sides , then x2 + x – 72

 

a =1, b =1, c = -72

From the general equation,

and

and

x = 8 and 9

or x = -9 and -8

The two consecutive numbers are 8 and 9 or -9 and -8.

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