QUADRATIC EQUATION

An equation of the type ax2+bx+c=0 is called a quadratic equation. [The constants a, b and c can

have any numerical values but a ≠ 0.]

SOLUTIONS BY FACTORS

Type 1: ax2+bx+c=0

Example; 2x2 – 7x – 15 = 0

Factorize 2x2 – 7x – 15 = 0 to get (2 x+3 ) ( x−5 )=0

Since the product of two factors (2 x+3 )and (x – 5) is zero then,

either (2x +3) =0 or (x – 5) =0

2 x=−3 or ( x−5 )=0

x=−32

or x=5

∴ The solutions are x=−32

,∧x=5

EXERCISE

Solve the following quadratic equation.

1) x2 + 7x + 12 = 02) x2 + x – 6 = 03) x2 – 5x + 6 = 0 4) 2x2 – 3x – 2 = 05) 3x2 + 10x – 8 = 06) 2x2 + 7x – 15 = 07) 6a2 – a – 1 = 08) 4a2 – 3a – 10 = 0

Type 2: ax2+bx=0

Example; 2x2 – 8x = 0

Factorize 2x2 – 8x = 0 to get 2 x (x−4)=0

Since the product of two factors 2x and (x – 4) is zero then,

either 2x = 0 or (x – 4) =0

x=0 or x=4

∴ The solutions are x=0 ,∧x=4

EXERCISE

Solve the following quadratic equation.

1) x2 – 3x = 0

2) x2 + 7x = 0

3) 2x2 – 2x = 0

4) x2 = x4

5)x5

– x2 = 0

Type 3: ax2−c=0

Example: x2−16=0

Solve the equation x2=16

Take the square root of both sides

x=±√16x=± 4

[Remember to write the + and – sign]

OR

Factorize using difference of two square

x2−16=0

( x−4 )(x+4)=0

Since the product of two factors ( x−4 )and (x + 4) is zero then,

either (x + 4) =0 or (x – 4) =0

x=−4 or x=4

EXERCISE

Solve the following quadratic equation.

1) 9x2 – 4 = 0

2) 4x2 – 1

36 = 0

3) x3 – x = 04) 144x2 – 225 = 05) 8x3 – 50x = 06) 27p3 – 48p = 0

SOLVING BY USING THE FORMULA

A question where the algebraic equation can not be factorized and the answer is required in

decimals, then it can be solved using the formula

i.e. x=−b±√b2−4 ac2a

Example: 3 x2−8 x+2=0

Step 1:

Comparing 3 x2−8 x+2=0 with ax2+bx+c=0, we have a=3 , b=−8 and c=2

Step 2:

Substitute these values in the formula

x=−b±√b2−4 ac2a

¿−(−8)±√(−8)2−4 ×3 ×2

2× 3

= 8 ±√64−246

¿ 8 ±√406

[The value in the root must be positive]

¿(8+√40)

6 or ¿

(8−√40)6

x=2.39 or = 0.28

EXERCISE

Solve the following quadratic equation giving your answer to 2 d.p.

1) 2x2 + 11x + 5 = 0

2) 3x2 – 7x – 20 = 0

3) 2x2 + 6x – 1 = 0

4) 3y2 – 2y – 5 = 0

5) 2 – x – 6x2 = 0

6) 12 – 5x2 – 11x = 0

7) 5x2 – 5x + 1 = 0

8) 2x2 – 7x – 15 = 0

WORD PROBLEM

1. (a) (i) The cost of a book is $x.Write down an expression in terms of x for the number of these books which are bought for $40. [1]

(ii) The cost of each book is increased by $2.The number of books which are bought for $40 is now one less than before.Write down an equation in x and show that it simplifies to x2 + 2x – 80 = 0 [4]

(iii) Solve the equation x2 + 2x – 80 = 0 [2](iv) Find the original cost of one book. [1]

2.

(a) When the area of triangle ABC is 48 cm2,(i) show that x2 + 4x − 96 = 0, [2](ii) solve the equation x2 + 4x − 96 = 0, [2](iii) write down the length of AB. [1]

(b) When tan y = 16

, find the value of x. [2]

(c) When the length of AC is 9 cm,(i) show that 2x2 + 8x − 65 = 0, [2](ii) solve the equation 2x2 + 8x − 65 = 0,(Show your working and give your answers correct to

2 decimal places.) [4](iii) calculate the perimeter of triangle ABC. [1]

3. (a) On 1st January 2000, Ashraf was x years old.Bukki was 5 years older than Ashraf and Claude was twice as old as Ashraf.

(i) Write down in terms of x, the ages of Bukki and Claude on 1st January 2000. [2](ii) Write down in terms of x, the ages of Ashraf, Bukki and Claude on 1st January 2002. [1](iii) The product of Claude’s age and Ashraf’s age on 1st January 2002 is the same as the

square of Bukki’s age on 1st January 2000.Write down an equation in x and show that it simplifies to x2 – 4x – 21 = 0. [4]

(iv) Solve the equation x2 – 4x – 21 = 0. [2](v) How old was Claude on 1st January 2002? [1]

(b) Claude’s height, h metres, is one of the solutions of h2 + 8h – 17 = 0.(i) Solve the equation h2 + 8h – 17 = 0.

Show all your working and give your answers correct to 2 decimal places. [4](ii) Write down Claude’s height, to the nearest centimetre. [1]

4.

The diagram shows two rectangles ABCD and PQRS.AB = (2x + 5) cm, AD = (x + 3) cm, PQ = (x + 4) cm and PS = x cm. (a) For one value of x, the area of rectangle ABCD is 59 cm2 more than the area of rectangle PQRS.

A

B

C

NOT TO SCALE

(i) Show that x2 + 7x − 44 = 0. [3](ii) Factorize x2 + 7x − 44. [2](iii) Solve the equation x2 + 7x − 44 = 0. [1](iv) Calculate the size of angle DBA. [2]

(b) For a different value of x, the rectangles ABCD and PQRS are similar.(i) Show that this value of x satisfies the equation x2 − 2x − 12 = 0. [3](ii) Solve the equation x2 − 2x − 12 = 0, giving your answers correct to 2 decimal places. [4](iii) Calculate the perimeter of the rectangle PQRS. [1]

5.

In triangle ABC, the line BD is perpendicular to AC.AD = (x + 6) cm, DC = (x + 2) cm and the height BD = (x + 1) cm.The area of triangle ABC is 40 cm2.

(i) Show that x2 + 5x – 36 = 0.[3](ii) Solve the equation x2 + 5x – 36 = 0. [2](iii) Calculate the length of BC. [2]

6.

The above diagram shows a triangle ABC. AB=3x cm, BC = x cm, and AC = (5x – 4) cm.

The angle ABC = 1200.

a) Show that 3x2 – 10x + 4 = 0 [7]

b) Solve the equation 3x2 – 10x + 4 = 0. Show all your working and give your answers

correct to two decimal places. [4]

c) Hence find the length the side AB and BC of the triangle ABC. [2]

d) Find the area of the triangle ABC. [2]