Chapter 2- Quadratic Equations

Chapter 2- Quadratic Equations

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

CHAPTER 2- QUADRATIC EQUATIONS

2.1 INTRODUCTION 1. General form of quadratic equation is ax 2 + bx + c = 0 where : (i) x is unknown (ii) a, b and c is constant (iii) a ≠ 0 (iv)The powers of x are positive integers up to a maximum value of 2. 2. Roots are the value of the unknown that satisfy the equation. Example 1:

x 2 − 2x − 3 = 0 ( x + 1)( x − 3) = 0 x + 1 = 0 or x − 3 = 0 x = −1 or x = 3 root

2.2 SOLVING QUADRATIC EQUATIONS 1. Factorization method Example 1:

x 2 − 2x − 3 = 0 ( x + 1)( x − 3) = 0 x + 1 = 0 or x − 3 = 0 x = −1 or x = 3

Example 2:

x 2 − 3x − 4 = 0 ( x + 1)( x − 4) = 0 x + 1 = 0 or x − 4 = 0 x = −1 or x = 4

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

2.Completing the square method

To solve the quadratic equation by completing the square method, the coefficient of x2 must be 1. If (−1) 2 − (−1) 2 , we

Example:

x 2 − 2x − 3 = 0

know that the solution is equal to zero. For this example the coefficient of x is -2 so -2 is divided by 2 and become -1.

x 2 − 2 x + ( −1) 2 − ( −1) 2 − 3 = 0 ( x − 1) 2 − 1 − 3 = 0

(−1) 2 − (−1) 2 is added between term bx and c.

( x − 1) 2 − 4 = 0

The concept of completing the square method is the coefficient of x which is b is divided by 2 and the number is squared

( x − 1) 2 = 4 x − 1 = ±2 x − 1 = 2 or x − 1 = −2 x = −1 or x=3

2

2

Part x − 2 x + (−1) is factorized and becomes

( x − 1) 2 while − (−1) 2 − 3 = 0 is solved.

If the coefficient of x is -4 so -4 is divided by 2 and become -2. So the equation will become like this x2 − 4x + ( − 2)2 − ( − 2)2 − 3=0. If the coefficient of x is 6 so 6 is divided by 2 and become 3. So the equation will become like this x2 + 6x + (3)2 − (3)2 − 3=0.

3. By Using formula

ax 2 + bx + c = 0 b c x2 + x + = 0 a a b b b c x 2 + x + ( )2 − ( )2 + = 0 a 2a 2a a 2 b b c (x + ) 2 − 2 + = 0 2a a 4a 2 b b c( 4a ) (x + ) 2 = 2 − 2a a( 4a ) 4a

(x +

How to obtain the formula? To obtain the formula is by using completing the square method.

b 2 b 2 − 4ac ) = 2a 4a 2

x+

b b 2 − 4ac =± 2a 4a 2

x+

b b 2 − 4ac =± 2a 4a 2

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

x+

b b 2 − 4ac =± 2a 2a

x=−

b b 2 − 4ac ± 2a 2a 2

− b ± b − 4ac 2a Example: x=

This is the formula. We can just substitute the value of a, b and c from the equation based on the general form ax 2 + bx + c = 0 to find the values of x

Example: Solve the equation x 2 − 2 x − 3 = 0 by using quadratic formula. Solution: From the equation, we know that a = 1 , b = −2 and c = −3 . So, we can just substitute the value into the formula,

x=

− (−2) ± ( −2) 2 − 4(1)(−3) 2(1)

x=

2 ± ( 4 + 12 2

x=

2 ± (16 2

x=

2±4 2

2+4 2−4 x= or x = 2 2

The value of ± 4 is 4 and -4. So convert the equation into two where the equation x =

2+4 and the other one 2

x = 3 or x = −1 EXERCISE 2.2 1. Find the roots of the quadratic equation 2 x 2 = 4 x − 1 by using completing the square method. Give your answer correct to 2 decimal places. 2. Solve the following quadratic equation using the quadratic formula. (a) x 2 − 5 x − 3 = 0 (b) 2 x 2 = 7 − 4 x 3. Factorize the following quadratic equations and hence, state their roots. (a) x 2 − 5 x − 3 = 0

(b) 2 x 2 = 7 − 4 x

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

2.3 FORMING QUADRATIC EQUATION FROM THE GIVEN ROOTS Given roots are − 1 and 4 , Sum of roots = 4 − 1 1 x = 4 or x = −1 2 =3 x − 4 = 0 or x + 1 = 0 Product of roots = 4 . − 1 ( x + 1)( x − 4) = 0

x 2 + x − 4x − 4 = 0 x 2 − 3x − 4 = 0

= −4 The general form is x 2 − ( S .O.R ) + ( P.O.R ) = 0 Substitute S.O.R = 3 and P.O.R = − 4

x 2 − (3) x + (−4) = 0 x 2 − 3x − 4 = 0 EXERCISE 2.3 1. Write quadratic equations with roots 3 and 5. 2. Form a quadratic equation whose roots are -3 and

3 4

3. Write quadratic equations with roots 1 and -2. 2.4 SUM OF ROOTS (S.O.R) AND PRODUCT OF ROOTS (P.O.R) If the roots are a and b, x = a or x = b x − a = 0 or x − b = 0

What is the general form?

( x − a )( x − b) = 0 x 2 − ax − bx + ab = 0 x 2 − (a + b ) + ab = 0 a and b is the roots so in the equation, a + b is the sum of roots and ab is the product of the roots Hence, the general form is

x 2 − ( S .O.R ) + ( P.O.R ) = 0

Example: The roots of the equation 2 x 2 − 4 x + 1 = 0 are m and n. Find the equation whose roots are 3m and 3n. Solution:

2x 2 − 4x + 1 = 0 1 x 2 − 2x + = 0 2

Make the equation in the general form

x 2 − ( S .O.R) + ( P.O.R) = 0 by divide all terms by 2. This is because in the general form, the value of a must be 1.

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

From the equation above, we know that S.O.R = 2

General form is x 2 − ( S .O.R) + ( P.O.R) = 0 . For this

1 P.O.R = 2

question, the equation is x 2 − 2 x + these two equations.

Given the roots are m and n. Hence, m+n=2 1 mn =

1 2

We know that the sum of roots of the equation is 2 and

2

1 . Given that m 2 1 and n is the roots, so m + n = 2 and mn = 2 the product of roots of the equation is

If the roots are 3m and 3n, S.O.R = 3m + 3n 3 = 3(m +n) P.O.R= 3m . 3n = 9mn Substitute S.O.R = 3(2) =6 P.O.R = 9( =

1

1 = 0 . Compare 2

4

into

3

and

2

into

4

,

1 ) 2

9 2

the equation whose roots are 3m and 3n is

9 =0 2 2 x 2 − 12 x + 9 = 0

x 2 − 6x +

We can leave the equation with x 2 − 6 x +

9 =0 2

but it is better to let the equation without fraction so we multiply all terms with 2. EXERCISE 2.4 1. Given that a and 3 are roots of the quadratic equation px 2 + 3 x + 18 = 0 , find the value of a and p. 2. One of the roots of the quadratic equation x 2 + px + 8 = 0 is half the value of the other root. Find the possible values of p. 3. Given that the value of one root is 3 times the other for the quadratic equation 3 x 2 − 2 x + p = 0 . find (a) the value of p (b) the two roots

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

2.5 CONDITIONS FOR THE TYPES OF ROOT OF QUADRATIC EQUATION

− b ± b 2 − 4ac , we know that the part b 2 − 4ac is called the discriminant of 2a quadratic equation ax 2 + bx + c = 0 .

1.From the formula x =

2. The value of the discriminate will determine the types of roots of a quadratic equation. 3. We can solve a quadratic equation by factorization if the value for b 2 − 4ac is a perfect square. Types of root of Quadratic Equation 1- If b 2 − 4ac > 0 , then the quadratic equation has two different roots(also known as two distinct roots)

x 2 − 2x − 3 = 0

b 2 − 4ac = (−2) 2 − 4(1)(−3)

( x + 1)( x − 3) = 0 x + 1 = 0 or x − 3 = 0 x = −1 or x = 3

= 16

b 2 − 4ac > 0

2- If b 2 − 4ac = 0 , then the quadratic equation has two equal roots

x 2 − 10 x + 25 = 0 ( x − 5)( x − 5) = 0 x−5 = 0 x=5

b 2 − 4ac = (−10) 2 − 4(1)(25) =0 2

b − 4ac = 0

3- If b 2 − 4ac < 0 , then the quadratic equation has no real roots(or no roots)

2 x 2 − 3x + 10 = 0

b 2 − 4ac = (−3) 2 − 4(2)(10) = 9 – 80 = - 71

b 2 − 4ac < 0 4- If b 2 − 4ac ≥ 0 , then the quadratic equation has real roots. Example 1: Given that 3 and k are roots of the quadratic equation x( x + 1) = 12 has two equal roots. Find the value of h. When compare the equation x 2 − ( S .O.R) + ( P.O.R) = 0 and Solution:

x 2 − (−1) x + (−12) = 0 , we would know sum of roots and

x 2 + x − 12 = 0

product of roots for the equation. Page | 25

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 2- Quadratic Equations

x 2 − (−1) x + (−12) = 0 From the equation above, we know that S.O.R = − 1 P.O.R = − 12 Given 3 and k are roots, S.O.R= k + 3 P.O.R= 3 . k = 3k Hence,

k + 3 = −1 or 3k = −12 k = −4 or k = −4 k = −4

From the equation, we know that S.O.R and P.O.R are − 1 and − 12 respectively. From the given roots, we know that S.O.R and P.O.R are k + 3 and 3k respectively. Hence compare both of them to find the value of k.

Example 2: Given that the equation x 2 − 4 x + k + 1 = 0 has two different roots, find the largest integer of k. Solution: From the equation x 2 − 4 x + k + 1 = 0 , we know that a = 1 , b = −4 and c = k + 1 . Two different roots:

b 2 − 4ac > 0 (−4) 2 − 4(1)(k + 1) > 0

16 − 4(k + 1) > 0 16 − 4k − 4 > 0 − 4k > −12 k