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ADDITIONAL MATHEMATICS FORM 4
NOTES, EXAMPLES & EXERCISE
2 : QUADRATIC EQUATIONS
2.1 Quadratic Equation and Its Roots
1. The genaral form of quadratics equation is ax2 + bx + c = 0, where a, b and c are
constants, with the condition a  0 and x is an unknown.
Example 1 : Express each of the following in the general form.
( x + 1 ) 2 = 16
Solution :
( x + 1 ) 2 = 16
x2 + 2x + 1 =16
x2 + 2x + 1 – 16 = 0
x2 + 2x – 15 = 0
Exercise 2.1
1. Express the following equation in the general form ax2 + bx + c = 0. Then , state the values
of a, b and c.
(a) x2 – 6x = 3
(b) ( x – 4 ) 2 = 1
2.
x cm
(x + 4 ) cm
The diagram above shows a rectangle with sides (x + 4 ) cm and x cm. Give that the area
of the rectangle is 60 cm2, find the value of x .
2.2 Solving a Quadratic Equation
1. A quadratic equation ax2 + bx + c = 0 can be solved by
(a) Factorisation
(b) completing the square
(C)using a formula
2. The formula for finding the roots of a quadratic equation ax2 + bx + c = 0 is
 b  b  4ac
x=
2a
2
-1-
3. Using S.O.R. and P.O.R. a quadratic equation can be written as
x2 – ( S.O.R ) x + ( P.O.R ) = 0 where
S.O.R = sum of roots
P.O.R = product of roots
4. For the quadratic equation ax2 + bx + c = 0 ,
b
c
S.O.R = and P.O.R =
a
a
Example 2 : Solve the following quadratic equation by factorisation.
x2 + 7x + 10 = 0
Solution :
x2 + 7x + 10 = 0
(x + 2 ) ( x + 5 ) = 0
x= -2 or x = -5
Example 3 : Solve the following quadratic equation by quadratic formula.
3x2 + 6x + 2 = 0
Solution :
a=3 , b=6 , c=2
 6  6  4(3)( 2)
x=
2(3)
2
- 6  12
6
x = -0.4226 or -1.577
x =
Example 4 : Solve the following quadratic equation by completing the square.
x2 + 8x – 6 = 0
Solution :
x2 + 8x – 6 = 0
x2 + 8x
= 6
2
(x+4)
= 6 + 42
x+4
=  22
x
= -4  22
x
= 0.6904 or -8.6904
-2-
Example 5 : Form a quadratic equation with the roots -2 and 4 .
Solution :
S.O.R = -2 + 4 = 2
P.O.R = -2 × 4 = -8
The quadratic equation is :
x2 – ( S.O.R ) x + ( P.O.R ) = 0
x2 – ( 2 ) x + ( -8 ) = 0
x2 – 2x -8 = 0
Example 6 : If m and n are the roots of the equation 2x2 + 8x – 3 = 0 , form the equation
whose roots are 2m and 2n .
Solution :
If m and n are the roots of the equation 2x2 + 8x – 3 = 0 , then
b
c
m+n=mn =
a
a
8
3
==2
2
=-4
For the roots 2m and 2n ,
Sum of roots = 2m + 2n
= 2 (m + n)
= 2 (-4)
= -8
Product of roots = (2m) (2n)
= 4mn
3
= 4 (- )
2
=-6
Therefore, the equation which has roots 2m and 2n is :
x2 – (sum of roots) x + product of roots = 0
x2 – (-8 ) x + ( -6 ) = 0
x2 + 8x – 6 = 0
Exercise 2.2
1. Solve the following quadratic equations by factorisation.
(a) x2 – 5x + 4 = 0
(b) 2x2 = 11x + 6
2. Solve the following quadratic equations by completing the square.
(a) x2 – x – 3 = 0
(b) 2x2 + 6x + 3 = 0
3. Solve the following quadratic equations by using the quadratic formula..
(a) x2 – 2x – 4 = 0
(b) 3x2 – 14x + 10 = 0
-3-
4. Form the quadratic equations which roots are :
1
(a) 1 , 7
(b) 4 ,
2
5. Given that m and n are roots of the quadratic equation 2x2 – 11x – 6 = 0 , form a quadratic
equation with roots 3m and 3n .
2.3 Types of the Roots of the Quadratic Equation
1. The conditions for the nature of the roots of quadratic equations are
( a ) two real and distinct roots : b2 – 4ac > 0
( b ) two real and equal roots : b2 – 4ac = 0
( c ) no real roots : b2 – 4ac < 0
( d ) two real roots : b2 – 4ac  0
2. ( a ) If a straight line intersects a curve at two distinct points, the condition
b2 – 4ac > 0 is applied.
( b ) If a straight line touches a curve at only one point, the condition b2 – 4ac = 0
is applied.
( c ) If a straight line does not meet a curve, the condition b2 – 4ac < 0 is applied.
b2 – 4ac > 0
b2 – 4ac = 0
b2 – 4ac < 0
Two real and
distinct roots
Two real and equal
roots
No real roots
b2 – 4ac  0
Two real roots
Example 7 : Determine the types of roots for the following quadratic equation.
x2 – 12x + 27 = 0
Solution :
x2 – 12x + 27 = 0
a = 1 , b = -12 , c = 27
b2 – 4ac = (-12) 2 – 4 (1) (27)
= 144 – 108
= 36 > 0
Thus, x2 -12x + 27 = 0 has two distinct roots
-4-
Example 8 : The quadratic equation x2 – 2px + 25 = 0 has two equal roots. Find the value
of p .
Solution :
x2 - 2px + 25 = 0
a = 1 , b = -2p , c = 25
Using b2 – 4ac = 0
(-2p)2 – 4(1)(25) = 0
4p2 – 100 = 0
4p2
= 100
p2
= 25
p
=± 5
Exercise 2.3
1. Determine the types of roots for each of the following quadratic equations.
(a) x2 + 9x + 3 = 0
(b) x2 – 12x + 36 = 0
2. Find the values of k for each of the following quadratic equations which has equal roots.
(a) x2 – kx + 36 = 0
(b) 4x2 + 12x + k = 0
3. Find the range of values of h for each of the following quadratic equations which has no
roots.
(a) x2 + 2x + h – 2 = 0
(b) (h+1)x2 + 8x + 6 = 0
4. The quadratic equation x2 + 2nx – 16 = 0 has two different roots. Find the range of values
of n .
5. The quadratic equation 9k + 6kx + ( k + 1 )x2 = 2 has two different roots. Find the range
of values of k.
SPM FOCUS PRACTICE
PAPER 1
1. Solve the quadratic equation x ( x – 4 ) = ( x – 1 ) ( 2 – x ). Give your answer correct to
four significant figures.
2. The quadratic equation x ( x + p ) = 2x –4 has two different roots. Find the range of values
of p .
3. Form the quadratic equation which has roots –2 and
ax2 + bx + c = 0, where a , b and c are constants.
-5-
1
. Give your answer in the form
3
PAPER 2
1. Given that the x-axis is the tangent to the curve x2 + 3px + 9q2 = 0, find p in terms of q .
2. Given p and q are the roots of the quadratic equation x2 – mx + 6 = 0 while 2p and 2q are
the roots of the quadratics equation 2x2 + 20x + n = 0 . Find the values of m and n .
3. The quadratic equation x2 + ( r + 3 ) x + t2 = 0 has two equal roots.
(a) Show that r = 2t – 3
(b) Hence, determine the root if r = 3 .
PAST YEAR QUESTION
1. Given that –1 and h are roots of the quadratic equation ( 3x – 1 ) ( x – 2 ) = p ( x – 1 ) ,
where p is a constant, find the values of h and p . ( 2001 P1 Q4 )
2. Solve the quadratic equation 2x (x – 4 ) = ( 1 – x ) ( x + 2 ) . Give your answer to four
significant figures.
( 2003 P1 Q3 )
1
. Give your answer in the form
2
( 2004 P1 Q4 )
3. From the quadratic equation which has the roots –3 and
ax2 + bx + c = 0 , where a, b and c are constants.
4. Solve the quadratic equation x ( 2x – 5 ) = 2x – 1. Give your answer correct to three
decimal places.
( 2005 P1 Q5 )
ANSWERS
4. n < -4 or n > 4
Exercise 2.1
1.(a) x2 – 6x – 3 = 0 ; a =1 , b = -6 , c = -3
2
5. k <
(b) x2 – 8x + 15 = 0 ; a =1 , b =-8 , c = 15
7
2. x = 6
SPM Focus Practice (Paper 1)
Exercise 2.2
1. x = 3.186 or 0.3139
1
2. p > 6 or p < -2
1. (a) 1 , 4
(b) 6 , 3. 3x2 + 5x – 2 = 0
2
2. (a) –1.303 , 2.303 (b)–2.366 , -0.634
Paper 2
3. (a) –1.236 , 3.236 (b) 0.880 , 3.786
1. p =  2q
4. (a) x2 –8x +7 =0
(b) 2x2 – 9x + 4 = 0
2. m = -5 , n = 48
5. 2x2 – 33x – 54 = 0
3. (b) x = - 3
Exercise 2.3
Past Year Question
1. (a) Two different roots
4
1. p = -6 h =
(b) Two equal roots
3
2. (a) k =  12
(b) k = 9
2. x = 2.591 or -0.2573
5
3. 2x2 + 5x – 3 = 0
3. (a) h > 3
(b) h >
4. x = 8.153 or 0.149
3
-6 CHNG/ AMF4/ C2