PART 2: PRELIMINARY ALGEBRA

UNIT 8 QUADRATIC EQUATIONS

Definition 8.1
The general form of a quadratic equation is 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 where a , b and c are
constants, and 𝑎 ≠ 0.
If 𝑎 = 0, 𝑓(𝑥) = 𝑏𝑥 + 𝑐 is a linear equation.
This could be written as 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
A quadratic equation has two solutions or roots.

8.1 YOU CAN SOLVE QUADRATIC EQUATIONS USING FACTORIZATION METHOD

Example 8.1 Solve the equation 𝑥 2 − 5𝑥 + 18 = 2 + 3𝑥

Solution

𝑥 2 − 5𝑥 + 18 = 2 + 3𝑥
𝑥 2 − 8𝑥 + 16 = 0
𝑥 2 − 4𝑥 − 4𝑥 + 16 = 0
𝑥(𝑥 − 4) − 4(𝑥 − 4) = 0
(𝑥 − 4)(𝑥 − 4) = 0
𝑥 = 4 ,𝑥 = 4
∴ 𝑥=4
8.2 YOU CAN WRITE QUADRATIC EXPRESSIONS IN ANOTHER FORM
BY COMPLETING THE SQUARE.
In general, completing the square:
𝑏 2 𝑏 2
𝑥 2 + 𝑏𝑥 = (𝑥 + 2) − (2)

Note:

make sure that the coefficient of 𝑥 2 is 1.

Example 8.2. Complete the square for the expression
(𝑎) 𝑥 2 + 12𝑥 (𝑏) 2𝑥 2 − 10𝑥
Solution

12 2 12 2
(𝑎) 𝑥 + 12𝑥 = (𝑥 + ) − ( ) = (𝑥 + 6)2 − (6)2 = (𝑥 + 6)2 − 36
2
2 2

2 2
5 2 5 2
(𝑏) 2𝑥 − 10𝑥 = 2[ 𝑥 − 5𝑥] = 2 [(𝑥 − ) − (− ) ]
2 2
5 2 25 5 2 25
= 2 [(𝑥 − ) − ] = 2 (𝑥 − ) −
2 4 2 2

8.3 YOU CAN SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE.

Example 8.3 Solve the equation 2𝑥 2 − 8𝑥 + 7 = 0

Solution

2𝑥 2 − 8𝑥 + 7 = 0(÷ 2)
7
𝑥 2 − 4𝑥 + = 0
2
7
𝑥 2 − 4𝑥 = −
2

4 2 −4 2 7
= (𝑥 − ) − ( ) = −
2 2 2
7
= (𝑥 − 2)2 − (−2)2 = −
2
7 1
= (𝑥 − 2)2 = − + 4 =
2 2
1
= (𝑥 − 2)2 =
2

1
= √(𝑥 − 2)2 = √
2

1
=𝑥 − 2 = ±√
2

√1
= 𝑥−2=±
√2
1 1 1
𝑥 = 2± 𝑥1 = 2 − or 𝑥2 = 2 +
√2 √2 √2
YOU CAN SOLVE QUADRATIC EQUATIONS 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 BY USING THE
FORMULA :
−𝑏±√𝑏2 −4𝑎𝑐
𝑥=
2𝑎
Example 8.4

Show that the solution of the equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐=0 are given by

−𝑏±√𝑏2 −4𝑎𝑐
𝑥=
2𝑎
Proof:

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

𝑏 2 𝑏 2 𝑐
(𝑥 + ) −( ) =−
2𝑎 2𝑎 𝑎

𝑏 2 𝑐 𝑏2
(𝑥 + ) =− +
2𝑎 𝑎 4𝑎2

𝑏 2 −4𝑎𝑐+𝑏2
± √(𝑥 + ) =±√
2𝑎 4𝑎2

𝑏 √𝑏2 − 4 𝑎 𝑐
𝑥+ =±
2𝑎 √4 𝑎2

𝑏 √𝑏2 −4𝑎𝑐
𝑥=− ±
2𝑎 2𝑎

−𝑏±√𝑏 2 −4𝑎𝑐
𝑥= 2𝑎
 Example 8.5
Solve the equation 4𝑥 2 − 3𝑥 − 2 = 0 by using the quadratic formula.

Solution

𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝑎 = 4, 𝑏 = 3 , 𝑐 = −2
−𝑏±√𝑏2 −4𝑎𝑐
𝑥=
2𝑎

−(3)±√(3)2 −4(4)(−2)
=
2(4)

−3±√9+32
=
8

−3±√41
𝑥=
8

−3+√41 −3 − √41
𝑥1 = or 𝑥2 =
8 8

Example 8.6 Solve the equation 𝑥 2 + 2𝑥 + 2 = 0 by using the quadratic formula.

solution

𝑎 = 1, 𝑏 = 2 , 𝑐 = 2
−𝑏±√𝑏2 −4𝑎𝑐
𝑥=
2𝑎

−2±√(2)2 −4(1)(2)
=
2(1)

−2±√4−8
=
2

−2±√−4
=
2
−2±2𝑖
= = −1 ± 𝑖
2

𝑥1 = −1+𝑖 𝑥2 = −1 − 𝑖
YOU NEED TO BE ABLE TO SKETCH GRAPHS OF QUADRATIC EQUATIONS AND
SOLVE PROBLEMS USING THE DISCRIMINANT.

Definition 8.1

𝑏 2 − 4𝑎𝑐 is called the discriminant of the quadratic equation.

The steps to help you sketch the graphs are
1) Decide on the shape of the graph

𝑖𝑓 𝑎 > 0, the curve will be ∪ shape

𝑖𝑓 𝑎 < 0, the curve will be ∩ 𝑠ℎ𝑎𝑝𝑒
2) Work out the points where the curve crosses the 𝑥 and 𝑦 -axes
i) Put 𝑦 = 0 to find the x-axis crossing points
ii) Put 𝑥 = 0 to find thy y-axis crossing points
3) Check the general shape of the curve by considering the discriminant 𝑏 2 − 4𝑎𝑐

Note:
You can use the discriminant to establish when a quadratic equation has;
1) Equal roots 𝑏 2 = 4𝑎𝑐
2) Real roots 𝑏 2 > 4𝑎𝑐
3) 𝑏 2 < 4𝑎𝑐

Example
Sketch the graphs of (a) 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4
(b) 𝑔(𝑥) = −𝑥 2 + 2𝑥 + 5
Solution

(1) 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4
𝑎 = 1 > 0 ∪ shape
(2) (i) 𝑦 = 0 𝑥 2 − 5𝑥 + 4 = 0
𝑥 2 − 4𝑥 − 𝑥 + 4 = 0
𝑥(𝑥 − 4) − 1(𝑥 − 4) = 0
(𝑥 − 4)(𝑥 − 1) = 0
𝑥 = 4 𝑜𝑟 𝑥 = 1
(4,0) (1,0)
2
(ii) 𝑥 = 0 𝑓(𝑥) = 𝑥 − 5𝑥 + 4
 𝑓(0) = (0)2 − 5(0) + 4 = 4 (0,4)
3) 𝑏 2 = 25 4𝑎𝑐 = 16
𝑏 2 > 4𝑎𝑐 and 𝑎 > 0 so we have real roots

(0,4)

x
(1,0) (4,0)

(b) 𝑔(𝑥) = −𝑥 2 + 2𝑥 + 5
𝑎<0 ∩
We cannot factorize so we use completing the square method
−(𝑥 2 − 2𝑥) + 5
−[(𝑥 − 1)2 − (−1)2 ] + 5
−(𝑥 − 1)2 + 1 + 5
−(𝑥 − 1)2 + 6

f(x)

6 y=−(𝑥 − 1)2 +6

0 1 x
2
y=−𝑥 2 . y=−(𝑥 − 1)
Example 8.7
Find the value of k for which 𝑥 2 + 𝑘𝑥 + 9 = 0 has equal roots
Solution
𝑥 2 + 𝑘𝑥 + 9 = 0
𝑎 = 1, 𝑏 = 𝑘 , 𝑐 = 9
𝑏 2 = 4𝑎𝑐 , 𝑘 2 = 4(1)(9)
𝑘 2 = 36
𝑘 = ±6

MAXIMUM AND MINIMUM VALUES OF QUADRATIC FUNCTIONS

The maximum or minimum of the function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 are values of 𝑓(𝑥) at the
top or bottom of the curve. These are called the turning points of the curve.
The turning points are found as follows:
𝑏 2 𝑏2
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎 (𝑥 + 2𝑎) − 4𝑎2 + 𝑐

where 𝑎 > 0
𝑏 2
𝑎 (𝑥 + 2𝑎) = 0,
𝑏 𝑏
then 𝑥 = − 2𝑎 ,𝑦 = 𝑓 (− 2𝑎)

𝑏 𝑏
Thus, turning point is (𝑥 = − 2𝑎 , 𝑦 = 𝑓 (− 2𝑎))

NOTE:
If 𝑎 > 0,then the turning point is a minimum
If 𝑎 < 0, the turning point is a maximum
Example 8.8

Determine whether the function 3𝑥 2 − 4𝑥 + 1 has a maximum or minimum value and

for what value of x does it occur.
Solution

𝑏 (−4) 2
𝑥 = − 2𝑎 = − 2(3) = 3

𝑎 = 3 > 0 ⟹ ∪ 𝑠ℎ𝑎𝑝𝑒
2 2 2 2 1
𝑓 (3) = 3 (3) − 4 (3) + 1 = − 3
 2
𝑥=3

1
𝑦 = −3

2
The line 𝑥 = 3 through the turning point is called the axis of the curve and the
curve is symmetry about this line.

Example
Express 5 − 𝑥 − 2𝑥 2 in the form 𝑎 − 𝑏(𝑥 + 𝑐)2 and hence or otherwise find its maximum
value and the value of x where this occurs and sketch the graph.
Solution
𝑥
5 − 𝑥 − 2𝑥 2 = 5 − 2 (𝑥 2 + 2)

1 2 1
5 − 2 [(𝑥 + 4) − 16]

1 2 1
5 − 2 (𝑥 + ) +
4 8

41 1 2
− 2 (𝑥 + 4)
8

1 2 −1
The least value is (𝑥 + 4) = 0 ⟹ 𝑥 = 4

1 1 −1 2 1
𝑓 (4) = 5 + 4 − 2 ( 4 ) = 5 8
 1
58

1
𝑥 = −4

RELATIONSHIP BETWEEN THE ROOTS AND THE COEFFICIENTS OF A

QUADRATIC EQUATION
We first consider the following formulas.

(i) (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2

(ii) (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2
(𝑖𝑖𝑖) (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2
(𝑖𝑖) (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3
(v)(𝑎 − 𝑏)3 = 𝑎3 − 3𝑎2 𝑏 + 3𝑎𝑏 2 − 𝑏 3

Let 𝛼 𝑎𝑛𝑑 𝛽 be the roots of the equation

𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Then it means that the quadratic equation can be factorized as:
(𝑥 − 𝛼)(𝑥 − 𝛽) = 0
𝑥(𝑥 − 𝛽) − 𝛼(𝑥 − 𝛽) = 0
𝑥 2 − 𝛼𝑥 − 𝛽𝑥 + 𝛼𝛽 = 0
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 0 (1)
Our general quadratic equation is given by
𝑎𝑥 2 + 𝑏𝑥 + 𝑐(÷ 𝑎)
𝑏 𝑐
𝑥2 + 𝑎 𝑥 + 𝑎 = 0 (2)

Comparing (1) and (2) we have ;

 𝑏 𝑏
−(𝛼 + 𝛽) = ⟹ 𝛼+𝛽 =−
𝑎 𝑎
Sum of roots
𝑐 𝑐
𝛼𝛽 = 𝑎 ⟹ 𝛼𝛽 = 𝑎
Product of roots

and the equation of roots may be written as

𝑥 2 − (𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠)𝑥 + (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0
Example 8.9
The equation 4𝑥 2 − 3𝑥 + 1 = 0 has roots 𝛼 𝑎𝑛𝑑 𝛽 , without evaluating 𝛼 and 𝛽 find

1 1
(i) 𝛼+𝛽 (ii) 𝛼𝛽 (iii) +𝛽
𝛼

(iv) 𝛼 2 + 𝛽 2 (v) 𝛼 3 + 𝛽 3 (vi) 𝛼 4 + 𝛽4

1 1
Hence write down the equation whose roots are 𝑎𝑛𝑑
𝛼 𝛽

Solution
𝑏
(i) 𝛼 + 𝛽 = −𝑎 4𝑥 2 − 3𝑥 + 1 = 0
−(3)
𝛼+𝛽 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
4
3
𝛼+𝛽 = 𝑎=4 𝑏 = −3 𝑐=1
4

𝑐 1
(ii) 𝛼𝛽 = 𝑎 , 𝛼𝛽 = 4
1 1 𝛽+𝛼 3 1
(𝑖𝑖𝑖) +𝛽 = = ÷4=3
𝛼 𝛼𝛽 4

(iv) 𝛼 2 + 𝛽 2
(𝛼 + 𝛽)2 = 𝛼 2 + 𝛽 2 + 2𝛼𝛽
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
3 2 1 1
= (4) − 2 (4) = 16

(v) (𝛼 + 𝛽)3
(𝛼 + 𝛽)3 = 𝛼 3 + 3𝛼 2 𝛽 + 3𝛼𝛽 2 + 𝛽 3
 (𝛼 + 𝛽)3 − 3𝛼 2 𝛽 − 3𝛼𝛽 2 = 𝛼 3 + 𝛽 3
(𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽) = 𝛼 3 + 𝛽 3
𝛼 3 + 𝛽 3 = (𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽)
3 3 1 3
= (4) − 3 (4) (4)
9
= − 16

(vi) 𝛼 4 + 𝛽 4 = ( 𝛼 2 )2 + (𝛽 2 )2
= ( 𝛼 2 + 𝛽 2 )2 − 2𝛼 2 𝛽 2
= ((𝛼 + 𝛽)2 − 2𝛼𝛽)2 − 2(𝛼𝛽)2
2
3 2 1 1 2
= ((4) − 2 (4)) − 2 (4)

9 2 2 1
= (16 − 4) − 2 (16)

9−8 2 1
= ( 16 ) − 8
1 1
= 256 − 8
31
= − 256

𝑥 2 − (𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠)𝑥 + (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0

1 1 1 1
𝑥 2 − (𝛼 + 𝛽) 𝑥 + 𝛼 . 𝛽 = 0

𝛽+𝛼 1
𝑥 2 − ( 𝛼𝛽 ) 𝑥 + 𝛼𝛽 = 0

3 1 1
𝑥 2 + (4 ÷ 4) 𝑥 + 1 ÷ 4 = 0

𝑥 2 + 3𝑥 + 4 = 0