Transcript Book 4 Chapter 2 Quadratic Equations

2
Quadratic Equations In One
Unknown
Case Study
2.1 Quadratic Equations
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
2.3 Solving Quadratic Equations by Quadratic
Formula
2.4 The Nature of Roots of Quadratic Equations
2.5 Simple Problems Leading to Quadratic Equations
2.6 Relations between the Roots and Coefficients
of Quadratic Equations
Chapter Summary
Case Study
We can answer this
question by solving a
quadratic equation.
Can 36 chocolate beans be
arranged in a triangle?
In junior forms, we learnt that
numbers which can be arranged
in a compact triangular pattern
are called triangular numbers.
n(n  1)
A triangular number p can be expressed as p 
, where n is
2
a positive integer.
It can be rewritten as n2  n  2p  0, which is called a quadratic
equation in n.
P. 2
2.1 Quadratic Equations
A quadratic equation in one unknown is an equation of
degree two with only one unknown, such as:
2x2  5x  3  0
and
4y2  7y  0
In general, a quadratic equation in x can be written in the form
ax2  bx  c  0
,
where a, b, and c are real constants and a  0.
This is called the general form of a quadratic equation.
P. 3
2.1 Quadratic Equations
Consider the quadratic equation x2  5x  6  0. ........... (*)
Try to substitute the following values of x into the equation to check if
any of them satisfies (*):
(1) x  6
L.H.S.  (6)2  5(6)  6  0  R.H.S.
(2) x  3
L.H.S.  (3)2  5(3)  6  12  R.H.S.
(3) x  0
L.H.S.  02  5(0)  6  6  R.H.S.
(4) x  1
L.H.S.  12  5(1)  6  0  R.H.S.
(5) x  2
L.H.S.  22  5(2)  6  8  R.H.S.
x  6 and x  1 satisfy (*) while the others do not.
Therefore, x  6 and x  1 are the roots (or solutions) of the equation
x2  5x  6  0.
P. 4
2.1 Quadratic Equations
To solve a quadratic equation means to find all the roots of
the equation.
It is not easy to find all the roots of a quadratic equation by substituting
different values of x.
We need systematic methods to solve quadratic equations.
Different methods in solving quadratic equations algebraically
(to be introduced in the Sections 2.2 and 2.3):



Factor Method
Method of Completing the Square
Quadratic Formula
P. 5
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
A. By Taking Square Root
This method can be used if a quadratic equation is written in
the form
x2  p
or (x  q)2  p,
where p and q are real numbers and p  0.
For example:
( x  2) 2  36  0
x 2  64  0
( x  2) 2  36
x 2  64
x  2  6
x  8
i.e., x  8 or 8
x  8 or 4
In general:
For x2  p, x   p .
For (x  q)2  p, x  q  p .
The method of rewriting a
quadratic equation in the
form ax2  bx  c  0 into
(x  q)2  p is called the
method of completing the
square, to be introduced in
Section 2.3.
If p  0, then p is not a real number.
We can represent the roots in terms of
complex number i.
For example, if x2  3, then x   3 i.
P. 6
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
A. By Taking Square Root
Example 2.1T
Solve the following quadratic equations.
(a) (x  1)2  81  0
(b) (2y  5)2  49  0
Solution:
(a)
(x  1)2  81  0
(b)
(x  1)2  81
Since 81  0, the quadratic equation
(x  1)2  81  0 has no real roots.
(2y  5)2  49  0
(2y  5)2  49
2y  5  7
2y  2 or 12
y  1 or 6
Consider the unreal roots of the equation:
x  1  9i
x  1  9i
P. 7
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
B. By Factor Method
For any real numbers A and B,
A  0  0 and 0  B  0.
We have the following result.
For any two real numbers A and B, if AB  0, then A  0 or B  0.
Consider a quadratic equation ax2  bx  c  0.
Suppose ax2  bx  c can be factorized into (px  q)(mx  n), where
p, q, m and n are real numbers, then
ax2  bx  c  0
(px  q)(mx  n)  0
px  q  0 or mx  n  0
n
q
x
or
m
p
This method is called the factor method.
P. 8
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
B. By Factor Method
Example 2.2T
Solve the following quadratic equations.
(a) x2  4x  3  0
(b) 72x2  96x  24  0
Solution:
(a)
(b)
x2  4x  3
(x  1)(x  3)
x1
x
0
0
 0 or x  3  0
 1 or 3
72x2  96x  24  0
3x2  4x  1  0
(3x  1)(x  1)  0
3x  1  0 or x  1  0
1
x  or 1
3
Using the cross method:
x
1
x
3
x
3x  4x
Divide both sides by 24.
P. 9
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
B. By Factor Method
Example 2.3T
Solve the following quadratic equations.
(a) 4x2  20x  25  0
(b) 2x2  8x  8
(c) 5x2  13x  0
(d) y(9y  5)  16y
Solution:
(a)
4x2  20x  25  0
(2x  5)(2x  5)  0
x
(b)
5
(repeated root)
2
2x2  8x  8
x2  4x  4
x2  4x  4  0
(x  2)2  0
x  2 (repeated root)
We can also use the identity
a2  2ab  b2  (a  b)2 to
solve (a), i.e.,
4x2  20x  25  0
(2x  5)2  0
5
x 
2
a2  2ab  b2  (a  b)2
P. 10
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
B. By Factor Method
Example 2.3T
Solve the following quadratic equations.
(a) 4x2  20x  25  0
(b) 2x2  8x  8
(c) 5x2  13x  0
(d) y(9y  5)  16y
Solution:
(c)
(d)
5x2  13x  0
x(5x  13)  0
x  0 or 5x  13  0
13
x  0 or 
5
y(9y  5)  16y
9y2  11y  0
y(9y  11)  0
y  0 or
Expand and simplify.
11
9
Do not cancel away x,
otherwise the root x  0 will
be missing.
Do not cancel y from both
sides of the equation,
otherwise the root y  0 will
be missing.
P. 11
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
B. By Factor Method
Example 2.4T
Solve the following quadratic equations.
(a) (4x  25)(x  4)  28
(b) (x  16)(11x  2)  2x(x  16)
Solution:
(a)
(b)
(4x  25)(x  4)  28
4x2  41x  100  28
4x2  41x  72  0
(4x  9)(x  8)  0
9
x  or 8
4
(x  16)(11x  2)
(x  16)(11x  2)  2x(x  16)
(x  16)[(11x  2)  2x]
(x  16)(13x  2)
x
To solve this kind of
quadratic equation, we
should first expand the
expression and rewrite it
into the general form.
 2x(x  16)
0
0
0
2
 16 or
13
Do not cancel x  16 from
both sides of the equation,
otherwise the root x  16
will be missing.
P. 12
2.3 Solving Quadratic Equations by
Quadratic Formula
Factor method may not be applicable in solving some quadratic
equations, such as
x2  3x  1  0.
We cannot factorize it using factor method.
In this case, we need to use other methods to solve the equation.
P. 13
2.3 Solving Quadratic Equations by
Quadratic Formula
A. The Method of Completing the Square
If a quadratic expression can be written as a product of two
identical linear factors, we call it a perfect square.
For example:


x2  4x  4  x2  2(x)(2)  22  (x  2)2
x2  6x  9  x2  2(x)(3)  32  (x  3)2
Notes:
The following identities are used to rewrite the above expressions:
x2  2ax  a2  (x  a)2
and
x2  2ax  a2  (x  a)2
Given a quadratic expression x2  2ax.
To turn it into a perfect square, we need to add an additional term a2.
This method is called the method of completing the square.
P. 14
2.3 Solving Quadratic Equations by
Quadratic Formula
A. The Method of Completing the Square
Examples of forming perfect squares:


x2  14x  49  (x  7 )2
x2  12x  36  (x  6 )2
x2  2ax  a2  (x  a)2
x2  2ax  a2  (x  a)2
In general, for a quadratic expression like x2  kx, we add the term
2
2
k
k

  to it in order to form a perfect square  x   .
2
2

2
2
k
k
k
k
That is, x  kx     x 2  2( x)       x  
2
2
2 2 
2
2
Further examples of forming perfect squares:
2
2
1
1


 x2  x 
 x 
5
25 
5 2
3
9
3
 x2  x 
  x  
4
64 
8
P. 15
2.3 Solving Quadratic Equations by
Quadratic Formula
B. Solving Quadratic Equations Using the
Method of Completing the Square
Using the method of completing the square, if we rewrite a
quadratic equation in the form ax2  bx  c  0 into (x  q)2  p,
then the equation can be solved easily by taking square root.
For example:
Consider the quadratic equation x2  10x  21  0.
If we are able to rewrite the quadratic equation into (x  q)2  p:
(x  5)2  4
then we can take square root on both sides to obtain
x  5  2
x  7 or 3
Remarks:
The roots obtained should be the same as in the factor method:
(x  7)(x  3)  0

x  7 or 3
P. 16
2.3 Solving Quadratic Equations by
Quadratic Formula
B. Solving Quadratic Equations Using the
Method of Completing the Square
Example 2.5T
Solve the following quadratic equations by the method of completing
the square.
(a) x2  2x  3  0
(b) 2x2  4x  1  0
Solution:
(a)
x2  2x  3
x2  2x
x2  2x  12
(x  1)2
x1
x
x
0
3
 3  12
4
 2
12
 1 or 3
Transpose the constant term to the R.H.S.
Add 12 on the L.H.S. to complete the square.
Add 12 on the R.H.S. to balance the equation.
Take square root on both sides.
P. 17
2.3 Solving Quadratic Equations by
Quadratic Formula
B. Solving Quadratic Equations Using the
Method of Completing the Square
Example 2.5T
Solve the following quadratic equations by the method of completing
the square.
(a) x2  2x  3  0
(b) 2x2  4x  1  0
Solution:
(b)
2x2  4x  1  0
2x2  4x  1
1
x2  2x  
2
1
x2  2x  12    12
2
1
(x  1)2 
2
1
x 1
2
Transpose the constant term to the R.H.S.
Divide the whole equation by 2 to change the
coefficient of x2 to 1.
Add 12 on the L.H.S. to complete the square;
Also add 12 on the R.H.S. to balance the equation.
 2 2
 or

2 

P. 18
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Another method of solving quadratic equations is to use the
quadratic formula.
Quadratic Formula
The roots of the quadratic equation ax2  bx  c  0 (a  0) are
given by
 b  b 2  4ac
.
x
2a
The quadratic formula is derived from the method of completing the
square as shown in the next page.
P. 19
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Derivation of the quadratic formula:
ax2  bx  c  0
ax2  bx  c
b
c
x2  x  
a
a
2
where a  0
b
c  b 
 b 
x2  x        
a
a  2a 
 2a 
2
Completing the square for the L.H.S.
2
b 
b 2  4ac

x  
2a 
4a 2

b  b 2  4ac
x

2a
2a
Taking square root on both sides.
 b  b 2  4ac
x
2a
P. 20
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.6T
Solve the following quadratic equations by the quadratic formula.
(a) x2  5x  14  0
(b) x2  9x  9  0
(c) 2x2  15  0
Solution:
(a) x2  5x  14  0
 5  52  4(1)(14)
x
2(1)
 5  81

2
59

2
59
59

or
2
2
 7 or 2
P. 21
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.6T
Solve the following quadratic equations by the quadratic formula.
(a) x2  5x  14  0
(b) x2  9x  9  0
(c) 2x2  15  0
Solution:
(b) x2  9x  9  0
 (9)  (9) 2  4(1)(9)
x
2(1)
9  45  9  3 5 
 or


2
2


(c) 2x2  15  0
 0  02  4(2)(15)
x
2(2)

30 
120
 or 


2 
4

Simplifying surds is not
required in the Foundation
part of the syllabus.
In (c), we put a  2, b  0
and c  15.
P. 22
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.7T
Solve the following quadratic equations by the quadratic formula.
(a) 4x2  12x  9  0
(b) x2  2x  2  0
Solution:
(a) 4x2  12x  9  0
 (12)  (12) 2  4(4)(9)
x
2(4)
12  0

8
3

(repeated root)
2
P. 23
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.7T
Solve the following quadratic equations by the quadratic formula.
(a) 4x2  12x  9  0
(b) x2  2x  2  0
Solution:
(b) x2  2x  2  0
 2  22  4(1)(2)
x
2(1)
2 4

2
Since  4 is not a real number, the equation has no real roots.
Consider the unreal roots of the equation:

 4  2i
 x  1  i
P. 24
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.8T
Solve the following quadratic equations by the quadratic formula.
(Give the answer correct to 3 significant figures.)
(a) 2x2  6x  3  0
(b) 3x2  7x  8  0
Solution:
(a) 2x2  6x  3  0
 6  62  4(2)(3)
x
2(2)
 6  12
4
 2.37 or 0.634

(cor. to 3 sig. fig.)
(b) 3x2  7x  8  0
 7  7 2  4(3)(8)
x
2(3)
 7  145
6
 3.17 or 0.840

(cor. to 3 sig. fig.)
P. 25
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Example 2.9T
Solve 12(x2  6)  8(x  7) by the quadratic formula.
Solution:
12(x2  6)
3(x2  6)
3x2  18
3x2  2x  4
 8(x  7)
 2(x  7)
 2x  14
0
Divide both sides by 4.
Expand and simplify.
Rewrite the given equation into the general form.
 (2)  (2) 2  4(3)(4)
x
2(3)
2  52  1  13 

 or

6
3 

Simplifying surds is not
required in the Foundation
part of the syllabus.
P. 26
2.3 Solving Quadratic Equations by
Quadratic Formula
C. Using Quadratic Formula to Solve
Quadratic Equations
Consider the quadratic equation x2  3x  2  0.

Factor Method
x 2  3x  2  0
( x  2)( x  1)  0
x  2 or  1


Method of Completing the Square
x 2  3x  2  0
x 2  3 x  2
2
2
3
3
x 2  3 x     2   
2
2
Quadratic Formula
2
3
1


x


x 2  3x  2  0


2
4


2
 3  3  4(1)(2)
3 1
x
x



2(1)
2 2
 2 or  1
 2 or  1
It is more efficient to solve the equation using the factor method.
Students should decide which method to use depending on the situation.
P. 27
2.4 The Nature of Roots of
Quadratic Equations
We have learnt solving quadratic equations in one unknown.
The roots of a quadratic equation ax2  bx  c  0 (a  0) are
 b  b 2  4ac
x
.........(*)
2a
A quadratic equation may have two unequal real roots, one double real
root or no real roots depending on the value of b2  4ac in (*).
We can determine the nature of roots of a quadratic equation by finding
the value of the expression b2  4ac.
Case 1:
If b2  4ac  0, then the equation has two unequal real roots.
Case 2:
If b2  4ac  0, then the equation has one double real root.
Case 3:
If b2  4ac  0, then the equation has no real roots.
P. 28
2.4 The Nature of Roots of
Quadratic Equations
The expression b2  4ac is called the discriminant of the
quadratic equation, denoted by the symbol D, i.e.,
D  b2  4ac.
Summarizing the nature of roots of quadratic equations for different
values of D:
Value of D
Nature of the roots of the equation
D0
Two unequal real roots
D0
One double real root
(Two equal real roots)
D0
No real roots
If D  0, then the equation is
also said to have two unreal
roots.
P. 29
2.4 The Nature of Roots of
Quadratic Equations
Example 2.10T
Consider the equation (p  1)x2  12x  9  0 where p  1. Find
the value of p or the range of values of p if the equation has
(a) two equal real roots;
(b) two unequal real roots;
(c) no real roots.
Solution:
D  (12)2  4(p  1)(9)
 144  36p  36
We can first find D in terms
of p first.
 180  36p
(a) Since the equation has two equal real roots, D  0.
180  36p  0
36p  180
p 5
P. 30
2.4 The Nature of Roots of
Quadratic Equations
Example 2.10T
Consider the equation (p  1)x2  12x  9  0 where p  1. Find
the value of p or the range of values of p if the equation has
(a) two equal real roots;
(b) two unequal real roots;
(c) no real roots.
Solution:
(b) Since the equation has two unequal real roots, D  0.
180  36p  0
p5
∴ The range of values of p is p  5.
(c) Since the equation has no real roots, D  0.
180  36p  0
p5
∴ The range of values of p is p  5.
P. 31
2.4 The Nature of Roots of
Quadratic Equations
Example 2.11T
If 2x2  12x  p  0 has two distinct real roots, find the range of
values of p.
Solution:
Since the equation has two distinct real roots, D  0.
(12)2  4(2)(p)  0
144  8p  0
8p  144
p  18
∴ The range of values of p is p  18.
P. 32
2.5 Simple Problems Leading to
Quadratic Equations
We are going to solve simple word problems related to quadratic
equations.
Strategies for solving word problems
Step 1: Study the problem carefully and understand the objective
of the problem. Draw a figure or diagram if necessary.
Step 2: First set a variable, say x, to one of the unknown
quantities. Then try to represent all the other unknown
quantities in terms of x.
Step 3: Set up a quadratic equation or an equation that can be
transformed into a quadratic equation.
Step 4: Solve the equation.
Step 5: Check all solutions in the context of the original problem.
P. 33
2.5 Simple Problems Leading to
Quadratic Equations
Example 2.12T
The product of two consecutive positive odd integers is 195.
Find the two numbers.
Solution:
Let x be the smaller number.
Then the larger number is x  2.
Set a variable to one of the unknown quantities.
Represent all the other unknown quantities in
terms of the variable.
x(x  2)  195
Set up an equation.
Solve the equation.
x2  2x  195  0
(x2  15)(x  13)  0
x  13 or 15 (rejected)
∴ The smaller number is 13.
∴ The two numbers are 13 and 15.
According to the question,
the odd integer is positive,
so 15 is rejected.
P. 34
2.5 Simple Problems Leading to
Quadratic Equations
Example 2.13T
The figure shows a right-angled triangle with AB  15 cm,
BC  5x cm and CA  (6x  1) cm. Find the value of x.
Solution:
Since B  90, we have
(Pyth. Theorem)
152  (5x)2  (6 x  1)2
225  25x 2  36 x 2  12 x  1
11x 2  12 x  224  0
( x  4)(11x  56)  0
56
x  4 or 
(rejected)
11
P. 35
2.5 Simple Problems Leading to
Quadratic Equations
Example 2.14T
Consider a rectangle with an area of 180 cm2. If its length is 3 cm
longer than its width, find the length of the rectangle.
Solution:
Let x cm be the length of the rectangle.
Then the width is (x  3) cm.
x( x  3)  180
x 2  3x  180  0
( x  15)( x  12)  0
x  15 or 12 (rejected)
∴ The length of the rectangle is 15 cm.
P. 36
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Consider the general form of the quadratic equation:
ax2  bx  c  0 (where a  0)
b
c
It can be rewritten as x 2  x   0 .................. (*)
a
a
Suppose a and b are the roots of the equation.
Then it can be written as (x  a)(x  b)  0 .......... (**)
Since (*) and (**) are the two forms of the same equation, we have
b
c
x 2  x   ( x  α )( x  β )
a
a
 x 2  αx  βx  αβ
 x 2  (α  β ) x  αβ
Sum of roots  a  b  
b
a
Product of roots  ab 
P. 37
c
a
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Example 2.15T
Suppose a and b are the roots of the equation 3x2  4x  5  0.
Find the values of the following expressions.
1
1
(a) (2a  1)(2b  1)
(b) 2  2
α
β
Solution:
ab
(a)
4
5
and ab   .
3
3
(2a  1)(2b  1)  4ab  2(a  b)  1
 5  4
 4    2    1
 3  3
25

3
P. 38
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Example 2.15T
Suppose a and b are the roots of the equation 3x2  4x  5  0.
Find the values of the following expressions.
1
1
(a) (2a  1)(2b  1)
(b) 2  2
α
β
Solution:
4
5
and ab   .
3
3
1
1
(b)

α2 β 2
β 2  α2
 2 2
α β
ab
(α  β ) 2  2αβ

(αβ ) 2
2
 4
 5


2


 
3
 3

2
 5
 
 3
16 10

9
3  46

25
25
9
Since a2  2ab  b2  (a  b)2,
we have the following result:
a 2  b 2  (a  b)2  2ab
P. 39
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Example 2.16T
Consider a quadratic equation (k  1)x2  (k  3)x  4k  0,
where k  1, find k if
(a) one of the roots is the reciprocal of the other,
(b) one of the roots is the negative of the other.
Solution:
(a)
Since one of the roots is the reciprocal of
the other, product of roots  1.
 4k
1
k 1
 4k  k  1
5k  1
1
k
5
Suppose the roots are a and b.
If one of the roots is the
reciprocal of the other, then
1
β
α
αβ  1 .
P. 40
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Example 2.16T
Consider a quadratic equation (k  1)x2  (k  3)x  4k  0,
where k  1, find k if
(a) one of the roots is the reciprocal of the other,
(b) one of the roots is the negative of the other.
Solution:
(b)
Since one of the roots is the negative of
the other, sum of roots  0.

k 3
0
k 1
k 3 0
k  3
Suppose the roots are a and b.
If one of the roots is the
negative of the other, then
a  b
a b  0.
P. 41
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
A. Sum and Product of the Roots of
Quadratic Equations
Example 2.17T
Consider a quadratic equation 4x2  (p  7)x  p  0. If the roots of
the equation are a and a  2, find the values of p.
Solution:
Sum of roots  a  (a  2)
 2a  2
p7
2a  2 
4
p 1
.......... (1)
a
8
Product of roots  a (a  2)
 a 2  2a
p
a 2  2a  ............... (2)
4
Substituting (1) into (2), we have
2
 p 1
 p 1 p

  2

 8 
 8  4
( p 1)2  16( p 1)  16 p
( p  1)2  16
p  1  4
p   3 or 5
P. 42
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
B. Forming Quadratic Equations Using
the Given Roots
We know that if a and b are two roots of a quadratic equation,
the equation can be expressed as:
(x  a)(x  b )  0 .......... (1)
x2  (a  b )x  ab  0
x2  (sum of roots)x  (products of roots)  0 .......... (2)
Therefore:

If the roots of a quadratic equation are known,
 form the corresponding equation according to (1).

If the sum and the product of the roots are known,
 form the corresponding equation according to (2).
For example:

Given x  2 or x  1.
∴ (x  2)(x  1)  0
x2  x  2  0

Given sum of roots  1,
product of roots  2.
∴ x2  x  2  0
P. 43
2.6 Relations between the Roots and
Coefficients of Quadratic Equations
B. Forming Quadratic Equations Using
the Given Roots
Example 2.18T
Suppose a and b are the roots of the equation x2  7x  2  0.
Form a quadratic equation with roots a 2 and b 2.
Solution:
Since a and b are the roots of the equation x2  7x  2, we have
a  b  7
and
ab  2.
For the quadratic equation with roots a 2 and b 2, we have
sum of roots  a 2  b 2
and product of roots  a 2b 2
 (a  b )2  2ab
 22
 (7)2  2(2)
4
 45
∴ The required equation is x2  45x  4  0.
P. 44
Chapter Summary
2.1 Quadratic Equations
A quadratic equation in one unknown is in the form
ax2  bx  c  0
where a, b and c are constants and a  0.
P. 45
Chapter Summary
2.2 Solving Quadratic Equations by Taking
Square Root or Factor Method
1.
If x2  p, then x  
2.
If (x  q)2  p, then x  q 
3.
If (px  q)(mx  n)  0, then
px  q  0 or mx  n  0
n
q
x
or
m
p
p.
p.
P. 46
Chapter Summary
2.3 Solving Quadratic Equations by
Quadratic Formula
If ax2  bx  c  0 where a  0, then
 b  b 2  4ac
.
x
2a
P. 47
Chapter Summary
2.4 The Nature of Roots of Quadratic Equations
D  b2  4ac
Number of real roots
of the equation
ax2  bx  c  0
D0
2
D0
1
D0
0
P. 48
Chapter Summary
2.5 Simple Problems Leading to Quadratic
Equations
Strategies for solving a word problem:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Study and understand the problem.
Set a variable to represent one of the unknown quantities.
Set up an equation.
Solve the equation.
Check all solutions in the context of the original problem.
P. 49
Chapter Summary
2.6 Relations between the Roots and Coefficients
of Quadratic Equations
1. If a and b are the roots of the equation ax2  bx  c  0 (where a  0),
then
ab
2.
b
a
and
c
a
ab  .
If we know a and b are the roots of a quadratic equation, we can
form the equation by the formula
(x  a)(x  b )  0 or
x2  (sum of roots)x  (products of roots)  0.
P. 50