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Quadratic Expressions and Equations

Algebraic Expressions
Algebraic expressions is a mathematical phrase in which numbers and letters are connected by mathematical symbols.
Examples of algebraic expressions are
a + b, 2c + 1, q + 3q + r, 3a + 2b + 2c ' 5a
When algebraic expressions involve fractions they are referred to as algebraic fractions. Examples of these fractions are (where the denominator )
Algebraic expressions can be expanded or factorized.

Algebraic Expressions
Algebraic expressions is a mathematical phrase in which numbers and letters are connected by mathematical symbols.
Examples of algebraic expressions are
a + b, 2c + 1, q + 3q + r, 3a + 2b + 2c ' 5a
When algebraic expressions involve fractions they are referred to as algebraic fractions. Examples of these fractions are 1 (where the denominator 1)
Algebraic expressions can be expanded or factorized. 

Expansion of algebraic expressions

Worked out example
Expand 3 (2x + 1)
Solution
3(2x + 1) = 3 x 2x + 3 x 1
= 6x + 3
Note:   that in expansion the brackets are removed by multiplying the value(s) outside the bracket by each of the values in the brackets.
Example 2
Expand
2x (x + 2a + 6b)

solution
2x ( x + 2a + 6b) = 2x x 2 + 2x x 2a + 2x + 6b
2x2 + 4xa + 12xb
Example 3
Expand and simplify(a + 2) (3a ' 5)
Solution
= a x 3a = a x 5 + 2 x 3a ' 2 x 5
= 3a2 ' 5a + 6a ' 10
= 3a2 + a ' 10

Factorization
In algebraic expressions factorization is writing the expression as a product of its factors.

We do this by introducing brackets and writing the common factor (s) outside the brackets.
 

Worked example

Example 1

Factorize (6x + 4y)

Solution
The common factor is 2
Therefore 6x + 4y = 2(3x + 2y)

Example 2

Factorise

6x + 12y ' 24n ' 36m
Solution
The common factor is 6
Therefore 6x + 12y ' 24n ' 36m
= 6(x + 2y ' 4n -6m)

Example 3
Factorise
Mn ' 2m + 5n ' 10
Solution
Mn ' 2m _ 5n ' 10
Group the terms with common factors together as follows
(mn ' 2m) + (5n ' 10)
M(n - 2) + 5(n - 2) (Factor out the common factors outside the brackets)
(n ' 2) is a common factor
Therefore when factored out we get (n-2) (m+5)
(animate (n ' 2) by blicking in different color)

Exercise
Use factorization to solve for x in
X2 ' 9x- 36 = 0
Solution
Step1
x2+3x-12x ' 36 = 0
x (x+3) ' 12 (x+3) = 0
(x-12)(x+3) = 0
 
Either (x+3) = 0 just voice x = -3
Or (x-12) 0 then x =12

2. Solve for y in the quadratic equation
         6x2 + 11x ' 20 =15
Solution
6x2 + 11x -35 = 0
6x2 + 21x ' 10x ' 35 = 0
3x(2x+7) ' 5 (2x+7) = 0
(3x-5) (2x+7) = 0

Either 3x-5 = 0 Voiced 3x = 5
                                and  X= 5/3
Or  2x + 7 = 0 Voiced  2x = -7
                             and    x= -7/2

3. The base of a right angled triangle is 7cm less than its height(h). If the area of the triangle is 99cm2.Find the dimensions of the triangle.

Solution
From the diagram A=  '  x h x b

                                                                                           

1 



                              h

                                              (h-7)
Therefore, ' x h x h-7= 99cm2
=h2 -7h = 198
=h2 -7h ' 198 = 0
=h2 +11h - 18(h+11) = 0
=(h ' 18) (h+11)= 0
h = 18    or     h = -11

The height of the triangle  = 18cm
Therefore, the base of the triangle  = 18-7
                                                          =11
One  thousand shillings is supposed to be shared equally among a number of people .If the number of people is reduced by 10 each of the remaining persons world receive fifty shillings more. Find how much each person world receive if nobody withdraws from the group.

Solution
Let the number of people by x.
x people receive 2
x -10 people receive   3 
4= 50

Multiplying both sides by the L.C.M  5
 1000x ' 1000(x-10) = 6
1000x ' 1000x + 10000 = 50x2 ' 500x
50x2 ' 500x ' 10,000 = 0
X2 ' 10x ' 200 =0

Solving the quadratic equation
X2 ' 20x + 10(x-20) = 0
X2 ' 20x +10x ' 200 = 0
X(x-20) + 10(x-20) = 0
(x-20) (x+10) = 0
x = 20 or   x= -10
Since the number of people cannot be negative we take x = 20
Therefore, if nobody withdraws from the group each person would receive 
7=sh 50

Objectives

By the end of the topic you should be able to

:

solve quadratic equations by factorization

form and solve quadratic equations

Introduction
Some problems have more than one solution and has to choose the most convenient answer depending on the situation.
Fig 1.
                                                                       
Animate movement form A to C
                                                                       
And from A through B and then to C

To move from A to C, one can decide to go direct or move from A to B and then to C.

Lesson development

Introduction

An algebraic expression that contain two terms is called a binomial expression e.g.
(a + b), (2 ' c) (x + y) etc.

Expansion of two binomial expressions with similar variables gives a quadratic expressions

An algebraic expression that contain two terms is called a binomial expression e.g.
(a + b), (2 ' c) (x + y) etc.

Example 1
Expand (x + 1) (x + 2)
Solution
(x + 1) (x + 2)
x (x + 2) + 1 (x + 2) Splitting the first bracket i.e. (x + 1)
x2 + 2x + x + 2 (Removing brackets)
x2 + 3x + 2 (Collecting like terms)
x2 + 3x + 2 (This is a quadratic expression)

Example 2
Expand (20 + 3) (p ' 2)
Solution
 (2p + 3) (p ' 2)
               = 2p (p -2) + 3 (p-2)
               = 2p2 ' 4p + 3p ' 6
               = 2p2 ' p ' 6
NB:     An expression of the form ax2 + bx + c where the highest power of the variable is 2 is called a quadratic expression
            a is the coefficient of x2
            b is the coefficient of x
            c is the constant term
The following quadratic identities are useful in expansion of binomial expressions.

  1. (a + b) 2 = (a + b) (a + b)

            = a (a + b) + b(a + b)
            = a2 + ab + ab + b2
            = a2+ 2ab + b2

  1. (a ' b) 2 = (a ' b) (a ' b)

            = a (a ' b) ' b(a ' b)
            = a2 ' ab ' ab + b2
            = a2 ' 2ab + b2

  1. (a2 + b2) (a ' b)

            = a (a ' b) + b (a ' b)
            = a2 ' ab + ab ' b2
            but -ab + ab = 0
Therefore (a + b) (a - b) = a2 ' b2 (This is called a difference of two squares)

Solving quadratic equations using factorization
When a quadratic expression is equated to zero we get a standard form of the quadratic equation i.e. (ax2 + bx + c = 0 (Highlight the equation)
However the expression can be equated to any other value e.g. 2x2 + 5x + 3 = 4.

Worked examples
Example 1
Solved the following quadratic equations
4x2 + x = 0
Solution
When 4x2 + x is factorized we get x(4x + 1)
Therefore for 4x + x = 0
x (4x + 1) = 0 which means
Either
=x = 0
or 4x + 1 = 0
4x + 1 = 0
1 subtracting 1 form both sides
12
Therefore the solutions of x in 4x2 + x = 0 are x = 0 and x = - '

Example 2
x2 + 8x + 12 = 0
Solution
x2 + 6x + 2x + 12 = 0 (splitting the middle term into two number whose and products is
1 x 12 = 12)
x (x + 6) + 2 (x + 6) = 0
(x + 2) (x + 6) = 0
Either x + 2 = 0  x = -2
Or x + 6 = 0         x = -6
The solutions of x in x2 + 8x + 12 = 0 are x = -2 and x = -6

Example 3
iii) 3y2 ' 19y ' 14 = 0                                  
Solution
3y2 + 2y ' 21y ' 14 = 02
= y (3y + 2) -7 (3y + 2) = 0
   (y - 7) (3y + 2) = 0
Either y ' 7 = 0  
y = 7
Or 3y + 2 = 0   3y = - 2 (dividing both sides by 3)
                            y = 2
Therefore the solution of x in 3y2 ' 19y ' 14 = 0
Are y = 7 and y = 22

Example 4
      ? k2 + 5/4 k + 2 = 0
      k2 + 10k + 16 = 0 (Multiplying through by the lcm of 8 and 4)
      k2 + 8k + 2k + 16=0
      k(k + 8) + 2(k + 8) = 0
      (k + 2) (k + 8) = 0
Either k + 2 = 0      and wiced k = - 2
Or      k + 8 = 0       and void k = -8
Therefore solutions of k in ? k2 + 5/4 k + 2 = 0
Are k = 2 and k = -8

Example 5
v) Solve for t in 8t2 + 14t + 1 = -2
Solution
8t2 + 14t + 3 = 0    (adding 2 on both sides to form a standard quadratic equation)
8t2 + 12t + 2t + 3 = 0   
4t(2t + 3) + 1 (2t + 3) = 0
(4t + 1) (2t + 3) = 0
Either 4t + 1 + 0
2                                            
Or 2t + 3 = 0 2
The solutions of t is 8t2 + 14t + 1 = -2 are t = - ' and t = - 1'   

Example 6
Solve for d in 1/9d2 ' 1 = 0
d2 ' 9 = 0 (multiplying all through by the LCM -9)
(d + 3) (d-3) = 0 (Difference of two squares)
Either d + 3 = 0 and voiced d = -3
Or d ' 3 = 0 and voiced d = +3
Therefore the difference for d in the equation 1/9d2-1 = 0
Are d = 3 and d = -3

Note
Generally from the examples above quadratic equations have two solutions. The solutions of quadratic equations are also called the root
s of the equation.

Forming and solving quadratic equation
We can form quadratic equations from their roots as follows : -
Solution
Either x ' 2 = 0 or x ' 3 = 0
Therefore (x ' 2) (x ' 3) = 0 (finding the product of the two binomials)
x (x ' 3) ' 2(x-3)=0(Expanding)
x2-3x-2x+=0 ( collecting the like terms)
x2 =5x+6=0
Therefore, our quadratic equation is  x2-5x+6=0

Example 2
Form the quadratic equation from the roots given.  P =1/2  as p=2/3
Solution
Either(P - 1/2) (P - 2/3)=0                      
P (P - 2/3)  -  1/2(P - 2/3)=0
P2 - 2/3p-1/2p+1/3 =0
P2 - 7/6p+1/3=0

Example 3
Find the quadratic equation whose roots are q= -10 and q = 1/3
Solution
Either q+10 = 0    or      q-1/3 = 0
2(q+10) (q - 1/3) = 0
= q(q -  1/3) +10 (q - 1/3) = 0
= q2 - 1/3q +10q - 10/3 = 0
= 3q2 ' q + 30q ' 10 = 0 (Multiplying through by 3)
= 3q2 + 29q ' 10 = 0

22The quadratic equation from the roots is
3q2+29q-10 = 0

Quadratic equation can also be formed from real life situations.

This can be illustrated by the examples given below.

Example 1
A piece of barbed wire is used to fence a rectangular field whose area is 300m2. Given that the length of the rectangular field is 10m more than twice its width (i) find the width of the field.
                 (ii)The length of the barbed wire.

Solution
                                                      (2y+10)
 

= A 300
            From the diagram                

                                                                                                      y

    1. Let the value of the width be y m.
    2. Then the length will be (2y+10)

Therefore, the area of the rectangle is LxW=A
y x (2y+10)=300
2y2+10y=300
2y2+10y-300=0           (simplifying the equation dividing through by 2)
y2 + 5y ' 150 = 0
Factorizing
y2+5y-150=0
y2+15y-10y-150 = 0
y2+15y-10y-150 = 0
y(y+15) -10(y+15) = 0
y(y -10) (y +15) = 0

Either  y -10=0   where y=10
Or
y + 15=0  where y= -15
We cannot have negative measurements and therefore the value of y= 10m
2The width of the field is 10m
The length of the field is 2(10)+10= 30m
The perimeter of the rectangular field is
= 2(L) + 2(W)
= 2(30) + 2(10)
= 60 + 20
= 80m
Therefore the length of the barbed wire is 80m

Example 2
The distance from Nairobi to Eldoret is 360km. A bus and a Nissan left Nairobi at 8.00am and travelled towards Eldoret. If the average speed of the Nissan was 20km/h more than that of the bus and that it arrived at Eldoret 3hours ealier than the Bus.

Find  (i) The speed of the            
         (ii)The time the bus arrived at Eldoret.

Solution

From the information given, let the speed of the Nissan be  km/h and the speed of the Bus will be (S-20) km/h since the distance from Nairobi to Eldoret=360km.
Then the time taken by the Nissan will be 2 
The time taken by the Bus will be   22
Difference in their time of arrival = 3hrs

Therefore, 2 
Multiplying by the L.C.M S(S ' 20) we get

360(S) - (360)(S-20) = 3(S)(S-20)
360S ' 360S+7200 =3(S2-20S)
360S-7200 = 3S2-60S
3S2-60S-7200 = 0 (forming a standard quadratic equation and simplify)
S2 ' 20S -2400 = 0
S(S-60) + 40(S-60) =0
(S+40)(S-60) =0

22Either S+40 = 0 where S= -40
Or S-60 = 0 where S=60
The speed cannot be negative therefore we take the positive value of S
The speed of the Nissan S = 60km/h
The speed of the bus will be 60km/h -20km/h
                                              = 40km/h
Time taken by the bus will be 2 
360 = 9hrs
 40

From 8.00 am we add 9hrs
8.00am + 9hrs = 17.00hrs
                       = 5.00p.m

 



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